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









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    1 Answer
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    1 Answer
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    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




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