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#
#
#           The Nim Compiler
#        (c) Copyright 2015 Andreas Rumpf
#
#    See the file "copying.txt", included in this
#    distribution, for details about the copyright.
#

import pathutils

template setX(k, field) {.dirty.} =
  var s: seq[TFullReg]
  move(s, cast[seq[TFullReg]](a.slots))
  if s[a.ra].kind != k:
    myreset(s[a.ra])
    s[a.ra].kind = k
  s[a.ra].field = v

proc setResult*(a: VmArgs; v: BiggestInt) = setX(rkInt, intVal)
proc setResult*(a: VmArgs; v: BiggestFloat) = setX(rkFloat, floatVal)
proc setResult*(a: VmArgs; v: bool) =
  let v = v.ord
  setX(rkInt, intVal)

proc setResult*(a: VmArgs; v: string) =
  var s: seq[TFullReg]
  move(s, cast[seq[TFullReg]](a.slots))
  if s[a.ra].kind != rkNode:
    myreset(s[a.ra])
    s[a.ra].kind = rkNode
  s[a.ra].node = newNode(nkStrLit)
  s[a.ra].node.strVal = v

proc setResult*(a: VmArgs; n: PNode) =
  var s: seq[TFullReg]
  move(s, cast[seq[TFullReg]](a.slots))
  if s[a.ra].kind != rkNode:
    myreset(s[a.ra])
    s[a.ra].kind = rkNode
  s[a.ra].node = n

proc setResult*(a: VmArgs; v: AbsoluteDir) = setResult(a, v.string)

proc setResult*(a: VmArgs; v: seq[string]) =
  var s: seq[TFullReg]
  move(s, cast[seq[TFullReg]](a.slots))
  if s[a.ra].kind != rkNode:
    myreset(s[a.ra])
    s[a.ra].kind = rkNode
  var n = newNode(nkBracket)
  for x in v: n.add newStrNode(nkStrLit, x)
  s[a.ra].node = n

template getX(k, field) {.dirty.} =
  doAssert i < a.rc-1
  let s = cast[seq[TFullReg]](a.slots)
  doAssert s[i+a.rb+1].kind == k
  result = s[i+a.rb+1].field

proc getInt*(a: VmArgs; i: Natural): BiggestInt = getX(rkInt, intVal)
proc getBool*(a: VmArgs; i: Natural): bool = getInt(a, i) != 0
proc getFloat*(a: VmArgs; i: Natural): BiggestFloat = getX(rkFloat, floatVal)
proc getString*(a: VmArgs; i: Natural): string =
  doAssert i < a.rc-1
  let s = cast[seq[TFullReg]](a.slots)
  doAssert s[i+a.rb+1].kind == rkNode
  result = s[i+a.rb+1].node.strVal

proc getNode*(a: VmArgs; i: Natural): PNode =
  doAssert i < a.rc-1
  let s = cast[seq[TFullReg]](a.slots)
  doAssert s[i+a.rb+1].kind == rkNode
  result = s[i+a.rb+1].node
='#n927'>927 928 929 930 931 932 933 934 935 936 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031 1032 1033 1034 1035 1036 1037 1038 1039 1040 1041 1042 1043 1044 1045 1046 1047 1048 1049 1050 1051 1052 1053 1054 1055 1056 1057 1058 1059 1060 1061 1062 1063 1064 1065 1066 1067 1068 1069 1070 1071 1072 1073 1074 1075 1076 1077 1078 1079 1080 1081 1082 1083 1084 1085 1086 1087 1088 1089 1090 1091 1092 1093 1094 1095 1096 1097 1098 1099 1100 1101 1102 1103 1104 1105 1106 1107 1108 1109 1110 1111 1112 1113 1114 1115 1116 1117 1118 1119 1120 1121 1122 1123 1124 1125 1126 1127 1128 1129 1130 1131 1132 1133 1134 1135 1136 1137 1138 1139 1140 1141 1142 1143 1144 1145 1146 1147 1148 1149 1150 1151 1152 1153 1154 1155 1156 1157 1158 1159 1160 1161 1162 1163 1164 1165 1166 1167 1168 1169 1170 1171 1172 1173 1174 1175 1176 1177 1178 1179 1180 1181 1182 1183 1184 1185 1186 1187 1188 1189 1190 1191 1192 1193 1194 1195 1196 1197 1198 1199 1200 1201 1202 1203 1204 1205 1206 1207 1208 1209 1210 1211 1212 1213 1214 1215 1216 1217 1218 1219 1220 1221 1222 1223 1224 1225 1226 1227 1228 1229 1230 1231 1232 1233 1234 1235 1236 1237 1238 1239 1240 1241 1242 1243 1244 1245 1246 1247 1248 1249 1250 1251 1252 1253 1254 1255 1256 1257 1258 1259 1260 1261 1262 1263 1264 1265 1266 1267 1268 1269 1270 1271 1272 1273 1274 1275 1276 1277 1278 1279 1280 1281 1282 1283 1284 1285 1286 1287 1288 1289 1290 1291 1292 1293 1294 1295 1296 1297 1298 1299 1300 1301 1302 1303 1304 1305 1306 1307 1308 1309 1310 1311 1312 1313 1314 1315 1316 1317 1318 1319 1320 1321 1322 1323 1324 1325 1326 1327 1328 1329 1330 1331 1332 1333 1334 1335 1336 1337 1338 1339 1340 1341 1342 1343 1344 1345 1346 1347 1348 1349 1350 1351 1352 1353 1354 1355 1356 1357 1358 1359 1360 1361 1362 1363 1364 1365 1366 1367 1368 1369 1370 1371 1372 1373 1374 1375 1376 1377 1378 1379 1380 1381 1382 1383 1384 1385 1386 1387 1388 1389 1390 1391 1392 1393 1394 1395 1396 1397 1398 1399 1400 1401 1402 1403 1404 1405 1406 1407 1408 1409 1410 1411 1412 1413 1414 1415 1416 1417 1418 1419 1420 1421 1422 1423 1424 1425 1426 1427 1428 1429 1430 1431 1432 1433 1434 1435 1436 1437 1438 1439 1440 1441 1442 1443 1444 1445 1446 1447 1448 1449 1450 1451 1452 1453 1454 1455 1456 1457 1458 1459 1460 1461 1462 1463 1464 1465 1466 1467 1468 1469 1470 1471 1472 1473 1474 1475 1476 1477 1478 1479 1480 1481 1482 1483 1484 1485 1486 1487 1488 1489 1490 1491 1492 1493 1494 1495 1496 1497 1498 1499 1500 1501 1502 1503 1504 1505 1506 1507 1508 1509 1510 1511 1512 1513 1514 1515 1516 1517 1518 1519 1520 1521 1522 1523 1524 1525 1526 1527 1528 1529 1530 1531 1532 1533 1534 1535 1536 1537 1538 1539 1540 1541 1542 1543 1544 1545 1546 1547 1548 1549 1550 1551 1552 1553 1554 1555 1556 1557 1558 1559 1560 1561 1562 1563 1564 1565 1566 1567 1568 1569 1570 1571 1572 1573 1574 1575 1576 1577 1578 1579 1580 1581 1582 1583 1584 1585 1586 1587 1588 1589 1590 1591 1592 1593 1594 1595 1596 1597 1598 1599 1600 1601 1602 1603 1604 1605 1606 1607 1608 1609 1610 1611 1612 1613 1614 1615 1616 1617 1618 1619 1620 1621 1622 1623 1624 1625 1626 1627 1628 1629 1630 1631 1632 1633 1634 1635 1636 1637 1638 1639 1640 1641 1642 1643 1644 1645 1646 1647 1648 1649 1650 1651 1652 1653 1654 1655 1656 1657 1658 1659 1660 1661 1662 1663 1664 1665 1666 1667 1668 1669 1670 1671 1672 1673 1674 1675 1676 1677 1678 1679 1680 1681 1682 1683 1684 1685 1686 1687 1688 1689 1690 1691 1692 1693 1694 1695 1696 1697 1698 1699 1700 1701 1702 1703 1704 1705 1706 1707 1708 1709 1710 1711 1712 1713 1714 1715 1716 1717 1718 1719 1720 1721 1722 1723 1724 1725 1726 1727 1728 1729 1730 1731 1732 1733 1734 1735 1736 1737 1738 1739 1740 1741 1742 1743 1744 1745 1746 1747 1748 1749 1750 1751 1752 1753 1754 1755 1756 1757 1758 1759 1760 1761 1762 1763 1764 1765 1766 1767 1768 1769 1770 1771 1772 1773 1774 1775 1776 1777 1778 1779 1780 1781 1782 1783 1784 1785 1786 1787 1788 1789 1790 1791 1792 1793 1794 1795 1796 1797 1798 1799 1800 1801 1802 1803 1804 1805 1806 1807 1808 1809 1810 1811 1812 1813 1814 1815 1816 1817 1818 1819 1820 1821 1822 1823 1824 1825 1826 1827 1828 1829 1830 1831 1832 1833 1834 1835 1836 1837 1838 1839 1840 1841 1842 1843 1844 1845 1846 1847 1848 1849 1850 1851 1852 1853 1854 1855 1856 1857 1858 1859 1860 1861 1862 1863 1864 1865 1866 1867 1868 1869 1870
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<title>Algebra 1</title>
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<div id="org-div-home-and-up">
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</div><div id="content" class="content">
<h1 class="title">Algebra 1</h1>
<div id="table-of-contents" role="doc-toc">
<h2>Table of Contents</h2>
<div id="text-table-of-contents" role="doc-toc">
<ul>
<li><a href="#org42f27fc">Contenu de la Matiére</a>
<ul>
<li><a href="#orgf20cf94">Rappels et compléments (11H)</a></li>
<li><a href="#orgf700058">Structures Algébriques (11H)</a></li>
<li><a href="#org7a29a82">Polynômes et fractions rationnelles</a></li>
</ul>
</li>
<li><a href="#org7207cb0">Premier cours : Logique mathématique et méthodes du raisonnement mathématique <i>Sep 25</i> :</a>
<ul>
<li><a href="#orgb936329">Properties:</a>
<ul>
<li><a href="#orgf5da498"><b>Absorption</b>:</a></li>
<li><a href="#org49dbf9d"><b>Commutativity</b>:</a></li>
<li><a href="#orge255044"><b>Associativity</b>:</a></li>
<li><a href="#org31cc6c8"><b>Distributivity</b>:</a></li>
<li><a href="#orgf861930"><b>Neutral element</b>:</a></li>
<li><a href="#org8cb6e02"><b>Negation of a conjunction &amp; a disjunction</b>:</a></li>
<li><a href="#orgfe01ac7"><b>Transitivity</b>:</a></li>
<li><a href="#org976f527"><b>Contraposition</b>:</a></li>
<li><a href="#org0865f2b">God only knows what this property is called:</a></li>
</ul>
</li>
<li><a href="#org316b141">Some exercices I found online :</a>
<ul>
<li><a href="#orga3825f4">USTHB 2022/2023 Section B :</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#org21d6c03">2éme cours <i>Oct 2</i></a>
<ul>
<li><a href="#orgd6c9f49">Quantifiers</a>
<ul>
<li><a href="#orgb332b43">Proprieties</a></li>
</ul>
</li>
<li><a href="#orged685c1">Multi-parameter proprieties :</a></li>
<li><a href="#org78d7ed0">Methods of mathematical reasoning :</a>
<ul>
<li><a href="#org7d21c38">Direct reasoning :</a></li>
<li><a href="#orgcfd8723">Reasoning by the Absurd:</a></li>
<li><a href="#org102d3fa">Reasoning by contraposition:</a></li>
<li><a href="#org81cb388">Reasoning by counter example:</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#orgc2178b8">3eme Cours : <i>Oct 9</i></a>
<ul>
<li>
<ul>
<li><a href="#org4855f6f">Reasoning by recurrence :</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#orgde6bfac">4eme Cours : Chapitre 2 : Sets and Operations</a>
<ul>
<li><a href="#orgfe8000a">Definition of a set :</a></li>
<li><a href="#orgfe04671">Belonging, inclusion, and equality :</a></li>
<li><a href="#orga2eb99d">Intersections and reunions :</a>
<ul>
<li><a href="#org560d563">Intersection:</a></li>
<li><a href="#org7147bc3">Union:</a></li>
<li><a href="#org16b5ab2">Difference between two sets:</a></li>
<li><a href="#orgdac190b">Complimentary set:</a></li>
<li><a href="#org4e0b111">Symmetrical difference</a></li>
</ul>
</li>
<li><a href="#org691c863">Proprieties :</a>
<ul>
<li><a href="#org9cc9f31">Commutativity:</a></li>
<li><a href="#org471083b">Associativity:</a></li>
<li><a href="#orge63be10">Distributivity:</a></li>
<li><a href="#orgfb01947">Lois de Morgan:</a></li>
<li><a href="#orge1a41eb">An other one:</a></li>
<li><a href="#org9939b0f">An other one:</a></li>
<li><a href="#org90bfdc4">And an other one:</a></li>
<li><a href="#org1e49001">And the last one:</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#org272eca3">5eme cours: L&rsquo;ensemble des parties d&rsquo;un ensemble <i>Oct 16</i></a>
<ul>
<li>
<ul>
<li><a href="#org5b29e32">Notes :</a></li>
<li><a href="#org5636bd8">Examples :</a></li>
</ul>
</li>
<li><a href="#orgd8cb2c3">Partition of a set :</a></li>
<li><a href="#orgf40404d">Cartesian products :</a>
<ul>
<li><a href="#orgd526cb8">Example :</a></li>
<li><a href="#org56dd088">Some proprieties:</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#org5ee4278">Binary relations in a set :</a>
<ul>
<li><a href="#orgddc9af6">Definition :</a></li>
<li><a href="#orge65424e">Proprieties :</a></li>
<li><a href="#orgd7877d3">Equivalence relationship :</a>
<ul>
<li><a href="#org85cf025">Equivalence class :</a></li>
</ul>
</li>
<li><a href="#orge18dcc7">Order relationship :</a>
<ul>
<li><a href="#org60d471a"><span class="todo TODO">TODO</span> Examples :</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#org77de6e3">TP exercices <i>Oct 20</i> :</a>
<ul>
<li><a href="#org3ca8006">Exercice 3 :</a>
<ul>
<li><a href="#orgad95ec3">Question 3</a></li>
</ul>
</li>
<li><a href="#org8180ae0">Exercice 4 :</a>
<ul>
<li><a href="#orgfe0b1e2"><span class="done DONE">DONE</span> Question 1 :</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#org2d6e0ba">Chapter 3 : Applications</a>
<ul>
<li><a href="#orga5be12f">3.1 Generalities about applications :</a>
<ul>
<li><a href="#org805d7bc">Definition :</a></li>
<li><a href="#org7947331">Restriction and prolongation of an application :</a></li>
<li><a href="#orgd94bc69">Composition of applications :</a></li>
</ul>
</li>
<li><a href="#org257d05a">3.2 Injection, surjection and bijection :</a>
<ul>
<li><a href="#org1612e09">Proposition :</a></li>
</ul>
</li>
<li><a href="#orgebdf518">3.3 Reciprocal applications :</a>
<ul>
<li><a href="#orgf072e42">Def :</a></li>
<li><a href="#org244b352">Theorem :</a></li>
<li><a href="#org1479c0e">Some proprieties :</a></li>
</ul>
</li>
<li><a href="#orgaf81bb3">3.4 Direct Image and reciprocal Image :</a>
<ul>
<li><a href="#org87b91e2">Direct Image :</a></li>
<li><a href="#org500bc40">Reciprocal image :</a></li>
</ul>
</li>
</ul>
</li>
</ul>
</div>
</div>
<div id="outline-container-org42f27fc" class="outline-2">
<h2 id="org42f27fc">Contenu de la Matiére</h2>
<div class="outline-text-2" id="text-org42f27fc">
</div>
<div id="outline-container-orgf20cf94" class="outline-3">
<h3 id="orgf20cf94">Rappels et compléments (11H)</h3>
<div class="outline-text-3" id="text-orgf20cf94">
<ul class="org-ul">
<li>Logique mathématique et méthodes du raisonnement mathématique<br /></li>
<li>Ensembles et Relations<br /></li>
<li>Applications<br /></li>
</ul>
</div>
</div>
<div id="outline-container-orgf700058" class="outline-3">
<h3 id="orgf700058">Structures Algébriques (11H)</h3>
<div class="outline-text-3" id="text-orgf700058">
<ul class="org-ul">
<li>Groupes et morphisme de groupes<br /></li>
<li>Anneaux et morphisme d&rsquo;anneaux<br /></li>
<li>Les corps<br /></li>
</ul>
</div>
</div>
<div id="outline-container-org7a29a82" class="outline-3">
<h3 id="org7a29a82">Polynômes et fractions rationnelles</h3>
<div class="outline-text-3" id="text-org7a29a82">
<ul class="org-ul">
<li>Notion du polynôme à une indéterminée á coefficients dans un anneau<br /></li>
<li>Opérations Algébriques sur les polynômes<br /></li>
<li>Arithmétique dans l&rsquo;anneau des polynômes<br /></li>
<li>Polynôme dérivé et formule de Taylor<br /></li>
<li>Notion de racine d&rsquo;un polynôme<br /></li>
<li>Notion de Fraction rationelle á une indéterminée<br /></li>
<li>Décomposition des fractions rationelles en éléments simples<br /></li>
</ul>
</div>
</div>
</div>
<div id="outline-container-org7207cb0" class="outline-2">
<h2 id="org7207cb0">Premier cours : Logique mathématique et méthodes du raisonnement mathématique <i>Sep 25</i> :</h2>
<div class="outline-text-2" id="text-org7207cb0">
<p>
Let <b>P</b> <b>Q</b> and <b>R</b> be propositions which can either be <b>True</b> or <b>False</b>. And let&rsquo;s also give the value <b>1</b> to each <b>True</b> proposition and <b>0</b> to each false one.<br />
</p>

<p>
<i>Ex:</i><br />
</p>
<ul class="org-ul">
<li><b>5 ≥ 2</b> is a proposition, a correct one !!!<br /></li>
<li><b>The webmaster is a girl</b> is also a proposition, which is also correct.<br /></li>
<li><b>x is always bigger than 5</b> is <b>not</b> a proposition, because we CAN&rsquo;T determine if it&rsquo;s correct or not as <b>x</b> changes.<br /></li>
</ul>
<p>
&#x2026;etc<br />
</p>

<p>
In order to avoid repetition, and rewriting the proposition over and over, we just assign a capital letter to them such as <b>P Q</b> or <b>R</b>.<br />
</p>

<p>
So now we could write :<br />
<b>Let the proposition P be 5 ≥ 2, we notice that P is always True, therefor its validity is 1</b><br />
</p>

<p>
We also have the opposite of <b>P</b>, which is <b>not(P)</b> but for simplicity we use <b></b> (A P with a bar on top, in case it doesn&rsquo;t load for you), now let&rsquo;s go back to the previous example:<br />
</p>

<p>
<b>Since we know that the proposition P is true, we can conclude that P̅ is false. As P and P̅ can NOT be true at the same time. It&rsquo;s like saying 5 is greater and also lesser than 2&#x2026;doesn&rsquo;t make sense, does it ?</b><br />
</p>

<p>
Now let&rsquo;s say we have two propositions, and we want to test the validity of their disjunction&#x2026;.. Okay what is this &ldquo;disjunction&rdquo; ? <b>Great Question Billy !!!</b> A disjunction is true if either propositions are true<br />
</p>

<p>
Ex:<br />
<b>Let proposition P be &ldquo;The webmaster is asleep&rdquo;, and Q be &ldquo;The reader loves pufferfishes&rdquo;. The disjunction of these two propositions can have 4 different values showed in this Table of truth (such a badass name):</b><br />
</p>

<table border="2" cellspacing="0" cellpadding="6" rules="groups" frame="hsides">


<colgroup>
<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />
</colgroup>
<thead>
<tr>
<th scope="col" class="org-right">P</th>
<th scope="col" class="org-right">Q</th>
<th scope="col" class="org-right">Disjunction</th>
</tr>
</thead>
<tbody>
<tr>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
</tr>

<tr>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
</tr>

<tr>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
</tr>

<tr>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
</tr>
</tbody>
</table>

<p>
<i>What the hell is this ?</i><br />
The first colomn is equivalent to saying : &ldquo;The webmaster is asleep AND The reader loves pufferfishes&rdquo;<br />
The second one means : &ldquo;The webmaster is asleep AND The reader DOESN&rsquo;T love pufferfishes (if you are in this case, then <b>I HATE YOU</b>)&rdquo;<br />
The third one&#x2026; <i>zzzzzzz</i><br />
</p>

<p>
You got the idea !!!<br />
And since we are talking about a disjunction here, <b>one of the propositions</b> need to be true in order for this disjunction to be true.<br />
</p>

<p>
You may be wondering&#x2026;. Crystal, can&rsquo;t we write a disjunction in magical math symbols ? And to this I respond with a big <b>YES</b>. A disjunction is symbolized by a <b></b> . So the disjunction between proposition <b>P &amp; Q</b> can be written this way : <b>P ∨ Q</b><br />
</p>

<p>
What if, we want to test whether or not two propositions are true AT THE SAME TIME ? Long story short, we can, it&rsquo;s called a conjunction, same concept, as before, only this time the symbol is <b>P ∧ Q</b>, and is only true if <b>P</b> and <b>Q</b> are true. So we get a Table like this :<br />
</p>

<table border="2" cellspacing="0" cellpadding="6" rules="groups" frame="hsides">


<colgroup>
<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />
</colgroup>
<thead>
<tr>
<th scope="col" class="org-right">P</th>
<th scope="col" class="org-right">Q</th>
<th scope="col" class="org-right">P ∨ Q</th>
<th scope="col" class="org-right">P ∧ Q</th>
</tr>
</thead>
<tbody>
<tr>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
</tr>

<tr>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
</tr>

<tr>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
</tr>

<tr>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
</tr>
</tbody>
</table>

<p>
<b>Always remember: 1 means true and 0 means false</b><br />
</p>

<p>
There are two more basics to cover here before going to some properties, the first one is implication symbolized by the double arrow <b></b><br />
</p>

<p>
Implication is kinda hard for my little brain to explain, so I will just say what it means:<br />
</p>

<p>
<b>If P implies Q, this means that either Q, or the opposite of P are correct</b><br />
</p>

<p>
or in math terms<br />
</p>

<p>
<b>P ⇒ Q translates to P̅ ∨ Q</b><br />
Let&rsquo;s illustrate :<br />
</p>

<table border="2" cellspacing="0" cellpadding="6" rules="groups" frame="hsides">


<colgroup>
<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />
</colgroup>
<thead>
<tr>
<th scope="col" class="org-right">P</th>
<th scope="col" class="org-right">Q</th>
<th scope="col" class="org-right"></th>
<th scope="col" class="org-right"></th>
<th scope="col" class="org-right">P ∨ Q</th>
<th scope="col" class="org-right">P ∧ Q</th>
<th scope="col" class="org-right">P ⇒ Q (P̅ ∨ Q)</th>
</tr>
</thead>
<tbody>
<tr>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
</tr>

<tr>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
</tr>

<tr>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
</tr>

<tr>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
</tr>
</tbody>
</table>

<p>
<b>If you look clearly, there is only one case where an implication is false. therefor you just need to find it, and blindly say that the others are correct. A rule of thumb is that: &ldquo;A correct never implies a false&rdquo;, or  &ldquo;If a 1 tries to imply a 0, the implication is a 0&rdquo;</b><br />
</p>

<p>
Aight, a last one and we are done!!! Equivalence, which is fairly easy, symbolized by a <b></b> symbol.<br />
</p>

<p>
A proposition is equivalent to another only when both of them have <b>the same value of truth</b> AKA: both true or both false. a little table will help demonstrate what i mean.<br />
</p>

<table border="2" cellspacing="0" cellpadding="6" rules="groups" frame="hsides">


<colgroup>
<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />

<col  class="org-right" />
</colgroup>
<thead>
<tr>
<th scope="col" class="org-right">P</th>
<th scope="col" class="org-right">Q</th>
<th scope="col" class="org-right"></th>
<th scope="col" class="org-right"></th>
<th scope="col" class="org-right">P ∨ Q</th>
<th scope="col" class="org-right">P ∧ Q</th>
<th scope="col" class="org-right">P ⇒ Q (P̅ ∨ Q)</th>
<th scope="col" class="org-right">P ⇔ Q</th>
</tr>
</thead>
<tbody>
<tr>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
</tr>

<tr>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
</tr>

<tr>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
</tr>

<tr>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
<td class="org-right">0</td>
<td class="org-right">0</td>
<td class="org-right">1</td>
<td class="org-right">1</td>
</tr>
</tbody>
</table>

<p>
<i>Note: P implying Q is equivalent to P̅ implying Q̅, or: (P ⇒ Q) ⇔ (P̅ ⇒ Q̅)</i><br />
</p>
</div>
<div id="outline-container-orgb936329" class="outline-3">
<h3 id="orgb936329">Properties:</h3>
<div class="outline-text-3" id="text-orgb936329">
</div>
<div id="outline-container-orgf5da498" class="outline-4">
<h4 id="orgf5da498"><b>Absorption</b>:</h4>
<div class="outline-text-4" id="text-orgf5da498">
<p>
(P ∨ P) ⇔ P<br />
</p>

<p>
(P ∧ P) ⇔ P<br />
</p>
</div>
</div>
<div id="outline-container-org49dbf9d" class="outline-4">
<h4 id="org49dbf9d"><b>Commutativity</b>:</h4>
<div class="outline-text-4" id="text-org49dbf9d">
<p>
(P ∧ Q) ⇔ (Q ∧ P)<br />
</p>

<p>
(P ∨ Q) ⇔ (Q ∨ P)<br />
</p>
</div>
</div>
<div id="outline-container-orge255044" class="outline-4">
<h4 id="orge255044"><b>Associativity</b>:</h4>
<div class="outline-text-4" id="text-orge255044">
<p>
P ∧ (Q ∧ R) ⇔ (P ∧ Q) ∧ R<br />
</p>

<p>
P ∨ (Q ∨ R) ⇔ (P ∨ Q) ∨ R<br />
</p>
</div>
</div>
<div id="outline-container-org31cc6c8" class="outline-4">
<h4 id="org31cc6c8"><b>Distributivity</b>:</h4>
<div class="outline-text-4" id="text-org31cc6c8">
<p>
P ∧ (Q ∨ R) ⇔ (P ∧ Q) ∨ (P ∧ R)<br />
</p>

<p>
P ∨ (Q ∧ R) ⇔ (P ∨ Q) ∧ (P ∨ R)<br />
</p>
</div>
</div>
<div id="outline-container-orgf861930" class="outline-4">
<h4 id="orgf861930"><b>Neutral element</b>:</h4>
<div class="outline-text-4" id="text-orgf861930">
<p>
<i>We define proposition <b>T</b> to be always <b>true</b> and <b>F</b> to be always <b>false</b></i><br />
</p>

<p>
P ∧ T ⇔ P<br />
</p>

<p>
P ∨ F ⇔ P<br />
</p>
</div>
</div>
<div id="outline-container-org8cb6e02" class="outline-4">
<h4 id="org8cb6e02"><b>Negation of a conjunction &amp; a disjunction</b>:</h4>
<div class="outline-text-4" id="text-org8cb6e02">
<p>
Now we won&rsquo;t use bars here because my lazy ass doesn&rsquo;t know how, so instead I will use not()!!!<br />
</p>

<p>
not(<b>P ∧ Q</b>) ⇔ P̅ ∨ Q̅<br />
</p>

<p>
not(<b>P ∨ Q</b>) ⇔ P̅ ∧ Q̅<br />
</p>

<p>
<b>A rule I really like to use here is: Break and Invert. Basically you break the bar into the three characters of the propositions, so you get not(P) not(∧ or ∨) <i>NOT AN ACTUAL MATH WRITING. DONT USE IT ANYWHERE ELSE OTHER THAN YOUR BRAIN</i> and not(Q)</b><br />
</p>
</div>
</div>
<div id="outline-container-orgfe01ac7" class="outline-4">
<h4 id="orgfe01ac7"><b>Transitivity</b>:</h4>
<div class="outline-text-4" id="text-orgfe01ac7">
<p>
[(P ⇒ Q) AND (Q ⇒ R)] ⇔ P ⇒ R<br />
</p>
</div>
</div>
<div id="outline-container-org976f527" class="outline-4">
<h4 id="org976f527"><b>Contraposition</b>:</h4>
<div class="outline-text-4" id="text-org976f527">
<p>
(P ⇒ Q) ⇔ (Q̅ ⇒ P̅)<br />
</p>
</div>
</div>
<div id="outline-container-org0865f2b" class="outline-4">
<h4 id="org0865f2b">God only knows what this property is called:</h4>
<div class="outline-text-4" id="text-org0865f2b">
<p>
<i>If</i><br />
</p>

<p>
(P ⇒ Q) is true<br />
</p>

<p>
and<br />
</p>

<p>
(P̅ ⇒ Q) is true<br />
</p>

<p>
then<br />
</p>

<p>
Q is always true<br />
</p>
</div>
</div>
</div>
<div id="outline-container-org316b141" class="outline-3">
<h3 id="org316b141">Some exercices I found online :</h3>
<div class="outline-text-3" id="text-org316b141">
</div>
<div id="outline-container-orga3825f4" class="outline-4">
<h4 id="orga3825f4">USTHB 2022/2023 Section B :</h4>
<div class="outline-text-4" id="text-orga3825f4">
</div>
<ul class="org-ul">
<li><a id="orge27aa8d"></a>Exercice 1: Démontrer les équivalences suivantes:<br />
<div class="outline-text-5" id="text-orge27aa8d">
<ol class="org-ol">
<li><p>
(P ⇒ Q) ⇔ (Q̅ ⇒ P̅)<br />
</p>

<p>
Basically we are asked to prove contraposition, so here we have ( P ⇒ Q ) which is equivalent to P̅ ∨ Q <b>By definition : (P ⇒ Q) ⇔  (P̅ ∨ Q)</b><br />
</p></li>
</ol>


<p>
So we end up with : <b>(P̅ ∨ Q) ⇔ (Q̅ ⇒ P̅)</b>, now we just do the same with the second part of the contraposition. <b>(Q̅ ⇒ P̅) ⇔ (Q ∨ P̅)</b> therefor :<br />
</p>


<p>
<b>(Q ∨ P̅) ⇔ (P̅ ∨ Q)</b>, which is true because of commutativity<br />
</p>

<ol class="org-ol">
<li>not(P ⇒ Q) ⇔  P ∧ Q̅<br /></li>
</ol>


<p>
Okaaaay so, let&rsquo;s first get rid of the implication, because I don&rsquo;t like it : <b>not(P̅ ∨ Q)</b><br />
</p>


<p>
Now that we got rid of it, we can negate the whole disjunction <b>not(P̅ ∨ Q) ⇔ (P ∧ Q̅)</b>. Which is the equivalence we needed to prove<br />
</p>

<ol class="org-ol">
<li><p>
P ⇒ (Q ∧ R) ⇔ (P ⇒ Q) ∧ (P ⇒ R)<br />
</p>

<p>
One might be tempted to replace P with P̅ to get rid of the implication&#x2026;sadly this isnt it. All we have to do here is resort to <b>Distributivity</b>, because yeah, we can distribute an implication across a {con/dis}junction<br />
</p></li>

<li><p>
P ∧ (Q ∨ R) ⇔ (P ∧ Q) ∨ (P ∧ R)<br />
</p>

<p>
Literally the same as above 🩷<br />
</p></li>
</ol>
</div>
</li>
<li><a id="orgd9c7023"></a>Exercice 2: Dire si les propositions suivantes sont vraies ou fausses, et les nier:<br />
<div class="outline-text-5" id="text-orgd9c7023">
<ol class="org-ol">
<li><p>
∀x ∈ ℝ ,∃y ∈ ℝ*+, tels que e^x = y<br />
</p>

<p>
For each x from the set of Real numbers, there exists a number y from the set of non-zero positive Real numbers that satisfies the equation : e^x = y<br />
</p></li>
</ol>


<p>
&ldquo;The function f(x)=e^x is always positive and non-null&rdquo;, the very definition of an exponential function !!!!<br />
</p>


<p>
<b>So the proposition is true</b><br />
</p>


<ol class="org-ol">
<li>∃x ∈ ℝ, tels que x^2 &lt; x &lt; x^3<br /></li>
</ol>


<p>
We just need to find a value that satisifies this condition&#x2026;thankfully its easy&#x2026;.<br />
</p>

<p>&lt; x &lt; x³ , we divide the three terms by x so we get :<br />
</p>


<p>
x &lt; 1 &lt; x² , or :<br />
</p>


<p>
<b>x &lt; 1</b> ; <b>1 &lt;</b><b>x &lt; 1</b> ; <b>1 &lt; x</b> <i>We square root both sides</i><br />
</p>


<p>
We end up with a contradiction, therefor its wrong<br />
</p>


<ol class="org-ol">
<li>∀x ∈ ℝ, ∃y ∈ ℝ tels que y = 3x - 8<br /></li>
</ol>


<p>
I dont really understand this one, so let me translate it &ldquo;For any value of x from the set of Real numbers, 3x - 8 is a Real number&rdquo;&#x2026;. i mean&#x2026;.yeah, we are substracting a Real number from an other real number&#x2026;<br />
</p>

<p>
<b>Since substraction is an  Internal composition law in ℝ, therefor all results of a substraction between two Real numbers is&#x2026;Real</b><br />
</p>

<ol class="org-ol">
<li><p>
∃x ∈ ℕ, ∀y ∈ ℕ, x &gt; y ⇒ x + y &lt; 8<br />
</p>

<p>
&ldquo;There exists a number x from the set of Natural numbers such as for all values of y from the set of Natural numbers, x &gt; y implies x + y &lt; 8&rdquo;<br />
</p></li>
</ol>


<p>
Let&rsquo;s get rid of the implication :<br />
</p>

<p>
∃x ∈ ℕ, ∀y ∈ ℕ, (y &gt; x) ∨ (x + y &lt; 8) <i>There exists a number x from the set of Natural numbers such as for all values of y from the set of Natural numbers y &gt; x OR x + y &lt; 8</i><br />
</p>

<p>
This proposition is true, because there exists a value of x that satisfies this condition, it&rsquo;s <b>all numbers under 8</b> let&rsquo;s take 3 as an example:<br />
</p>


<p>
<b>x = 3 , if y &gt; 3 then the first condition is true ; if y &lt; 3 then the second one is true</b><br />
</p>


<p>
Meaning that the two propositions CAN NOT BE WRONG TOGETHER, either one is wrong, or the other<br />
</p>


<p>
y &gt; x<br />
</p>


<p>
<b>y - x &gt; 0</b><br />
</p>


<p>
y + x &lt; 8<br />
</p>


<p>
<b>y &lt; 8 - x</b> <i>This one is always true for all values of x below 8, since we are working in the set ℕ</i><br />
</p>


<ol class="org-ol">
<li><p>
∀x ∈ ℝ, x² ≥ 1 ⇔  x ≥ 1<br />
</p>

<p>
&#x2026;.This is getting stupid. of course it&rsquo;s true it&rsquo;s part of the definition of the power of 2<br />
</p></li>
</ol>
</div>
</li>
</ul>
</div>
</div>
</div>
<div id="outline-container-org21d6c03" class="outline-2">
<h2 id="org21d6c03">2éme cours <i>Oct 2</i></h2>
<div class="outline-text-2" id="text-org21d6c03">
</div>
<div id="outline-container-orgd6c9f49" class="outline-3">
<h3 id="orgd6c9f49">Quantifiers</h3>
<div class="outline-text-3" id="text-orgd6c9f49">
<p>
A propriety P can depend on a parameter x<br />
</p>


<p>
∀ is the universal quantifier which stands for &ldquo;For any value of&#x2026;&rdquo;<br />
</p>


<p>
∃ is the existential quantifier which stands for &ldquo;There exists at least one&#x2026;&rdquo;<br />
</p>
</div>
<ul class="org-ul">
<li><a id="orge92d880"></a>Example<br />
<div class="outline-text-6" id="text-orge92d880">
<p>
P(x) : x+1≥0<br />
</p>

<p>
P(X) is True or False depending on the values of x<br />
</p>
</div>
</li>
</ul>
<div id="outline-container-orgb332b43" class="outline-4">
<h4 id="orgb332b43">Proprieties</h4>
<div class="outline-text-4" id="text-orgb332b43">
</div>
<ul class="org-ul">
<li><a id="org8587885"></a>Propriety Number 1:<br />
<div class="outline-text-5" id="text-org8587885">
<p>
The negation of the universal quantifier is the existential quantifier, and vice-versa :<br />
</p>

<ul class="org-ul">
<li>not(∀x ∈ E , P(x)) ⇔ ∃ x ∈ E, not(P(x))<br /></li>
<li>not(∃x ∈ E , P(x)) ⇔ ∀ x ∈ E, not(P(x))<br /></li>
</ul>
</div>
<ul class="org-ul">
<li><a id="org3a19f5f"></a>Example:<br />
<div class="outline-text-6" id="text-org3a19f5f">
<p>
∀ x ≥ 1  x² &gt; 5 ⇔ ∃ x ≥ 1 x² &lt; 5<br />
</p>
</div>
</li>
</ul>
</li>
<li><a id="orgab7b647"></a>Propriety Number 2:<br />
<div class="outline-text-5" id="text-orgab7b647">
<p>
<b>∀x ∈ E, [P(x) ∧ Q(x)] ⇔ [∀ x ∈ E, P(x)] ∧ [∀ x ∈ E, Q(x)]</b><br />
</p>


<p>
The propriety &ldquo;For any value of x from a set E , P(x) and Q(x)&rdquo; is equivalent to &ldquo;For any value of x from a set E, P(x) AND for any value of x from a set E, Q(x)&rdquo;<br />
</p>
</div>
<ul class="org-ul">
<li><a id="org8ba49ff"></a>Example :<br />
<div class="outline-text-6" id="text-org8ba49ff">
<p>
P(x) : sqrt(x) &gt; 0 ;  Q(x) : x ≥ 1<br />
</p>


<p>
∀x ∈ ℝ*+, [sqrt(x) &gt; 0 , x ≥ 1] ⇔ [∀x ∈ R*+, sqrt(x) &gt; 0] ∧ [∀x ∈ R*+, x ≥ 1]<br />
</p>


<p>
<b>Which is true</b><br />
</p>
</div>
</li>
</ul>
</li>
<li><a id="org91796f9"></a>Propriety Number 3:<br />
<div class="outline-text-5" id="text-org91796f9">
<p>
<b>∃ x ∈ E, [P(x) ∧ Q(x)] <i></i> [∃ x ∈ E, P(x)] ∧ [∃ x ∈ E, Q(x)]</b><br />
</p>


<p>
<i>Here its an implication and not an equivalence</i><br />
</p>
</div>
<ul class="org-ul">
<li><a id="org1f20a27"></a>Example of why it&rsquo;s NOT an equivalence :<br />
<div class="outline-text-6" id="text-org1f20a27">
<p>
P(x) : x &gt; 5  ;  Q(x) : x &lt; 5<br />
</p>


<p>
Of course there is no value of x such as its inferior and superior to 5 at the same time, so obviously the proposition is false. However, the two propositions separated are correct on their own, because there is a value of x such as its superior to 5, and there is also a value of x such as its inferior to 5. This is why it&rsquo;s an implication and NOT AN EQUIVALENCE!!!<br />
</p>
</div>
</li>
</ul>
</li>
<li><a id="org2b9f54b"></a>Propriety Number 4:<br />
<div class="outline-text-5" id="text-org2b9f54b">
<p>
<b>[∀ x ∈ E, P(x)] ∨ [∀ x ∈ E, Q(x)] <i></i> ∀x ∈ E, [P(x) ∨ Q(x)]</b><br />
</p>


<p>
<i>Same here, implication and NOT en equivalence</i><br />
</p>
</div>
</li>
</ul>
</div>
</div>
<div id="outline-container-orged685c1" class="outline-3">
<h3 id="orged685c1">Multi-parameter proprieties :</h3>
<div class="outline-text-3" id="text-orged685c1">
<p>
A propriety P can depend on two or more parameters, for convenience we call them x,y,z&#x2026;etc<br />
</p>
</div>
<ul class="org-ul">
<li><a id="org747b217"></a>Example :<br />
<div class="outline-text-6" id="text-org747b217">
<p>
P(x,y): x+y &gt; 0<br />
</p>


<p>
P(0,1) is a True proposition<br />
</p>


<p>
P(-2,-1) is a False one<br />
</p>
</div>
</li>
<li><a id="org5d93eaf"></a>WARNING :<br />
<div class="outline-text-6" id="text-org5d93eaf">
<p>
∀x ∈ E, ∃y ∈ F , P(x,y)<br />
</p>


<p>
∃y ∈ F, ∀x ∈ E , P(x,y)<br />
</p>


<p>
Are different because in the first one y depends on x, while in the second one, it doesn&rsquo;t<br />
</p>
</div>
<ul class="org-ul">
<li><a id="orgc60c61d"></a>Example :<br />
<div class="outline-text-7" id="text-orgc60c61d">
<p>
∀ x ∈ ℕ , ∃ y ∈ ℕ y &gt; x -&#x2013;&#x2014; True<br />
</p>


<p>
∃ y ∈ ℕ , ∀ x ∈ ℕ y &gt; x -&#x2013;&#x2014; False<br />
</p>
</div>
</li>
</ul>
</li>
</ul>
<li><a id="orgda9f614"></a>Proprieties :<br />
<div class="outline-text-5" id="text-orgda9f614">
<ol class="org-ol">
<li>not(∀x ∈ E ,∃y ∈ F P(x,y)) ⇔ ∃x ∈ E, ∀y ∈ F not(P(x,y))<br /></li>
<li>not(∃x ∈ E ,∀y ∈ F P(x,y)) ⇔ ∀x ∈ E, ∃y ∈ F not(P(x,y))<br /></li>
</ol>
</div>
</li>
</ul>
</div>
<div id="outline-container-org78d7ed0" class="outline-3">
<h3 id="org78d7ed0">Methods of mathematical reasoning :</h3>
<div class="outline-text-3" id="text-org78d7ed0">
</div>
<div id="outline-container-org7d21c38" class="outline-4">
<h4 id="org7d21c38">Direct reasoning :</h4>
<div class="outline-text-4" id="text-org7d21c38">
<p>
To show that an implication P ⇒ Q is true, we suppose that P is true and we show that Q is true<br />
</p>
</div>
<ul class="org-ul">
<li><a id="org59d34b3"></a>Example:<br />
<div class="outline-text-5" id="text-org59d34b3">
<p>
Let a,b be two Real numbers, we have to prove that <b>a² + b² = 1 ⇒ |a + b| ≤ 2</b><br />
</p>


<p>
We suppose that a²+b² = 1 and we prove that |a + b| ≤ 2<br />
</p>


<p>
a²+b²=1 ⇒  b² = 1 - a² ; a² = 1 - b²<br />
</p>


<p>
a²+b²=1 ⇒  1 - a² ≥ 0 ; 1 - b² ≥ 0<br />
</p>


<p>
a²+b²=1 ⇒  a² ≤ 1 ; b² ≤ 1<br />
</p>


<p>
a²+b²=1 ⇒ -1 ≤ a ≤ 1 ; -1 ≤ b ≤ 1<br />
</p>


<p>
a²+b²=1 ⇒ -2 ≤ a + b ≤ 2<br />
</p>


<p>
a²+b²=1 ⇒ |a + b| ≤ 2 <b>Which is what we wanted to prove, therefor the implication is correct</b><br />
</p>
</div>
</li>
</ul>
</div>
<div id="outline-container-orgcfd8723" class="outline-4">
<h4 id="orgcfd8723">Reasoning by the Absurd:</h4>
<div class="outline-text-4" id="text-orgcfd8723">
<p>
To prove that a proposition is True, we suppose that it&rsquo;s False and we must come to a contradiction<br />
</p>


<p>
And to prove that an implication P ⇒ Q is true using the reasoning by the absurd, we suppose that  P ∧ not(Q) is true, and then we come to a contradiction as well<br />
</p>
</div>
<ul class="org-ul">
<li><a id="orga4a0e2d"></a>Example:<br />
<div class="outline-text-5" id="text-orga4a0e2d">
<p>
Prove that this proposition is correct using the reasoning by the absurd : ∀x ∈ ℝ* , sqrt(1+x²) ≠ 1 + x²/2<br />
</p>


<p>
We assume that ∃ x ℝ* , sqrt(1+x²) = 1 + x²/2<br />
</p>


<p>
sqrt(1+x²) = 1 + x²/2 ; 1 + x² = (1+x²/2)² ; 1 + x² = 1 + x^4/4 + x²  ;  x^(4)/4 = 0 &#x2026; Which contradicts with our proposition, since x = 4 and we are working on the ℝ* set<br />
</p>
</div>
</li>
</ul>
</div>
<div id="outline-container-org102d3fa" class="outline-4">
<h4 id="org102d3fa">Reasoning by contraposition:</h4>
<div class="outline-text-4" id="text-org102d3fa">
<p>
If an implication P ⇒ Q is too hard to prove, we just have to prove not(Q) ⇒ not(P) is true !!! or in other words that both not(P) and not(Q) are true<br />
</p>
</div>
</div>
<div id="outline-container-org81cb388" class="outline-4">
<h4 id="org81cb388">Reasoning by counter example:</h4>
<div class="outline-text-4" id="text-org81cb388">
<p>
To prove that a proposition ∀x ∈ E, P(x) is false, all we have to do is find a single value of x from E such as not(P(x)) is true<br />
</p>
</div>
</div>
</div>
</div>
<div id="outline-container-orgc2178b8" class="outline-2">
<h2 id="orgc2178b8">3eme Cours : <i>Oct 9</i></h2>
<div class="outline-text-2" id="text-orgc2178b8">
</div>
<div id="outline-container-org4855f6f" class="outline-4">
<h4 id="org4855f6f">Reasoning by recurrence :</h4>
<div class="outline-text-4" id="text-org4855f6f">
<p>
P is a propriety dependent of <b>n ∈ ℕ</b>. If for n0 ∈ ℕ P(n0) is true, and if for n ≥ n0 (P(n) ⇒ P(n+1)) is true. Then P(n) is true for n ≥ n0<br />
</p>
</div>
<ul class="org-ul">
<li><a id="orga792d9c"></a>Example:<br />
<div class="outline-text-5" id="text-orga792d9c">
<p>
Let&rsquo;s prove that ∀ n ≥ 1 , (n,k=1)Σk = [n(n+1)]/2<br />
</p>


<p>
P(n) : (n,k=1)Σk = [n(n+1)]/2<br />
</p>



<p>
<b>Pour n = 1:</b> (1,k=1)Σk = 1 ; [n(n+1)]/2 = 1 . <b>So P(1) is true</b><br />
</p>



<p>
For n ≥ 1. We assume that P(n) is true, OR : <b>(n, k=1)Σk = n(n+1)/2</b>. We now have to prove that P(n+1) is true, Or : <b>(n+1, k=1)Σk = (n+1)(n+2)/2</b><br />
</p>


<p>
(n+1, k=1)Σk = 1 + 2 + &#x2026;. + n + (n+1) ; (n+1, k=1)Σk = (n, k=1)Σk + (n+1) ; = n(n+1)/2 + (n+1) ; = [n(n+1) + 2(n+1)]/2 ; = <b>[(n+2)(n+1)]/2</b> <i>WHICH IS WHAT WE NEEDED TO FIND</i><br />
</p>


<p>
<b>Conclusion: ∀n ≥ 1 , (n,k=1)Σk = n(n+1)/2</b><br />
</p>
</div>
</li>
</ul>
</div>
</div>
<div id="outline-container-orgde6bfac" class="outline-2">
<h2 id="orgde6bfac">4eme Cours : Chapitre 2 : Sets and Operations</h2>
<div class="outline-text-2" id="text-orgde6bfac">
</div>
<div id="outline-container-orgfe8000a" class="outline-3">
<h3 id="orgfe8000a">Definition of a set :</h3>
<div class="outline-text-3" id="text-orgfe8000a">
<p>
A set is a collection of objects that share the sane propriety<br />
</p>
</div>
</div>
<div id="outline-container-orgfe04671" class="outline-3">
<h3 id="orgfe04671">Belonging, inclusion, and equality :</h3>
<div class="outline-text-3" id="text-orgfe04671">
<ol class="org-ol">
<li>Let E be a set. If x is an element of E, we say that x belongs to E we write <b>x ∈ E</b>, and if it doesn&rsquo;t, we write <b>x ∉ E</b><br /></li>
<li>A set E is included in a set F if all elements of E are elements of F and we write <b>E ⊂ F ⇔ (∀x , x ∈ E ⇒ x ∈ F)</b>. We say that E is a subset of F, or a part of F. The negation of this propriety is : <b>E ⊄ F ⇔ ∃x , x ∈ E and x ⊄ F</b><br /></li>
<li>E and F are equal if E is included in F and F is included in E, and we write <b>E = F ⇔ (E ⊂ F) et (F ⊂ E)</b><br /></li>
<li>The empty set (symbolized by ∅) is a set without elements, and is included in all sets (by convention) : <b>∅ ⊂ E</b><br /></li>
</ol>
</div>
</div>
<div id="outline-container-orga2eb99d" class="outline-3">
<h3 id="orga2eb99d">Intersections and reunions :</h3>
<div class="outline-text-3" id="text-orga2eb99d">
</div>
<div id="outline-container-org560d563" class="outline-4">
<h4 id="org560d563">Intersection:</h4>
<div class="outline-text-4" id="text-org560d563">
<p>
E ∩ F = {x / x ∈ E AND x ∈ F} ; x ∈ E ∩ F ⇔ x ∈ F AND x ∈ F<br />
</p>


<p>
x ∉ E ∩ F ⇔ x ∉ E OR x ∉ F<br />
</p>
</div>
</div>
<div id="outline-container-org7147bc3" class="outline-4">
<h4 id="org7147bc3">Union:</h4>
<div class="outline-text-4" id="text-org7147bc3">
<p>
E ∪ F = {x / x ∈ E OR x ∈ F} ;  x ∈ E ∪ F ⇔ x ∈ F OR x ∈ F<br />
</p>


<p>
x ∉ E ∪ F ⇔ x ∉ E AND x ∉ F<br />
</p>
</div>
</div>
<div id="outline-container-org16b5ab2" class="outline-4">
<h4 id="org16b5ab2">Difference between two sets:</h4>
<div class="outline-text-4" id="text-org16b5ab2">
<p>
E(Which is also written as : E - F) = {x / x ∈ E and x ∉ F}<br />
</p>
</div>
</div>
<div id="outline-container-orgdac190b" class="outline-4">
<h4 id="orgdac190b">Complimentary set:</h4>
<div class="outline-text-4" id="text-orgdac190b">
<p>
If F ⊂ E. E - F is the complimentary of F in E.<br />
</p>


<p>
FCE = {x /x ∈ E AND x ∉ F} <b>ONLY WHEN F IS A SUBSET OF E</b><br />
</p>
</div>
</div>
<div id="outline-container-org4e0b111" class="outline-4">
<h4 id="org4e0b111">Symmetrical difference</h4>
<div class="outline-text-4" id="text-org4e0b111">
<p>
E Δ F = (E - F) ∪ (F - E) ; = (E ∪ F) - (E ∩ F)<br />
</p>
</div>
</div>
</div>
<div id="outline-container-org691c863" class="outline-3">
<h3 id="org691c863">Proprieties :</h3>
<div class="outline-text-3" id="text-org691c863">
<p>
Let E,F and G be 3 sets. We have :<br />
</p>
</div>
<div id="outline-container-org9cc9f31" class="outline-4">
<h4 id="org9cc9f31">Commutativity:</h4>
<div class="outline-text-4" id="text-org9cc9f31">
<p>
E ∩ F = F ∩ E<br />
E ∪ F = F ∪ E<br />
</p>
</div>
</div>
<div id="outline-container-org471083b" class="outline-4">
<h4 id="org471083b">Associativity:</h4>
<div class="outline-text-4" id="text-org471083b">
<p>
E ∩ (F ∩ G) = (E ∩ F) ∩ G<br />
E ∪ (F ∪ G) = (E ∪ F) ∪ G<br />
</p>
</div>
</div>
<div id="outline-container-orge63be10" class="outline-4">
<h4 id="orge63be10">Distributivity:</h4>
<div class="outline-text-4" id="text-orge63be10">
<p>
E ∩ (F ∪ G) = (E ∩ F) ∪ (E ∩ G)<br />
E ∪ (F ∩ G) = (E ∪ F) ∩ (E ∪ G)<br />
</p>
</div>
</div>
<div id="outline-container-orgfb01947" class="outline-4">
<h4 id="orgfb01947">Lois de Morgan:</h4>
<div class="outline-text-4" id="text-orgfb01947">
<p>
If E ⊂ G and F ⊂ G ;<br />
</p>

<p>
(E ∩ F)CG = ECG ∪ FCG ; (E ∪ F)CG = ECG ∩ FCG<br />
</p>
</div>
</div>
<div id="outline-container-orge1a41eb" class="outline-4">
<h4 id="orge1a41eb">An other one:</h4>
<div class="outline-text-4" id="text-orge1a41eb">
<p>
E - (F ∩ G) = (E-F) ∪ (E-G) ;  E - (F ∪ G) = (E-F) ∩ (E-G)<br />
</p>
</div>
</div>
<div id="outline-container-org9939b0f" class="outline-4">
<h4 id="org9939b0f">An other one:</h4>
<div class="outline-text-4" id="text-org9939b0f">
<p>
E ∩ ∅ = ∅ ; E ∪ ∅ = E<br />
</p>
</div>
</div>
<div id="outline-container-org90bfdc4" class="outline-4">
<h4 id="org90bfdc4">And an other one:</h4>
<div class="outline-text-4" id="text-org90bfdc4">
<p>
E ∩ (F Δ G) = (E ∩ F) Δ (E ∩ G)<br />
</p>
</div>
</div>
<div id="outline-container-org1e49001" class="outline-4">
<h4 id="org1e49001">And the last one:</h4>
<div class="outline-text-4" id="text-org1e49001">
<p>
E Δ ∅ = E ; E Δ E = ∅<br />
</p>
</div>
</div>
</div>
</div>
<div id="outline-container-org272eca3" class="outline-2">
<h2 id="org272eca3">5eme cours: L&rsquo;ensemble des parties d&rsquo;un ensemble <i>Oct 16</i></h2>
<div class="outline-text-2" id="text-org272eca3">
<p>
Let E be a set. We define P(E) as the set of all parts of E : <b>P(E) = {X/X ⊂ E}</b><br />
</p>
</div>
<div id="outline-container-org5b29e32" class="outline-4">
<h4 id="org5b29e32">Notes :</h4>
<div class="outline-text-4" id="text-org5b29e32">
<p>
∅ ∈ P(E) ; E ∈ P(E)<br />
</p>


<p>
cardinal E = n <i>The number of terms in E</i> , cardinal P(E) = 2^n <i>The number of all parts of E</i><br />
</p>
</div>
</div>
<div id="outline-container-org5636bd8" class="outline-4">
<h4 id="org5636bd8">Examples :</h4>
<div class="outline-text-4" id="text-org5636bd8">
<p>
E = {a,b,c} ; P(E)={∅, {a}, {b}, {c}, {a,b}, {b,c}, {a,c}, {a,b,c}}<br />
</p>
</div>
</div>
<div id="outline-container-orgd8cb2c3" class="outline-3">
<h3 id="orgd8cb2c3">Partition of a set :</h3>
<div class="outline-text-3" id="text-orgd8cb2c3">
<p>
We say that <b>A</b> is a partition of E if:<br />
</p>
<ol class="org-ol">
<li>∀ x ∈ A , x ≠ 0<br /></li>
<li>All the elements of <b>A</b> are two by two disjoint. Or in other terms, there should not be two elements that intersects with each other.<br /></li>
<li>The reunion of all elements of <b>A</b> is equal to E<br /></li>
</ol>
</div>
</div>
<div id="outline-container-orgf40404d" class="outline-3">
<h3 id="orgf40404d">Cartesian products :</h3>
<div class="outline-text-3" id="text-orgf40404d">
<p>
Let E and F be two sets, the set EXF = {(x,y)/ x ∈ E AND y ∈ F} is called the Cartesian product of E and F<br />
</p>
</div>
<div id="outline-container-orgd526cb8" class="outline-4">
<h4 id="orgd526cb8">Example :</h4>
<div class="outline-text-4" id="text-orgd526cb8">
<p>
A = {4,5} ; B= {4,5,6} ; AxB = {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6)}<br />
</p>


<p>
BxA = {(4,4), (4,5), (5,4), (5,5), (6,4), (6,5)} ; Therefore AxB ≠ BxA<br />
</p>
</div>
</div>
<div id="outline-container-org56dd088" class="outline-4">
<h4 id="org56dd088">Some proprieties:</h4>
<div class="outline-text-4" id="text-org56dd088">
<ol class="org-ol">
<li>ExF = ∅ ⇔ E=∅ OR F=∅<br /></li>
<li>ExF = FxE ⇔ E=F OR E=∅ OR F=∅<br /></li>
<li>E x (F∪G) = (ExF) ∪ (ExG)<br /></li>
<li>(E∪F) x G = (ExG) ∪ (FxG)<br /></li>
<li>(E∪F) ∩ (GxH) = (E ∩ G) x (F ∩ H)<br /></li>
<li>Generally speaking : (ExF) ∪ (GxH) ≠ (E∪G) x (F∪H)<br /></li>
</ol>
</div>
</div>
</div>
</div>
<div id="outline-container-org5ee4278" class="outline-2">
<h2 id="org5ee4278">Binary relations in a set :</h2>
<div class="outline-text-2" id="text-org5ee4278">
</div>
<div id="outline-container-orgddc9af6" class="outline-3">
<h3 id="orgddc9af6">Definition :</h3>
<div class="outline-text-3" id="text-orgddc9af6">
<p>
Let E be a set and x,y ∈ E. If there exists a link between x and y, we say that they are tied by a relation <b>R</b> and we write <b>xRy</b><br />
</p>
</div>
</div>
<div id="outline-container-orge65424e" class="outline-3">
<h3 id="orge65424e">Proprieties :</h3>
<div class="outline-text-3" id="text-orge65424e">
<p>
Let E be a set and R a relation defined in E<br />
</p>
<ol class="org-ol">
<li>We say that R is reflexive if ∀ x ∈ E, xRx (for any element x in E,x is related to itself)<br /></li>
<li>We say that R is symmetrical if ∀ x,y ∈ E , xRy ⇒ yRx<br /></li>
<li>We say that R is transitive if ∀ x,y,z ∈ E (xRy , yRz) ⇒ xRz<br /></li>
<li>We say that R is anti-symmetrical if ∀ x,y ∈ E xRy AND yRx ⇒ x = y<br /></li>
</ol>
</div>
</div>
<div id="outline-container-orgd7877d3" class="outline-3">
<h3 id="orgd7877d3">Equivalence relationship :</h3>
<div class="outline-text-3" id="text-orgd7877d3">
<p>
We say that R is a relation of equivalence in E if its reflexive, symetrical and transitive<br />
</p>
</div>
<div id="outline-container-org85cf025" class="outline-4">
<h4 id="org85cf025">Equivalence class :</h4>
<div class="outline-text-4" id="text-org85cf025">
<p>
Let R be a relation of equivalence in E and a ∈ E, we call equivalence class of <b>a</b>, and we write ̅a or ȧ, or cl a the following set :<br />
</p>


<p>
<b>a̅ = {y ∈ E/ y R a}</b><br />
</p>
</div>
<ul class="org-ul">
<li><a id="orga316a01"></a>The quotient set :<br />
<div class="outline-text-5" id="text-orga316a01">
<p>
E/R = {̅a , a ∈ E}<br />
</p>
</div>
</li>
</ul>
</div>
</div>
<div id="outline-container-orge18dcc7" class="outline-3">
<h3 id="orge18dcc7">Order relationship :</h3>
<div class="outline-text-3" id="text-orge18dcc7">
<p>
Let E be a set and R be a relation defined in E. We say that R is a relation of order if its reflexive, anti-symetrical and transitive.<br />
</p>
<ol class="org-ol">
<li>The order R is called total if ∀ x,y ∈ E xRy OR yRx<br /></li>
<li>The order R is called partial if ∃ x,y ∈ E xR̅y AND yR̅x<br /></li>
</ol>
</div>
<div id="outline-container-org60d471a" class="outline-4">
<h4 id="org60d471a"><span class="todo TODO">TODO</span> Examples :</h4>
<div class="outline-text-4" id="text-org60d471a">
<p>
∀x,y ∈ ℝ , xRy ⇔ x²-y²=x-y<br />
</p>
<ol class="org-ol">
<li>Prove that R is an equivalence relation<br /></li>
<li>Let a ∈ ℝ, find ̅a<br /></li>
</ol>
</div>
</div>
</div>
</div>
<div id="outline-container-org77de6e3" class="outline-2">
<h2 id="org77de6e3">TP exercices <i>Oct 20</i> :</h2>
<div class="outline-text-2" id="text-org77de6e3">
</div>
<div id="outline-container-org3ca8006" class="outline-3">
<h3 id="org3ca8006">Exercice 3 :</h3>
<div class="outline-text-3" id="text-org3ca8006">
</div>
<div id="outline-container-orgad95ec3" class="outline-4">
<h4 id="orgad95ec3">Question 3</h4>
<div class="outline-text-4" id="text-orgad95ec3">
<p>
Montrer par l&rsquo;absurde que P : ∀x ∈ ℝ*, √(4+x³) ≠ 2 + x³/4 est vraies<br />
</p>

<p class="verse">
On suppose que ∃ x ∈ ℝ* , √(4+x³) = 2 + x³/4<br />
4+x³ = (2 + x³/4)²<br />
4+x³ = 4 + x⁶/16 + 4*(x³/4)<br />
4+x³ = 4 + x⁶/16 + x³<br />
x⁶/16 = 0<br />
x⁶ = 0<br />
x = 0 . Or, x appartiens a ℝ\{0}, donc P̅ est fausse. Ce qui est equivalent a dire que P est vraie<br />
</p>
</div>
</div>
</div>
<div id="outline-container-org8180ae0" class="outline-3">
<h3 id="org8180ae0">Exercice 4 :</h3>
<div class="outline-text-3" id="text-org8180ae0">
</div>
<div id="outline-container-orgfe0b1e2" class="outline-4">
<h4 id="orgfe0b1e2"><span class="done DONE">DONE</span> Question 1 :</h4>
<div class="outline-text-4" id="text-orgfe0b1e2">
<p class="verse">
∀ n ∈ ℕ* , (n ,k=1)Σ1/k(k+1) = 1 - 1/1+n<br />
P(n) : (n ,k=1)Σ1/k(k+1) = 1 - 1/1+n<br />
1. <b>On vérifie P(n) pour n = 1</b><br />
(1 ,k=1)Σ1/k(k+1) = 1/1(1+1)<br />
&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;= 1/2 &#x2014; (1)<br />
1 - 1/1+1         = 1 - 1/2<br />
&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;= 1/2 &#x2014; (2)<br />
De (1) et (2), P(0) est vraie -&#x2014; (a)<br />
<br />
2. <b>On suppose que P(n) est vraie pour n ≥ n1 puis on vérifie pour n+1</b><br />
(n ,k=1)Σ1/k(k+1) = 1 - 1/1+n<br />
(n ,k=1)Σ1/k(k+1) + 1/(n+1)(n+2) = 1 - (1/(1+n)) + 1/(n+1)(n+2)<br />
(n+1 ,k=1)Σ1/k(k+1) = 1 - 1/(n+1) + 1/[(n+1)(n+2)]<br />
(n+1 ,k=1)Σ1/k(k+1) = 1 + 1/[(n+1)(n+2)] - (n+2)/[(n+1)(n+2)]<br />
(n+1 ,k=1)Σ1/k(k+1) = 1 + [1-(n+2)]/[(n+1)(n+2)]<br />
(n+1 ,k=1)Σ1/k(k+1) = 1 + [-n-1]/[(n+1)(n+2)]<br />
(n+1 ,k=1)Σ1/k(k+1) = 1 - [n+1]/[(n+1)(n+2)]<br />
(n+1 ,k=1)Σ1/k(k+1) = 1 - 1/(n+1+1) <b>CQFD</b><br />
<br />
Donc P(n+1) est vraie. -&#x2014; (b)<br />
De (a) et (b) on conclus que la proposition de départ est vraie<br />
</p>
</div>
</div>
</div>
</div>
<div id="outline-container-org2d6e0ba" class="outline-2">
<h2 id="org2d6e0ba">Chapter 3 : Applications</h2>
<div class="outline-text-2" id="text-org2d6e0ba">
</div>
<div id="outline-container-orga5be12f" class="outline-3">
<h3 id="orga5be12f">3.1 Generalities about applications :</h3>
<div class="outline-text-3" id="text-orga5be12f">
</div>
<div id="outline-container-org805d7bc" class="outline-4">
<h4 id="org805d7bc">Definition :</h4>
<div class="outline-text-4" id="text-org805d7bc">
<p>
Let E and F be two sets.<br />
</p>
<ol class="org-ol">
<li>We call a function of the set E to the set F any relation from E to F such as for any element of E, we can find <span class="underline">at most one</span> element of F that corresponds to it.<br /></li>
<li>We call an application of the set E to the set F a relation from E to F such as for any element of E, we can find <span class="underline">one and only one</span> element of F that corresponds to it.<br /></li>
<li><p>
f: E<sub>1</sub> &#x2014;&gt; F<sub>1</sub> ; g: E<sub>2</sub> &#x2014;&gt; F<sub>2</sub> ; f ≡ g ⇔ [E<sub>1 </sub>= E<sub>2</sub> ; F<sub>1</sub> = F<sub>2</sub> ; f(x) = g(x) ∀x ∈ E<sub>1</sub><br />
</p>

<p>
Generally speaking, we schematize a function or an application by this writing :<br />
</p>
<p class="verse">
f : E &#x2014;&gt; F<br />
&#xa0;&#xa0;&#xa0;&#xa0;x &#x2014;&gt; f(x)=y<br />
&#xa0;&#xa0;&#xa0;Γ = {(x , f(x))/ x ∈ E ; f(x) ∈ F} is the graph of f<br />
</p></li>
</ol>
</div>
<ul class="org-ul">
<li><a id="org2936c19"></a>Some examples :<br />
<ul class="org-ul">
<li><a id="orgd77c836"></a>Ex1:<br />
<div class="outline-text-6" id="text-orgd77c836">
<p class="verse">
f : ℝ &#x2014;&gt;<br />
&#xa0;&#xa0;&#xa0;&#xa0;x &#x2014;&gt; f(x) = (x-1)/x<br />
is a function, because 0 does NOT have a corresponding element using that relation.<br />
</p>
</div>
</li>
<li><a id="orga45fd32"></a>Ex2:<br />
<div class="outline-text-6" id="text-orga45fd32">
<p class="verse">
f : ℝ<sup>*</sup> &#x2014;&gt;<br />
&#xa0;&#xa0;&#xa0;&#xa0;x &#x2014;&gt; f(x)= (x-1)/x<br />
is, however, an application<br />
</p>
</div>
</li>
</ul>
</li>
</ul>
</div>
<div id="outline-container-org7947331" class="outline-4">
<h4 id="org7947331">Restriction and prolongation of an application :</h4>
<div class="outline-text-4" id="text-org7947331">
<p>
Let f : E -&gt; F an application and E<sub>1</sub> ⊂ E therefore :<br />
</p>
<p class="verse">
g : E<sub>1</sub> -&gt; F<br />
g(x) = f(x) ∀x ∈ E<sub>1</sub><br />
<br />
g is called the <b>restriction</b> of f to E<sub>1</sub>. And f is called the <b>prolongation</b> of g to E.<br />
</p>
</div>
<ul class="org-ul">
<li><a id="org7c848c6"></a>Example<br />
<div class="outline-text-5" id="text-org7c848c6">
<p class="verse">
f : ℝ &#x2014;&gt;<br />
&#xa0;&#xa0;&#xa0;&#xa0;x &#x2014;&gt; f(x) = x<sup>2</sup><br />
<br />
g : [0 , <del>∞[ &#x2014;&gt;<br />
&#xa0;&#xa0;&#xa0;&#xa0;x &#x2014;&gt; g(x) = x²<br />
<br />
g is called the <b>restriction</b> of f to ℝ^{</del>}. And f is called the <b>prolongation</b> of g to ℝ.<br />
</p>
</div>
</li>
</ul>
</div>
<div id="outline-container-orgd94bc69" class="outline-4">
<h4 id="orgd94bc69">Composition of applications :</h4>
<div class="outline-text-4" id="text-orgd94bc69">
<p>
Let E,F, and G be three sets, f: E -&gt; F and g: F -&gt; G are two applications. We define their composition, symbolized by g<sub>o</sub>f as follow :<br />
</p>


<p>
g<sub>o</sub>f : E -&gt; G . ∀x ∈ E (g<sub>o</sub>f)<sub>(x)</sub>= g(f(x))<br />
</p>
</div>
</div>
</div>
<div id="outline-container-org257d05a" class="outline-3">
<h3 id="org257d05a">3.2 Injection, surjection and bijection :</h3>
<div class="outline-text-3" id="text-org257d05a">
<p>
Let f: E -&gt; F be an application :<br />
</p>
<ol class="org-ol">
<li>We say that f is injective if : ∀x,x&rsquo; ∈ E : f(x) = f(x&rsquo;) ⇒ x = x&rsquo;<br /></li>
<li>We say that f is surjective if : ∀ y ∈ F , ∃ x ∈ E : y = f(x)<br /></li>
<li>We say that if is bijective if it&rsquo;s both injective and surjective at the same time.<br /></li>
</ol>
</div>
<div id="outline-container-org1612e09" class="outline-4">
<h4 id="org1612e09">Proposition :</h4>
<div class="outline-text-4" id="text-org1612e09">
<p>
Let f : E -&gt; F be an application. Therefore:<br />
</p>
<ol class="org-ol">
<li>f is injective ⇔ y = f(x) has at most one solution.<br /></li>
<li>f is surjective ⇔ y = f(x) has at least one solution.<br /></li>
<li>f is bijective ⇔ y = f(x) has a single and unique solution.<br /></li>
</ol>
</div>
</div>
</div>
<div id="outline-container-orgebdf518" class="outline-3">
<h3 id="orgebdf518">3.3 Reciprocal applications :</h3>
<div class="outline-text-3" id="text-orgebdf518">
</div>
<div id="outline-container-orgf072e42" class="outline-4">
<h4 id="orgf072e42">Def :</h4>
<div class="outline-text-4" id="text-orgf072e42">
<p>
Let f : E -&gt; F a bijective application. So there exists an application named f<sup>-1</sup> : F -&gt; E such as : y = f(x) ⇔ x = f<sup>-1</sup>(y)<br />
</p>
</div>
</div>
<div id="outline-container-org244b352" class="outline-4">
<h4 id="org244b352">Theorem :</h4>
<div class="outline-text-4" id="text-org244b352">
<p>
Let f : E -&gt; F be a bijective application. Therefore its reciprocal f<sup>-1</sup> verifies : f<sup>-1</sup><sub>o</sub>f=Id<sub>E </sub>; f<sub>o</sub>f<sup>-1</sup>=Id<sub>F</sub> Or :<br />
</p>


<p>
Id<sub>E</sub> : E -&gt; E ; x -&gt; Id<sub>E</sub>(x) = x<br />
</p>
</div>
</div>
<div id="outline-container-org1479c0e" class="outline-4">
<h4 id="org1479c0e">Some proprieties :</h4>
<div class="outline-text-4" id="text-org1479c0e">
<ol class="org-ol">
<li>(f<sup>-1</sup>)<sup>-1</sup> = f<br /></li>
<li>(g<sub>o</sub>f)⁻¹ = f⁻¹<sub>o</sub>g⁻¹<br /></li>
<li>The graphs of f and f⁻¹ are symmetrical to each other by the first bis-sectrice of the equation y = x<br /></li>
</ol>
</div>
</div>
</div>
<div id="outline-container-orgaf81bb3" class="outline-3">
<h3 id="orgaf81bb3">3.4 Direct Image and reciprocal Image :</h3>
<div class="outline-text-3" id="text-orgaf81bb3">
</div>
<div id="outline-container-org87b91e2" class="outline-4">
<h4 id="org87b91e2">Direct Image :</h4>
<div class="outline-text-4" id="text-org87b91e2">
<p>
Let f: E-&gt; F be an application and A ⊂ E. We call a direct image of A by f, and we symbolize as f(A) the subset of F defined by :<br />
</p>


<p>
f(A) = {f(x)/ x ∈ A} ; = { y ∈ F ∃ x ∈ A  y=f(x)}<br />
</p>
</div>
<ul class="org-ul">
<li><a id="orgb5bc08c"></a>Example :<br />
<div class="outline-text-5" id="text-orgb5bc08c">
<p class="verse">
f: ℝ -&gt;<br />
&#xa0;&#xa0;&#xa0;x -&gt; f(x) = x²<br />
A = {0,4}<br />
f(A) = {f(0), f(4)} = {0, 16}<br />
</p>
</div>
</li>
</ul>
</div>
<div id="outline-container-org500bc40" class="outline-4">
<h4 id="org500bc40">Reciprocal image :</h4>
<div class="outline-text-4" id="text-org500bc40">
<p>
Let f: E -&gt; F be an application and B ⊂ F. We call the reciprocal image of E by F the subset f<sup>-1</sup>(B) :<br />
</p>


<p>
f<sup>-1</sup>(B) = {x ∈ E/f(x) ∈ B} ; x ∈ f<sup>-1</sup>(B) ⇔ f(x) ∈ B<br />
</p>
</div>
<ul class="org-ul">
<li><a id="org885d21d"></a>Example :<br />
<div class="outline-text-5" id="text-org885d21d">
<p class="verse">
f: ℝ -&gt;<br />
&#xa0;&#xa0;&#xa0;x -&gt; f(x) = x²<br />
B = {1,9,4}<br />
f<sup>-1</sup>(B) = {1,-1,2,-2,3,-3}<br />
&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;&#xa0;= {x ∈ ℝ/x² ∈ {1,4,9}}<br />
</p>
</div>
</li>
</ul>
</div>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="author">Author: Crystal</p>
<p class="date">Created: 2023-11-01 Wed 20:17</p>
</div>
</body>
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