Why is my proof for “if $0 leqslant x leqslant 2$, then $-x^3 + 4x + 1 > 0$” is false?












2












$begingroup$



$$text{if $0 leqslant x leqslant 2$, then $-x^3 + 4x + 1 > 0$}$$




$$x(4-x^2)>-1$$
$$x>dfrac{1}{x^2-4}$$ if the last statement is smaller than $x$ then it is also smaller than $2$ because of first statement ($0 leqslant x leqslant 2$):
$$dfrac{1}{x^2-4}<2$$
$$1<2x^2-8$$
$$4.5<x^2$$
$$-2.121<x<2.121$$
Which contradicts the first statement because $x$ cannot be greater than $2$. My proof is wrong but why? (I took that question from Math for Computer Science book p. 12)






discrete-mathematics proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    You divide by 0 when x = 2.
    $endgroup$
    – William Elliot
    Dec 15 '18 at 10:42










  • $begingroup$
    In your third line you (must have) assumed $4-x^2>0$
    $endgroup$
    – lcv
    Dec 15 '18 at 10:55
















2












$begingroup$



$$text{if $0 leqslant x leqslant 2$, then $-x^3 + 4x + 1 > 0$}$$




$$x(4-x^2)>-1$$
$$x>dfrac{1}{x^2-4}$$ if the last statement is smaller than $x$ then it is also smaller than $2$ because of first statement ($0 leqslant x leqslant 2$):
$$dfrac{1}{x^2-4}<2$$
$$1<2x^2-8$$
$$4.5<x^2$$
$$-2.121<x<2.121$$
Which contradicts the first statement because $x$ cannot be greater than $2$. My proof is wrong but why? (I took that question from Math for Computer Science book p. 12)






discrete-mathematics proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    You divide by 0 when x = 2.
    $endgroup$
    – William Elliot
    Dec 15 '18 at 10:42










  • $begingroup$
    In your third line you (must have) assumed $4-x^2>0$
    $endgroup$
    – lcv
    Dec 15 '18 at 10:55














2












2








2





$begingroup$



$$text{if $0 leqslant x leqslant 2$, then $-x^3 + 4x + 1 > 0$}$$




$$x(4-x^2)>-1$$
$$x>dfrac{1}{x^2-4}$$ if the last statement is smaller than $x$ then it is also smaller than $2$ because of first statement ($0 leqslant x leqslant 2$):
$$dfrac{1}{x^2-4}<2$$
$$1<2x^2-8$$
$$4.5<x^2$$
$$-2.121<x<2.121$$
Which contradicts the first statement because $x$ cannot be greater than $2$. My proof is wrong but why? (I took that question from Math for Computer Science book p. 12)






discrete-mathematics proof-verification







share|cite|improve this question











$endgroup$





$$text{if $0 leqslant x leqslant 2$, then $-x^3 + 4x + 1 > 0$}$$




$$x(4-x^2)>-1$$
$$x>dfrac{1}{x^2-4}$$ if the last statement is smaller than $x$ then it is also smaller than $2$ because of first statement ($0 leqslant x leqslant 2$):
$$dfrac{1}{x^2-4}<2$$
$$1<2x^2-8$$
$$4.5<x^2$$
$$-2.121<x<2.121$$
Which contradicts the first statement because $x$ cannot be greater than $2$. My proof is wrong but why? (I took that question from Math for Computer Science book p. 12)






discrete-mathematics proof-verification




discrete-mathematics proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 11:33









egreg

181k1485203




181k1485203










asked Dec 15 '18 at 10:28









Ahmet SoyyiğitAhmet Soyyiğit

133




133












  • $begingroup$
    You divide by 0 when x = 2.
    $endgroup$
    – William Elliot
    Dec 15 '18 at 10:42










  • $begingroup$
    In your third line you (must have) assumed $4-x^2>0$
    $endgroup$
    – lcv
    Dec 15 '18 at 10:55


















  • $begingroup$
    You divide by 0 when x = 2.
    $endgroup$
    – William Elliot
    Dec 15 '18 at 10:42










  • $begingroup$
    In your third line you (must have) assumed $4-x^2>0$
    $endgroup$
    – lcv
    Dec 15 '18 at 10:55
















$begingroup$
You divide by 0 when x = 2.
$endgroup$
– William Elliot
Dec 15 '18 at 10:42




$begingroup$
You divide by 0 when x = 2.
$endgroup$
– William Elliot
Dec 15 '18 at 10:42












$begingroup$
In your third line you (must have) assumed $4-x^2>0$
$endgroup$
– lcv
Dec 15 '18 at 10:55




$begingroup$
In your third line you (must have) assumed $4-x^2>0$
$endgroup$
– lcv
Dec 15 '18 at 10:55










4 Answers
4






active

oldest

votes


















0












$begingroup$

You made two mistakes and recovered:
$$dfrac{1}{x^2-4}<2 notRightarrow 1<2(x^2-4) (text{it must be} 1color{red}>2(x^2-4), text{because} x^2-4le 0)\
4.5<x^2 notRightarrow -2.121<x<2.121 (text{it must be} 4.5color{red}>x^2 Rightarrow -2.121<x<2.121)$$



Alternatively, you can prove it as follows:
$$-x^3 + 4x + 1 > 0 iff x^3-4x-1<0 iff x(x-2)(x+2)<1 iff \
x(x-2)(x+2)le 0<1 text{for} 0le xle 2 text{(why?)}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You are correct that $-3/sqrt2<x<3/sqrt2$ because $x$ still lies between $0$ and $2$. You have found a wider bound for $x$, but it is not wrong. Note that the larger interval does not mean that $x$ takes all values between $pm3/sqrt2$.



    An analogy could be that $x=3$, and we know that $2<x<4$. But it is also true that $-1<x<7$, or even $-infty<x<infty$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So, does the last statement mean "x can be any number between -2.121 and 2.121" or "x is one of the numbers between -2.121 and 2.121 but we don't know which number is it" ?
      $endgroup$
      – Ahmet Soyyiğit
      Dec 15 '18 at 11:15










    • $begingroup$
      We know that $0le xle2$. Therefore the last statement is saying that $x$ can be a set of numbers between $pm2.121$ but is restricted to $[0,2]$ due to the criterion at the start.
      $endgroup$
      – TheSimpliFire
      Dec 15 '18 at 11:17



















    0












    $begingroup$

    Stating that a statement is smaller than a number makes no sense.



    And from the fact that $dfrac1{x^2-4}<2$ together with $0leqslant xleqslant2$, you can't deduce that $dfrac1{x^2-4}<x$.



    Finally, from $dfrac1{x^2-4}<2$, what you deduce is that $1>x(x^2-4)$, since $x^2-4<0$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      1 + 4x - x$^3$ = 1 + x(4 - x$^2$).

      For all x in [0,2], 0 < x, 0 < 4 - x$^2$.

      Desired conclusion quickly follows.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

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        active

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        active

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        0












        $begingroup$

        You made two mistakes and recovered:
        $$dfrac{1}{x^2-4}<2 notRightarrow 1<2(x^2-4) (text{it must be} 1color{red}>2(x^2-4), text{because} x^2-4le 0)\
        4.5<x^2 notRightarrow -2.121<x<2.121 (text{it must be} 4.5color{red}>x^2 Rightarrow -2.121<x<2.121)$$



        Alternatively, you can prove it as follows:
        $$-x^3 + 4x + 1 > 0 iff x^3-4x-1<0 iff x(x-2)(x+2)<1 iff \
        x(x-2)(x+2)le 0<1 text{for} 0le xle 2 text{(why?)}$$






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          You made two mistakes and recovered:
          $$dfrac{1}{x^2-4}<2 notRightarrow 1<2(x^2-4) (text{it must be} 1color{red}>2(x^2-4), text{because} x^2-4le 0)\
          4.5<x^2 notRightarrow -2.121<x<2.121 (text{it must be} 4.5color{red}>x^2 Rightarrow -2.121<x<2.121)$$



          Alternatively, you can prove it as follows:
          $$-x^3 + 4x + 1 > 0 iff x^3-4x-1<0 iff x(x-2)(x+2)<1 iff \
          x(x-2)(x+2)le 0<1 text{for} 0le xle 2 text{(why?)}$$






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            You made two mistakes and recovered:
            $$dfrac{1}{x^2-4}<2 notRightarrow 1<2(x^2-4) (text{it must be} 1color{red}>2(x^2-4), text{because} x^2-4le 0)\
            4.5<x^2 notRightarrow -2.121<x<2.121 (text{it must be} 4.5color{red}>x^2 Rightarrow -2.121<x<2.121)$$



            Alternatively, you can prove it as follows:
            $$-x^3 + 4x + 1 > 0 iff x^3-4x-1<0 iff x(x-2)(x+2)<1 iff \
            x(x-2)(x+2)le 0<1 text{for} 0le xle 2 text{(why?)}$$






            share|cite|improve this answer









            $endgroup$



            You made two mistakes and recovered:
            $$dfrac{1}{x^2-4}<2 notRightarrow 1<2(x^2-4) (text{it must be} 1color{red}>2(x^2-4), text{because} x^2-4le 0)\
            4.5<x^2 notRightarrow -2.121<x<2.121 (text{it must be} 4.5color{red}>x^2 Rightarrow -2.121<x<2.121)$$



            Alternatively, you can prove it as follows:
            $$-x^3 + 4x + 1 > 0 iff x^3-4x-1<0 iff x(x-2)(x+2)<1 iff \
            x(x-2)(x+2)le 0<1 text{for} 0le xle 2 text{(why?)}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 15 '18 at 11:32









            farruhotafarruhota

            20.2k2738




            20.2k2738























                0












                $begingroup$

                You are correct that $-3/sqrt2<x<3/sqrt2$ because $x$ still lies between $0$ and $2$. You have found a wider bound for $x$, but it is not wrong. Note that the larger interval does not mean that $x$ takes all values between $pm3/sqrt2$.



                An analogy could be that $x=3$, and we know that $2<x<4$. But it is also true that $-1<x<7$, or even $-infty<x<infty$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  So, does the last statement mean "x can be any number between -2.121 and 2.121" or "x is one of the numbers between -2.121 and 2.121 but we don't know which number is it" ?
                  $endgroup$
                  – Ahmet Soyyiğit
                  Dec 15 '18 at 11:15










                • $begingroup$
                  We know that $0le xle2$. Therefore the last statement is saying that $x$ can be a set of numbers between $pm2.121$ but is restricted to $[0,2]$ due to the criterion at the start.
                  $endgroup$
                  – TheSimpliFire
                  Dec 15 '18 at 11:17
















                0












                $begingroup$

                You are correct that $-3/sqrt2<x<3/sqrt2$ because $x$ still lies between $0$ and $2$. You have found a wider bound for $x$, but it is not wrong. Note that the larger interval does not mean that $x$ takes all values between $pm3/sqrt2$.



                An analogy could be that $x=3$, and we know that $2<x<4$. But it is also true that $-1<x<7$, or even $-infty<x<infty$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  So, does the last statement mean "x can be any number between -2.121 and 2.121" or "x is one of the numbers between -2.121 and 2.121 but we don't know which number is it" ?
                  $endgroup$
                  – Ahmet Soyyiğit
                  Dec 15 '18 at 11:15










                • $begingroup$
                  We know that $0le xle2$. Therefore the last statement is saying that $x$ can be a set of numbers between $pm2.121$ but is restricted to $[0,2]$ due to the criterion at the start.
                  $endgroup$
                  – TheSimpliFire
                  Dec 15 '18 at 11:17














                0












                0








                0





                $begingroup$

                You are correct that $-3/sqrt2<x<3/sqrt2$ because $x$ still lies between $0$ and $2$. You have found a wider bound for $x$, but it is not wrong. Note that the larger interval does not mean that $x$ takes all values between $pm3/sqrt2$.



                An analogy could be that $x=3$, and we know that $2<x<4$. But it is also true that $-1<x<7$, or even $-infty<x<infty$.






                share|cite|improve this answer









                $endgroup$



                You are correct that $-3/sqrt2<x<3/sqrt2$ because $x$ still lies between $0$ and $2$. You have found a wider bound for $x$, but it is not wrong. Note that the larger interval does not mean that $x$ takes all values between $pm3/sqrt2$.



                An analogy could be that $x=3$, and we know that $2<x<4$. But it is also true that $-1<x<7$, or even $-infty<x<infty$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 15 '18 at 10:35









                TheSimpliFireTheSimpliFire

                12.4k62460




                12.4k62460












                • $begingroup$
                  So, does the last statement mean "x can be any number between -2.121 and 2.121" or "x is one of the numbers between -2.121 and 2.121 but we don't know which number is it" ?
                  $endgroup$
                  – Ahmet Soyyiğit
                  Dec 15 '18 at 11:15










                • $begingroup$
                  We know that $0le xle2$. Therefore the last statement is saying that $x$ can be a set of numbers between $pm2.121$ but is restricted to $[0,2]$ due to the criterion at the start.
                  $endgroup$
                  – TheSimpliFire
                  Dec 15 '18 at 11:17


















                • $begingroup$
                  So, does the last statement mean "x can be any number between -2.121 and 2.121" or "x is one of the numbers between -2.121 and 2.121 but we don't know which number is it" ?
                  $endgroup$
                  – Ahmet Soyyiğit
                  Dec 15 '18 at 11:15










                • $begingroup$
                  We know that $0le xle2$. Therefore the last statement is saying that $x$ can be a set of numbers between $pm2.121$ but is restricted to $[0,2]$ due to the criterion at the start.
                  $endgroup$
                  – TheSimpliFire
                  Dec 15 '18 at 11:17
















                $begingroup$
                So, does the last statement mean "x can be any number between -2.121 and 2.121" or "x is one of the numbers between -2.121 and 2.121 but we don't know which number is it" ?
                $endgroup$
                – Ahmet Soyyiğit
                Dec 15 '18 at 11:15




                $begingroup$
                So, does the last statement mean "x can be any number between -2.121 and 2.121" or "x is one of the numbers between -2.121 and 2.121 but we don't know which number is it" ?
                $endgroup$
                – Ahmet Soyyiğit
                Dec 15 '18 at 11:15












                $begingroup$
                We know that $0le xle2$. Therefore the last statement is saying that $x$ can be a set of numbers between $pm2.121$ but is restricted to $[0,2]$ due to the criterion at the start.
                $endgroup$
                – TheSimpliFire
                Dec 15 '18 at 11:17




                $begingroup$
                We know that $0le xle2$. Therefore the last statement is saying that $x$ can be a set of numbers between $pm2.121$ but is restricted to $[0,2]$ due to the criterion at the start.
                $endgroup$
                – TheSimpliFire
                Dec 15 '18 at 11:17











                0












                $begingroup$

                Stating that a statement is smaller than a number makes no sense.



                And from the fact that $dfrac1{x^2-4}<2$ together with $0leqslant xleqslant2$, you can't deduce that $dfrac1{x^2-4}<x$.



                Finally, from $dfrac1{x^2-4}<2$, what you deduce is that $1>x(x^2-4)$, since $x^2-4<0$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Stating that a statement is smaller than a number makes no sense.



                  And from the fact that $dfrac1{x^2-4}<2$ together with $0leqslant xleqslant2$, you can't deduce that $dfrac1{x^2-4}<x$.



                  Finally, from $dfrac1{x^2-4}<2$, what you deduce is that $1>x(x^2-4)$, since $x^2-4<0$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Stating that a statement is smaller than a number makes no sense.



                    And from the fact that $dfrac1{x^2-4}<2$ together with $0leqslant xleqslant2$, you can't deduce that $dfrac1{x^2-4}<x$.



                    Finally, from $dfrac1{x^2-4}<2$, what you deduce is that $1>x(x^2-4)$, since $x^2-4<0$.






                    share|cite|improve this answer









                    $endgroup$



                    Stating that a statement is smaller than a number makes no sense.



                    And from the fact that $dfrac1{x^2-4}<2$ together with $0leqslant xleqslant2$, you can't deduce that $dfrac1{x^2-4}<x$.



                    Finally, from $dfrac1{x^2-4}<2$, what you deduce is that $1>x(x^2-4)$, since $x^2-4<0$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 15 '18 at 10:38









                    José Carlos SantosJosé Carlos Santos

                    160k22127232




                    160k22127232























                        0












                        $begingroup$

                        1 + 4x - x$^3$ = 1 + x(4 - x$^2$).

                        For all x in [0,2], 0 < x, 0 < 4 - x$^2$.

                        Desired conclusion quickly follows.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          1 + 4x - x$^3$ = 1 + x(4 - x$^2$).

                          For all x in [0,2], 0 < x, 0 < 4 - x$^2$.

                          Desired conclusion quickly follows.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            1 + 4x - x$^3$ = 1 + x(4 - x$^2$).

                            For all x in [0,2], 0 < x, 0 < 4 - x$^2$.

                            Desired conclusion quickly follows.






                            share|cite|improve this answer









                            $endgroup$



                            1 + 4x - x$^3$ = 1 + x(4 - x$^2$).

                            For all x in [0,2], 0 < x, 0 < 4 - x$^2$.

                            Desired conclusion quickly follows.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 15 '18 at 11:02









                            William ElliotWilliam Elliot

                            8,0862720




                            8,0862720






























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                                Journaliste

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                                Pour les articles homonymes, voir Journaliste (homonymie). Journaliste Journalistes Appellation Journaliste Secteur d'activité Média - Communication Professions voisines Journaliste sportif - Reporter-photographe - Éditorialiste Codes CITP 2451, 2642 IDEO (France) 10197 ROME (France) E1106 modifier - modifier le code - modifier Wikidata Un journaliste est une personne qui recueille, écrit ou distribue des informations. L'activité d'un journaliste est le journalisme. Un journaliste peut travailler sur des sujets généraux ou se spécialiser dans certaines questions. La plupart des journalistes ont tendance à se spécialiser et, en collaborant avec d'autres journalistes, rédigent un article ou mettent en forme un reportage qui couvre l'actualité. Par exemple, un journaliste sportif couvre les nouvelles dans le monde du sport, mais ce journaliste peut aussi faire partie d'un journal qui couvre d'autres su...
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