Proving the contradiction/negation of a statement












1












$begingroup$


Is proving the contradiction and proving the negation the same thing? Proving it true is proving the original statement false and proving it's false proves the statement is true?



So if a statement is an implication we just assume that the hypothesis is false?
What if there's a quantifier in front?



What would proving the negation of this statement look like?:



$forall x in Bbb R$, if $x > 2$, then $x^2 + 3 > 0$?






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Proving the negation of that statement would be gibberish, because the statement is true.
    $endgroup$
    – Steve B
    Dec 8 '18 at 2:02










  • $begingroup$
    What do you mean, “proving the contradiction”? Do you mean a proof by contradiction?
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:03










  • $begingroup$
    The negation of the statement you write is “there exists $xinmathbb{R}$ such that $xgt 2$, and $x^2+3leq 0$.” But that statement is false.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:03






  • 1




    $begingroup$
    @ming: You are saying things in a confusing manner, which is why you are getting confusing answers. The statement “prove the contradiction” is incorrect. A “proof by contradiction” is a method by which you prove a given statement; it is not what you are proving. By contrast, the negation of a statement is itself a statement (which is obtained by negating the given statement). They are different things. Proving the negation shows that the given statement is false. A proof by contradiction establishes the truth of the given statement.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:51






  • 1




    $begingroup$
    Stop using the wrong words, you may stop getting confusing answers. In the meantime, perhaps see math.stackexchange.com/questions/112774/…
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:52
















1












$begingroup$


Is proving the contradiction and proving the negation the same thing? Proving it true is proving the original statement false and proving it's false proves the statement is true?



So if a statement is an implication we just assume that the hypothesis is false?
What if there's a quantifier in front?



What would proving the negation of this statement look like?:



$forall x in Bbb R$, if $x > 2$, then $x^2 + 3 > 0$?






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Proving the negation of that statement would be gibberish, because the statement is true.
    $endgroup$
    – Steve B
    Dec 8 '18 at 2:02










  • $begingroup$
    What do you mean, “proving the contradiction”? Do you mean a proof by contradiction?
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:03










  • $begingroup$
    The negation of the statement you write is “there exists $xinmathbb{R}$ such that $xgt 2$, and $x^2+3leq 0$.” But that statement is false.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:03






  • 1




    $begingroup$
    @ming: You are saying things in a confusing manner, which is why you are getting confusing answers. The statement “prove the contradiction” is incorrect. A “proof by contradiction” is a method by which you prove a given statement; it is not what you are proving. By contrast, the negation of a statement is itself a statement (which is obtained by negating the given statement). They are different things. Proving the negation shows that the given statement is false. A proof by contradiction establishes the truth of the given statement.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:51






  • 1




    $begingroup$
    Stop using the wrong words, you may stop getting confusing answers. In the meantime, perhaps see math.stackexchange.com/questions/112774/…
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:52














1












1








1





$begingroup$


Is proving the contradiction and proving the negation the same thing? Proving it true is proving the original statement false and proving it's false proves the statement is true?



So if a statement is an implication we just assume that the hypothesis is false?
What if there's a quantifier in front?



What would proving the negation of this statement look like?:



$forall x in Bbb R$, if $x > 2$, then $x^2 + 3 > 0$?






linear-algebra







share|cite|improve this question









$endgroup$




Is proving the contradiction and proving the negation the same thing? Proving it true is proving the original statement false and proving it's false proves the statement is true?



So if a statement is an implication we just assume that the hypothesis is false?
What if there's a quantifier in front?



What would proving the negation of this statement look like?:



$forall x in Bbb R$, if $x > 2$, then $x^2 + 3 > 0$?






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 1:26









mingming

3205




3205












  • $begingroup$
    Proving the negation of that statement would be gibberish, because the statement is true.
    $endgroup$
    – Steve B
    Dec 8 '18 at 2:02










  • $begingroup$
    What do you mean, “proving the contradiction”? Do you mean a proof by contradiction?
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:03










  • $begingroup$
    The negation of the statement you write is “there exists $xinmathbb{R}$ such that $xgt 2$, and $x^2+3leq 0$.” But that statement is false.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:03






  • 1




    $begingroup$
    @ming: You are saying things in a confusing manner, which is why you are getting confusing answers. The statement “prove the contradiction” is incorrect. A “proof by contradiction” is a method by which you prove a given statement; it is not what you are proving. By contrast, the negation of a statement is itself a statement (which is obtained by negating the given statement). They are different things. Proving the negation shows that the given statement is false. A proof by contradiction establishes the truth of the given statement.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:51






  • 1




    $begingroup$
    Stop using the wrong words, you may stop getting confusing answers. In the meantime, perhaps see math.stackexchange.com/questions/112774/…
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:52


















  • $begingroup$
    Proving the negation of that statement would be gibberish, because the statement is true.
    $endgroup$
    – Steve B
    Dec 8 '18 at 2:02










  • $begingroup$
    What do you mean, “proving the contradiction”? Do you mean a proof by contradiction?
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:03










  • $begingroup$
    The negation of the statement you write is “there exists $xinmathbb{R}$ such that $xgt 2$, and $x^2+3leq 0$.” But that statement is false.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:03






  • 1




    $begingroup$
    @ming: You are saying things in a confusing manner, which is why you are getting confusing answers. The statement “prove the contradiction” is incorrect. A “proof by contradiction” is a method by which you prove a given statement; it is not what you are proving. By contrast, the negation of a statement is itself a statement (which is obtained by negating the given statement). They are different things. Proving the negation shows that the given statement is false. A proof by contradiction establishes the truth of the given statement.
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:51






  • 1




    $begingroup$
    Stop using the wrong words, you may stop getting confusing answers. In the meantime, perhaps see math.stackexchange.com/questions/112774/…
    $endgroup$
    – Arturo Magidin
    Dec 8 '18 at 2:52
















$begingroup$
Proving the negation of that statement would be gibberish, because the statement is true.
$endgroup$
– Steve B
Dec 8 '18 at 2:02




$begingroup$
Proving the negation of that statement would be gibberish, because the statement is true.
$endgroup$
– Steve B
Dec 8 '18 at 2:02












$begingroup$
What do you mean, “proving the contradiction”? Do you mean a proof by contradiction?
$endgroup$
– Arturo Magidin
Dec 8 '18 at 2:03




$begingroup$
What do you mean, “proving the contradiction”? Do you mean a proof by contradiction?
$endgroup$
– Arturo Magidin
Dec 8 '18 at 2:03












$begingroup$
The negation of the statement you write is “there exists $xinmathbb{R}$ such that $xgt 2$, and $x^2+3leq 0$.” But that statement is false.
$endgroup$
– Arturo Magidin
Dec 8 '18 at 2:03




$begingroup$
The negation of the statement you write is “there exists $xinmathbb{R}$ such that $xgt 2$, and $x^2+3leq 0$.” But that statement is false.
$endgroup$
– Arturo Magidin
Dec 8 '18 at 2:03




1




1




$begingroup$
@ming: You are saying things in a confusing manner, which is why you are getting confusing answers. The statement “prove the contradiction” is incorrect. A “proof by contradiction” is a method by which you prove a given statement; it is not what you are proving. By contrast, the negation of a statement is itself a statement (which is obtained by negating the given statement). They are different things. Proving the negation shows that the given statement is false. A proof by contradiction establishes the truth of the given statement.
$endgroup$
– Arturo Magidin
Dec 8 '18 at 2:51




$begingroup$
@ming: You are saying things in a confusing manner, which is why you are getting confusing answers. The statement “prove the contradiction” is incorrect. A “proof by contradiction” is a method by which you prove a given statement; it is not what you are proving. By contrast, the negation of a statement is itself a statement (which is obtained by negating the given statement). They are different things. Proving the negation shows that the given statement is false. A proof by contradiction establishes the truth of the given statement.
$endgroup$
– Arturo Magidin
Dec 8 '18 at 2:51




1




1




$begingroup$
Stop using the wrong words, you may stop getting confusing answers. In the meantime, perhaps see math.stackexchange.com/questions/112774/…
$endgroup$
– Arturo Magidin
Dec 8 '18 at 2:52




$begingroup$
Stop using the wrong words, you may stop getting confusing answers. In the meantime, perhaps see math.stackexchange.com/questions/112774/…
$endgroup$
– Arturo Magidin
Dec 8 '18 at 2:52










1 Answer
1






active

oldest

votes


















1












$begingroup$

What you are actually asking about, according to the comments, is “proof by contradiction” (which is not “prove the contradiction”).



A proof by contradiction is a method of proof in which one assumes the negation of what you want to prove, and deduce a statement that is impossible. In classical logic, this means that the original statement must be true, because of the law of the excluded middle: it cannot be false (because if it were false that would lead to a contradiction), and if it is cannot be false, then it must be true.



In order to do a proof by contradiction, you must know the negation of the statement; but you are not trying to prove that the negation is true. You are assuming that the negation is true, and trying to deduce a statement known to be false/impossible.



In the case you give, the statement you want to prove is
$$forall xinmathbb{R}Bigl( xgt 2 rightarrow x^2+3gt 0Bigr)$$



The negation of this statement is
$$begin{align*}
&negBiggl( forall xinmathbb{R}Bigl(xgt 2 rightarrow x^2+3gt 0Bigr)Biggr)\
&exists xinmathbb{R}Biggl(negBigl( xgt 2rightarrow x^2+3gt 0Bigr)Biggr)\
&exists xinmathbb{R}Biggl( xgt 2 text{ and } neg(x^2+3gt 0)Biggr)\
&exists xinmathbb{R}Bigl( xgt 2 text{ and }x^2+3leq 0Bigr)
end{align*}
$$
So, to do a proof by contradiction, you would start by assuming that there is a real number $x$ that is both greater than 2, and also has the property that $x^2+3leq 0$. From this, you would want to deduce something utterly impossible.



If you were trying to prove the negation, you would be trying to prove that there is a real number $x$ that is both greater than 2 and also has the property that $x^2+3leq 0$.



Different things entirely.



If you do a proof by contradiction successfully, you have established that the given statement is true.



If you successfully prove the negation, you have established that the given statement is false.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030595%2fproving-the-contradiction-negation-of-a-statement%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    What you are actually asking about, according to the comments, is “proof by contradiction” (which is not “prove the contradiction”).



    A proof by contradiction is a method of proof in which one assumes the negation of what you want to prove, and deduce a statement that is impossible. In classical logic, this means that the original statement must be true, because of the law of the excluded middle: it cannot be false (because if it were false that would lead to a contradiction), and if it is cannot be false, then it must be true.



    In order to do a proof by contradiction, you must know the negation of the statement; but you are not trying to prove that the negation is true. You are assuming that the negation is true, and trying to deduce a statement known to be false/impossible.



    In the case you give, the statement you want to prove is
    $$forall xinmathbb{R}Bigl( xgt 2 rightarrow x^2+3gt 0Bigr)$$



    The negation of this statement is
    $$begin{align*}
    &negBiggl( forall xinmathbb{R}Bigl(xgt 2 rightarrow x^2+3gt 0Bigr)Biggr)\
    &exists xinmathbb{R}Biggl(negBigl( xgt 2rightarrow x^2+3gt 0Bigr)Biggr)\
    &exists xinmathbb{R}Biggl( xgt 2 text{ and } neg(x^2+3gt 0)Biggr)\
    &exists xinmathbb{R}Bigl( xgt 2 text{ and }x^2+3leq 0Bigr)
    end{align*}
    $$
    So, to do a proof by contradiction, you would start by assuming that there is a real number $x$ that is both greater than 2, and also has the property that $x^2+3leq 0$. From this, you would want to deduce something utterly impossible.



    If you were trying to prove the negation, you would be trying to prove that there is a real number $x$ that is both greater than 2 and also has the property that $x^2+3leq 0$.



    Different things entirely.



    If you do a proof by contradiction successfully, you have established that the given statement is true.



    If you successfully prove the negation, you have established that the given statement is false.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      What you are actually asking about, according to the comments, is “proof by contradiction” (which is not “prove the contradiction”).



      A proof by contradiction is a method of proof in which one assumes the negation of what you want to prove, and deduce a statement that is impossible. In classical logic, this means that the original statement must be true, because of the law of the excluded middle: it cannot be false (because if it were false that would lead to a contradiction), and if it is cannot be false, then it must be true.



      In order to do a proof by contradiction, you must know the negation of the statement; but you are not trying to prove that the negation is true. You are assuming that the negation is true, and trying to deduce a statement known to be false/impossible.



      In the case you give, the statement you want to prove is
      $$forall xinmathbb{R}Bigl( xgt 2 rightarrow x^2+3gt 0Bigr)$$



      The negation of this statement is
      $$begin{align*}
      &negBiggl( forall xinmathbb{R}Bigl(xgt 2 rightarrow x^2+3gt 0Bigr)Biggr)\
      &exists xinmathbb{R}Biggl(negBigl( xgt 2rightarrow x^2+3gt 0Bigr)Biggr)\
      &exists xinmathbb{R}Biggl( xgt 2 text{ and } neg(x^2+3gt 0)Biggr)\
      &exists xinmathbb{R}Bigl( xgt 2 text{ and }x^2+3leq 0Bigr)
      end{align*}
      $$
      So, to do a proof by contradiction, you would start by assuming that there is a real number $x$ that is both greater than 2, and also has the property that $x^2+3leq 0$. From this, you would want to deduce something utterly impossible.



      If you were trying to prove the negation, you would be trying to prove that there is a real number $x$ that is both greater than 2 and also has the property that $x^2+3leq 0$.



      Different things entirely.



      If you do a proof by contradiction successfully, you have established that the given statement is true.



      If you successfully prove the negation, you have established that the given statement is false.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        What you are actually asking about, according to the comments, is “proof by contradiction” (which is not “prove the contradiction”).



        A proof by contradiction is a method of proof in which one assumes the negation of what you want to prove, and deduce a statement that is impossible. In classical logic, this means that the original statement must be true, because of the law of the excluded middle: it cannot be false (because if it were false that would lead to a contradiction), and if it is cannot be false, then it must be true.



        In order to do a proof by contradiction, you must know the negation of the statement; but you are not trying to prove that the negation is true. You are assuming that the negation is true, and trying to deduce a statement known to be false/impossible.



        In the case you give, the statement you want to prove is
        $$forall xinmathbb{R}Bigl( xgt 2 rightarrow x^2+3gt 0Bigr)$$



        The negation of this statement is
        $$begin{align*}
        &negBiggl( forall xinmathbb{R}Bigl(xgt 2 rightarrow x^2+3gt 0Bigr)Biggr)\
        &exists xinmathbb{R}Biggl(negBigl( xgt 2rightarrow x^2+3gt 0Bigr)Biggr)\
        &exists xinmathbb{R}Biggl( xgt 2 text{ and } neg(x^2+3gt 0)Biggr)\
        &exists xinmathbb{R}Bigl( xgt 2 text{ and }x^2+3leq 0Bigr)
        end{align*}
        $$
        So, to do a proof by contradiction, you would start by assuming that there is a real number $x$ that is both greater than 2, and also has the property that $x^2+3leq 0$. From this, you would want to deduce something utterly impossible.



        If you were trying to prove the negation, you would be trying to prove that there is a real number $x$ that is both greater than 2 and also has the property that $x^2+3leq 0$.



        Different things entirely.



        If you do a proof by contradiction successfully, you have established that the given statement is true.



        If you successfully prove the negation, you have established that the given statement is false.






        share|cite|improve this answer









        $endgroup$



        What you are actually asking about, according to the comments, is “proof by contradiction” (which is not “prove the contradiction”).



        A proof by contradiction is a method of proof in which one assumes the negation of what you want to prove, and deduce a statement that is impossible. In classical logic, this means that the original statement must be true, because of the law of the excluded middle: it cannot be false (because if it were false that would lead to a contradiction), and if it is cannot be false, then it must be true.



        In order to do a proof by contradiction, you must know the negation of the statement; but you are not trying to prove that the negation is true. You are assuming that the negation is true, and trying to deduce a statement known to be false/impossible.



        In the case you give, the statement you want to prove is
        $$forall xinmathbb{R}Bigl( xgt 2 rightarrow x^2+3gt 0Bigr)$$



        The negation of this statement is
        $$begin{align*}
        &negBiggl( forall xinmathbb{R}Bigl(xgt 2 rightarrow x^2+3gt 0Bigr)Biggr)\
        &exists xinmathbb{R}Biggl(negBigl( xgt 2rightarrow x^2+3gt 0Bigr)Biggr)\
        &exists xinmathbb{R}Biggl( xgt 2 text{ and } neg(x^2+3gt 0)Biggr)\
        &exists xinmathbb{R}Bigl( xgt 2 text{ and }x^2+3leq 0Bigr)
        end{align*}
        $$
        So, to do a proof by contradiction, you would start by assuming that there is a real number $x$ that is both greater than 2, and also has the property that $x^2+3leq 0$. From this, you would want to deduce something utterly impossible.



        If you were trying to prove the negation, you would be trying to prove that there is a real number $x$ that is both greater than 2 and also has the property that $x^2+3leq 0$.



        Different things entirely.



        If you do a proof by contradiction successfully, you have established that the given statement is true.



        If you successfully prove the negation, you have established that the given statement is false.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 3:01









        Arturo MagidinArturo Magidin

        261k34586906




        261k34586906






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030595%2fproving-the-contradiction-negation-of-a-statement%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc
            Read more

            Fiat S.p.A.

            Image
            Cet article court présente un sujet plus développé dans : Fiat et Fiat Chrysler Automobiles. Fiat S.p.A. Fiat S.p.A. ( Fabbrica Italiana Automobili Torino , acronyme FIAT) est un groupe, notamment connu comme société-mère du constructeur automobile italien, Fiat. Portail des entreprises Portail de Turin Portail de l’automobile This page is only for reference, If you need detailed information, please check here
            Read more

            Type 'String' is not a subtype of type 'int' of 'index'

            Image
            up vote 0 down vote favorite My App can authenticate successfully when debugging against the Android emulator, but if I try to authenticate using debugging against the physical device (with same OS version), an error message appears after more than one minute waiting: Exception has occurred. _TypeError (type 'String' is not a subtype of type 'int' of 'index') The error message points to the following code: if (responseJson[AuthUtils.authTokenKey] != null) { AuthUtils.insertDetails( userNameController.text, passwordController.text, responseJson); ... } else { ... }; And in the DEBUG CONSOLE I get the following: I/flutter ( 9531): Auth: SocketException: OS Error: Connection timed out, errno = 110, address = example.com, port = 3897
            Read more