{"id":204,"date":"2023-02-01T00:03:35","date_gmt":"2023-02-01T00:03:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/ct-state-quantitative-reasoning\/chapter\/demorgans-laws\/"},"modified":"2024-01-10T15:32:41","modified_gmt":"2024-01-10T15:32:41","slug":"demorgans-laws","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ct-state-quantitative-reasoning\/chapter\/demorgans-laws\/","title":{"raw":"DeMorgan's Laws","rendered":"DeMorgan&#8217;s Laws"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use DeMorgan\u2019s laws to define logical equivalences of a statement<\/li>\r\n<\/ul>\r\n<\/div>\r\nThere are two pairs of logically equivalent statements that come up\u00a0again and again in logic. They are prevalent\u00a0enough to be dignified by a special name: <b>DeMorgan\u2019s laws.<\/b>\r\n\r\nThe laws are named after Augustus De Morgan (1806\u20131871),\u00a0who introduced a formal version of the laws to classical propositional logic. De Morgan's formulation was influenced by algebraization of logic undertaken by George Boole, which later cemented De Morgan's claim to the find. Nevertheless, a similar observation was made by Aristotle, and was known to Greek and Medieval logicians.\u00a0For example, in the 14th century, William of Ockham wrote down the words that would result by reading the laws out. Jean Buridan, in his <i>Summulae de Dialectica<\/i>, also describes rules of conversion that follow the lines of De Morgan's laws.\u00a0Still, De Morgan is given credit for stating the laws in the terms of modern formal logic, and incorporating them into the language of logic. De Morgan's laws can be proved easily, and may even seem trivial.\u00a0Nonetheless, these laws are helpful in making valid inferences in proofs and deductive arguments.\r\n<div class=\"textbox examples\">\r\n<h3>success strategy<\/h3>\r\nGet plenty of practice and repetition with the ideas in this page! The notation will become more familiar as you do. Remember to get help if you need it!\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>DeMorgan\u2019s Laws<\/h3>\r\n<ol>\r\n \t<li>[latex]\\sim\\left(P{\\wedge}Q\\right)=({\\sim}P)\\vee\\left(\\sim{Q}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\sim\\left(P\\vee{Q}\\right)=\\left(\\sim{P}\\right)\\wedge\\left(\\sim{Q}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\nThe first of DeMorgan\u2019s laws is verified by the following table. You are\u00a0asked to verify the second in an\u00a0exercise.\r\n<table cellspacing=\"0\" cellpadding=\"0\">\r\n<tbody>\r\n<tr>\r\n<td><b>[latex]P[\/latex]<\/b><\/td>\r\n<td><b>[latex]Q[\/latex]<\/b><\/td>\r\n<td><b>[latex]\\sim{P}[\/latex]<\/b><\/td>\r\n<td><b>[latex]\\sim{Q}[\/latex]<\/b><\/td>\r\n<td>[latex]P\\wedge{Q}[\/latex]<\/td>\r\n<td><strong>[latex]\\sim\\left(P\\wedge{Q}\\right)[\/latex]<\/strong><\/td>\r\n<td><b>[latex]\\left(\\sim{P}\\right)\\vee\\left(\\sim{Q}\\right)[\/latex]<\/b><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>T<\/td>\r\n<td>T<\/td>\r\n<td>F<\/td>\r\n<td>F<\/td>\r\n<td>T<\/td>\r\n<td>F<\/td>\r\n<td>F<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>T<\/td>\r\n<td>F<\/td>\r\n<td>F<\/td>\r\n<td>T<\/td>\r\n<td>F<\/td>\r\n<td>T<\/td>\r\n<td>T<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>F<\/td>\r\n<td>T<\/td>\r\n<td>T<\/td>\r\n<td>F<\/td>\r\n<td>F<\/td>\r\n<td>T<\/td>\r\n<td>T<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>F<\/td>\r\n<td>F<\/td>\r\n<td>T<\/td>\r\n<td>T<\/td>\r\n<td>F<\/td>\r\n<td>T<\/td>\r\n<td>T<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nDeMorgan\u2019s laws are actually very natural and intuitive. Consider the\u00a0statement [latex]\\sim\\left(P\\wedge{Q}\\right)[\/latex], which we can interpret as meaning that it is not the\u00a0case that both <i>P<\/i> and <i>Q<\/i> are true. If it is not the case that both <i>P<\/i> and <i>Q<\/i>\u00a0are true, then at least one of <i>P<\/i> or <i>Q<\/i> is false, in which case [latex]\\left(\\sim{P}\\right)\\vee\\left(\\sim{Q}\\right)[\/latex]\u00a0is\u00a0true. Thus [latex]\\sim\\left(P\\wedge{Q}\\right)[\/latex]\u00a0means the same thing as [latex]\\left(\\sim{P}\\right)\\vee\\left(\\sim{Q}\\right)[\/latex].\r\n\r\nDeMorgan\u2019s laws can be very useful. Suppose we happen to know that\u00a0some statement having form [latex]\\sim\\left(P\\vee{Q}\\right)[\/latex]\u00a0is true. The second of DeMorgan\u2019s\u00a0laws tells us that [latex]\\left(\\sim{Q}\\right)\\wedge\\left(\\sim{P}\\right)[\/latex]\u00a0is also true, hence [latex]\\sim{P}[\/latex] and [latex]\\sim{Q}[\/latex] are both true\u00a0as well. Being able to quickly obtain such additional pieces of information\u00a0can be extremely useful.\r\n\r\nHere is a summary of some significant logical equivalences. Those that\u00a0are not immediately obvious can be verified with a truth table.\r\n\r\n[latex]\\text{Contrapositive law}\\begin{array}{c}P\\rightarrow{Q}=(\\sim{Q})\\rightarrow(\\sim{P})\\end{array}[\/latex]\r\n\r\n[latex]\\text{DeMorgan's laws}\\begin{array}{c}{\\sim(P\\land{Q})=\\sim{P}\\lor\\sim{Q}}\\\\{\\sim(P\\lor{Q})=\\sim{P}\\land\\sim{Q}}\\end{array}[\/latex]\r\n\r\n[latex]\\text{Commutative laws}\\begin{array}{c}{(P\\land{Q})={P}\\land{Q}}\\\\{(P\\lor{Q})={P}\\lor{Q}}\\end{array}[\/latex]\r\n\r\n[latex]\\text{Distributive laws}\\begin{array}{c}{{P}\\land(Q\\lor{R})=({P}\\land{Q})\\lor(P\\land{R})}\\\\{P\\lor(Q\\land{R})=({P}\\lor{Q})\\land(P\\lor{R})}\\end{array}[\/latex]\r\n\r\n[latex]\\text{Associative laws}\\begin{array}{c}{P\\land(Q\\land{R})=(P\\land{Q})\\land{R}}\\\\{P\\lor(Q\\lor{R})=(P\\lor{Q})\\lor{R}}\\end{array}[\/latex]\r\n\r\nNotice how the distributive law [latex]P\\wedge\\left(Q\\vee{R}\\right)=\\left(P\\wedge{Q}\\right)\\vee\\left(P\\wedge{Q}\\right)\\vee\\left(P\\wedge{R}\\right)[\/latex]\u00a0has the\u00a0same structure as the distributive law [latex]p\\left(q+r\\right)=p\\cdot{q}+p\\cdot{r}[\/latex]\u00a0from algebra.\u00a0Concerning the associative laws, the fact that [latex]P\\wedge\\left(Q\\wedge{R}\\right)=\\left(P\\wedge{Q}\\right)\\wedge{R}[\/latex]\u00a0means\u00a0that the position of the parentheses is irrelevant, and we can write this as [latex]P\\wedge{Q}\\wedge{R}[\/latex]\u00a0without ambiguity. Similarly, we may drop the parentheses in\u00a0an expression such as [latex]P\\vee\\left(Q\\vee{R}\\right)[\/latex].\r\n\r\nBut parentheses are essential when there is a mix of [latex]\\wedge[\/latex]\u00a0and [latex]\\vee[\/latex], as in [latex]P\\vee\\left(Q\\wedge{R}\\right)[\/latex]. Indeed, [latex]P\\vee\\left(Q\\wedge{R}\\right)[\/latex] and [latex]P\\vee\\left(Q\\wedge{R}\\right)[\/latex] and [latex]P\\vee\\left({Q}\\right)\\wedge{R}[\/latex]<i><\/i> are <b>not<\/b> logically equivalent.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]109604[\/ohm_question]\r\n\r\n<\/div>\r\n<h3><b>Negating Statements<\/b><\/h3>\r\nGiven a statement <i>R<\/i>, the statement [latex]\\sim{R}[\/latex] is called the <b>negation<\/b> of <i>R<\/i>. If <i>R<\/i>\u00a0is a complex statement, then it is often the case that its negation [latex]\\sim{R}[\/latex]\u00a0can\u00a0be written in a simpler or more useful form. The process of finding this\u00a0form is called <b>negating<\/b> <i>R<\/i>. In proving theorems it is often necessary to\u00a0negate certain statements. We now investigate how to do this.\r\n\r\nWe have already examined part of this topic. <b>DeMorgan\u2019s laws<\/b>\r\n\r\n[latex]\\sim\\left(P\\wedge{Q}\\right)=\\left(\\sim{P}\\right)\\vee\\left(\\sim{Q}\\right)\\\\\\sim\\left(P\\vee{Q}\\right)=\\left(\\sim{P}\\right)\\wedge\\left(\\sim{Q}\\right)[\/latex]\r\n\r\n(from \"Logical Equivalence\") can be viewed as rules that tell us how to negate the\u00a0statements [latex]P\\wedge{Q}[\/latex]\u00a0and [latex]P\\vee{Q}[\/latex]. Here are some examples that illustrate how\u00a0DeMorgan\u2019s laws are used to negate statements involving \u201cand\u201d or \u201cor.\u201d\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nConsider negating the following statement.\r\n\r\n<i>R<\/i> : You can solve it by factoring or with the quadratic formula.\r\n\r\n[reveal-answer q=\"102469\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"102469\"]\r\n\r\nNow, <i>R<\/i> means (You can solve it by factoring) [latex]\\vee[\/latex]\u00a0(You can solve it with Q.F.),\u00a0which we will denote as [latex]P\\vee{Q}[\/latex]. The negation of this is [latex]\\sim\\left(P\\vee{Q}\\right)=\\left(\\sim{P}\\right)\\wedge\\left(\\sim{Q}\\right)[\/latex].\r\n\r\nTherefore, in words, the negation of <i>R<\/i> is\r\n\r\n[latex]\\sim{R}[\/latex] : You can\u2019t solve it by factoring and you can\u2019t solve it with\u00a0the quadratic formula.\r\n\r\nMaybe you can find [latex]\\sim{R}[\/latex]\u00a0without invoking DeMorgan\u2019s laws. That is good;\u00a0you have internalized DeMorgan\u2019s laws and are using them unconsciously.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWe will negate the following sentence.\r\n\r\n<i>R<\/i> : The numbers x and y are both odd.\r\n\r\n[reveal-answer q=\"441993\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"441993\"]\r\n\r\nThis statement means [latex]\\left(x\\text{ is odd}\\right)\\wedge\\left(y\\text{ is odd}\\right)[\/latex], so its negation is\r\n<p style=\"text-align: center;\">[latex]\\sim\\left[\\left(x\\text{ is odd}\\right)\\wedge\\left(y\\text{ is odd}\\right)\\right]=\\sim\\left(x\\text{ is odd}\\right)\\vee\\sim\\left(y\\text{ is odd}\\right)\\\\\\left(x\\text{ is odd}\\right)\\wedge\\left(y\\text{ is odd}\\right)=\\left(x\\text{ is even}\\right)\\vee\\left(y\\text{ is even}\\right)[\/latex]<\/p>\r\nTherefore the negation of <i>R<\/i> can be expressed in the following ways:\r\n\r\n[latex]\\sim{R}[\/latex]: The number x is even or the number y is even.\r\n[latex]\\sim{R}[\/latex]: At least one of x and y is even.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]109608[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Time to Practice!<\/h3>\r\n<p style=\"text-align: center;\">Complete the homework assigned by your teacher before moving on to the next section.<\/p>\r\n\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use DeMorgan\u2019s laws to define logical equivalences of a statement<\/li>\n<\/ul>\n<\/div>\n<p>There are two pairs of logically equivalent statements that come up\u00a0again and again in logic. They are prevalent\u00a0enough to be dignified by a special name: <b>DeMorgan\u2019s laws.<\/b><\/p>\n<p>The laws are named after Augustus De Morgan (1806\u20131871),\u00a0who introduced a formal version of the laws to classical propositional logic. De Morgan&#8217;s formulation was influenced by algebraization of logic undertaken by George Boole, which later cemented De Morgan&#8217;s claim to the find. Nevertheless, a similar observation was made by Aristotle, and was known to Greek and Medieval logicians.\u00a0For example, in the 14th century, William of Ockham wrote down the words that would result by reading the laws out. Jean Buridan, in his <i>Summulae de Dialectica<\/i>, also describes rules of conversion that follow the lines of De Morgan&#8217;s laws.\u00a0Still, De Morgan is given credit for stating the laws in the terms of modern formal logic, and incorporating them into the language of logic. De Morgan&#8217;s laws can be proved easily, and may even seem trivial.\u00a0Nonetheless, these laws are helpful in making valid inferences in proofs and deductive arguments.<\/p>\n<div class=\"textbox examples\">\n<h3>success strategy<\/h3>\n<p>Get plenty of practice and repetition with the ideas in this page! The notation will become more familiar as you do. Remember to get help if you need it!<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>DeMorgan\u2019s Laws<\/h3>\n<ol>\n<li>[latex]\\sim\\left(P{\\wedge}Q\\right)=({\\sim}P)\\vee\\left(\\sim{Q}\\right)[\/latex]<\/li>\n<li>[latex]\\sim\\left(P\\vee{Q}\\right)=\\left(\\sim{P}\\right)\\wedge\\left(\\sim{Q}\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<p>The first of DeMorgan\u2019s laws is verified by the following table. You are\u00a0asked to verify the second in an\u00a0exercise.<\/p>\n<table cellpadding=\"0\" style=\"border-spacing: 0px;\">\n<tbody>\n<tr>\n<td><b>[latex]P[\/latex]<\/b><\/td>\n<td><b>[latex]Q[\/latex]<\/b><\/td>\n<td><b>[latex]\\sim{P}[\/latex]<\/b><\/td>\n<td><b>[latex]\\sim{Q}[\/latex]<\/b><\/td>\n<td>[latex]P\\wedge{Q}[\/latex]<\/td>\n<td><strong>[latex]\\sim\\left(P\\wedge{Q}\\right)[\/latex]<\/strong><\/td>\n<td><b>[latex]\\left(\\sim{P}\\right)\\vee\\left(\\sim{Q}\\right)[\/latex]<\/b><\/td>\n<\/tr>\n<tr>\n<td>T<\/td>\n<td>T<\/td>\n<td>F<\/td>\n<td>F<\/td>\n<td>T<\/td>\n<td>F<\/td>\n<td>F<\/td>\n<\/tr>\n<tr>\n<td>T<\/td>\n<td>F<\/td>\n<td>F<\/td>\n<td>T<\/td>\n<td>F<\/td>\n<td>T<\/td>\n<td>T<\/td>\n<\/tr>\n<tr>\n<td>F<\/td>\n<td>T<\/td>\n<td>T<\/td>\n<td>F<\/td>\n<td>F<\/td>\n<td>T<\/td>\n<td>T<\/td>\n<\/tr>\n<tr>\n<td>F<\/td>\n<td>F<\/td>\n<td>T<\/td>\n<td>T<\/td>\n<td>F<\/td>\n<td>T<\/td>\n<td>T<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>DeMorgan\u2019s laws are actually very natural and intuitive. Consider the\u00a0statement [latex]\\sim\\left(P\\wedge{Q}\\right)[\/latex], which we can interpret as meaning that it is not the\u00a0case that both <i>P<\/i> and <i>Q<\/i> are true. If it is not the case that both <i>P<\/i> and <i>Q<\/i>\u00a0are true, then at least one of <i>P<\/i> or <i>Q<\/i> is false, in which case [latex]\\left(\\sim{P}\\right)\\vee\\left(\\sim{Q}\\right)[\/latex]\u00a0is\u00a0true. Thus [latex]\\sim\\left(P\\wedge{Q}\\right)[\/latex]\u00a0means the same thing as [latex]\\left(\\sim{P}\\right)\\vee\\left(\\sim{Q}\\right)[\/latex].<\/p>\n<p>DeMorgan\u2019s laws can be very useful. Suppose we happen to know that\u00a0some statement having form [latex]\\sim\\left(P\\vee{Q}\\right)[\/latex]\u00a0is true. The second of DeMorgan\u2019s\u00a0laws tells us that [latex]\\left(\\sim{Q}\\right)\\wedge\\left(\\sim{P}\\right)[\/latex]\u00a0is also true, hence [latex]\\sim{P}[\/latex] and [latex]\\sim{Q}[\/latex] are both true\u00a0as well. Being able to quickly obtain such additional pieces of information\u00a0can be extremely useful.<\/p>\n<p>Here is a summary of some significant logical equivalences. Those that\u00a0are not immediately obvious can be verified with a truth table.<\/p>\n<p>[latex]\\text{Contrapositive law}\\begin{array}{c}P\\rightarrow{Q}=(\\sim{Q})\\rightarrow(\\sim{P})\\end{array}[\/latex]<\/p>\n<p>[latex]\\text{DeMorgan's laws}\\begin{array}{c}{\\sim(P\\land{Q})=\\sim{P}\\lor\\sim{Q}}\\\\{\\sim(P\\lor{Q})=\\sim{P}\\land\\sim{Q}}\\end{array}[\/latex]<\/p>\n<p>[latex]\\text{Commutative laws}\\begin{array}{c}{(P\\land{Q})={P}\\land{Q}}\\\\{(P\\lor{Q})={P}\\lor{Q}}\\end{array}[\/latex]<\/p>\n<p>[latex]\\text{Distributive laws}\\begin{array}{c}{{P}\\land(Q\\lor{R})=({P}\\land{Q})\\lor(P\\land{R})}\\\\{P\\lor(Q\\land{R})=({P}\\lor{Q})\\land(P\\lor{R})}\\end{array}[\/latex]<\/p>\n<p>[latex]\\text{Associative laws}\\begin{array}{c}{P\\land(Q\\land{R})=(P\\land{Q})\\land{R}}\\\\{P\\lor(Q\\lor{R})=(P\\lor{Q})\\lor{R}}\\end{array}[\/latex]<\/p>\n<p>Notice how the distributive law [latex]P\\wedge\\left(Q\\vee{R}\\right)=\\left(P\\wedge{Q}\\right)\\vee\\left(P\\wedge{Q}\\right)\\vee\\left(P\\wedge{R}\\right)[\/latex]\u00a0has the\u00a0same structure as the distributive law [latex]p\\left(q+r\\right)=p\\cdot{q}+p\\cdot{r}[\/latex]\u00a0from algebra.\u00a0Concerning the associative laws, the fact that [latex]P\\wedge\\left(Q\\wedge{R}\\right)=\\left(P\\wedge{Q}\\right)\\wedge{R}[\/latex]\u00a0means\u00a0that the position of the parentheses is irrelevant, and we can write this as [latex]P\\wedge{Q}\\wedge{R}[\/latex]\u00a0without ambiguity. Similarly, we may drop the parentheses in\u00a0an expression such as [latex]P\\vee\\left(Q\\vee{R}\\right)[\/latex].<\/p>\n<p>But parentheses are essential when there is a mix of [latex]\\wedge[\/latex]\u00a0and [latex]\\vee[\/latex], as in [latex]P\\vee\\left(Q\\wedge{R}\\right)[\/latex]. Indeed, [latex]P\\vee\\left(Q\\wedge{R}\\right)[\/latex] and [latex]P\\vee\\left(Q\\wedge{R}\\right)[\/latex] and [latex]P\\vee\\left({Q}\\right)\\wedge{R}[\/latex]<i><\/i> are <b>not<\/b> logically equivalent.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm109604\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109604&theme=oea&iframe_resize_id=ohm109604&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h3><b>Negating Statements<\/b><\/h3>\n<p>Given a statement <i>R<\/i>, the statement [latex]\\sim{R}[\/latex] is called the <b>negation<\/b> of <i>R<\/i>. If <i>R<\/i>\u00a0is a complex statement, then it is often the case that its negation [latex]\\sim{R}[\/latex]\u00a0can\u00a0be written in a simpler or more useful form. The process of finding this\u00a0form is called <b>negating<\/b> <i>R<\/i>. In proving theorems it is often necessary to\u00a0negate certain statements. We now investigate how to do this.<\/p>\n<p>We have already examined part of this topic. <b>DeMorgan\u2019s laws<\/b><\/p>\n<p>[latex]\\sim\\left(P\\wedge{Q}\\right)=\\left(\\sim{P}\\right)\\vee\\left(\\sim{Q}\\right)\\\\\\sim\\left(P\\vee{Q}\\right)=\\left(\\sim{P}\\right)\\wedge\\left(\\sim{Q}\\right)[\/latex]<\/p>\n<p>(from &#8220;Logical Equivalence&#8221;) can be viewed as rules that tell us how to negate the\u00a0statements [latex]P\\wedge{Q}[\/latex]\u00a0and [latex]P\\vee{Q}[\/latex]. Here are some examples that illustrate how\u00a0DeMorgan\u2019s laws are used to negate statements involving \u201cand\u201d or \u201cor.\u201d<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Consider negating the following statement.<\/p>\n<p><i>R<\/i> : You can solve it by factoring or with the quadratic formula.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q102469\">Show Solution<\/span><\/p>\n<div id=\"q102469\" class=\"hidden-answer\" style=\"display: none\">\n<p>Now, <i>R<\/i> means (You can solve it by factoring) [latex]\\vee[\/latex]\u00a0(You can solve it with Q.F.),\u00a0which we will denote as [latex]P\\vee{Q}[\/latex]. The negation of this is [latex]\\sim\\left(P\\vee{Q}\\right)=\\left(\\sim{P}\\right)\\wedge\\left(\\sim{Q}\\right)[\/latex].<\/p>\n<p>Therefore, in words, the negation of <i>R<\/i> is<\/p>\n<p>[latex]\\sim{R}[\/latex] : You can\u2019t solve it by factoring and you can\u2019t solve it with\u00a0the quadratic formula.<\/p>\n<p>Maybe you can find [latex]\\sim{R}[\/latex]\u00a0without invoking DeMorgan\u2019s laws. That is good;\u00a0you have internalized DeMorgan\u2019s laws and are using them unconsciously.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>We will negate the following sentence.<\/p>\n<p><i>R<\/i> : The numbers x and y are both odd.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q441993\">Show Solution<\/span><\/p>\n<div id=\"q441993\" class=\"hidden-answer\" style=\"display: none\">\n<p>This statement means [latex]\\left(x\\text{ is odd}\\right)\\wedge\\left(y\\text{ is odd}\\right)[\/latex], so its negation is<\/p>\n<p style=\"text-align: center;\">[latex]\\sim\\left[\\left(x\\text{ is odd}\\right)\\wedge\\left(y\\text{ is odd}\\right)\\right]=\\sim\\left(x\\text{ is odd}\\right)\\vee\\sim\\left(y\\text{ is odd}\\right)\\\\\\left(x\\text{ is odd}\\right)\\wedge\\left(y\\text{ is odd}\\right)=\\left(x\\text{ is even}\\right)\\vee\\left(y\\text{ is even}\\right)[\/latex]<\/p>\n<p>Therefore the negation of <i>R<\/i> can be expressed in the following ways:<\/p>\n<p>[latex]\\sim{R}[\/latex]: The number x is even or the number y is even.<br \/>\n[latex]\\sim{R}[\/latex]: At least one of x and y is even.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm109608\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109608&theme=oea&iframe_resize_id=ohm109608&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Time to Practice!<\/h3>\n<p style=\"text-align: center;\">Complete the homework assigned by your teacher before moving on to the next section.<\/p>\n<p>&nbsp;<\/p><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-204\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Math in Society. <strong>Authored by<\/strong>: Lippman, David. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>DeMorgan&#039;s Laws. <strong>Authored by<\/strong>: Wikipedia. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/en.wikipedia.org\/wiki\/De_Morgan%27s_laws\">https:\/\/en.wikipedia.org\/wiki\/De_Morgan%27s_laws<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Question ID 109608, 109064. <strong>Authored by<\/strong>: Hartley,Josiah. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":678587,"menu_order":20,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"Lippman, 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