What is the next step in the prove? (Mathematical Induction) $left(x^{n}+1right)<left(x+1right)^{n}$












4












$begingroup$


I have to prove this preposition by mathematical induction:



$$left(x^n+1right)<left(x+1right)^n quad forall ngeq 2 quad text{and}quad x>0,,, n in mathbb{N}$$



I started the prove with $n=2$:



$left(x^{2}+1right)<left(x+1right)^{2}$



$x^{2}+1<x^{2}+2x+1$



We see that;



$x^{2}+1-x^{2}-1<2x$



$0<2x$



Then



$x>0$



And this one carries out for $n=2$



Now for $quad n=k quad$ (Hypothesis)



$left(x^{k}+1right)<left(x+1right)^{k}$



We have



$displaystyle x^{k}<left(x+1right)^{k}-1ldots quad (1)$



Then, we must prove for $quad n= k+1 quad$ (Thesis):



$x^{k+1}+1<left(x+1right)^{k+1}$



We develop before expression as:



$x^{k+1}<left(x+1right)^{k+1}-1ldots quad (2)$



According to the steps of mathematical induction, the next stpe would be use the hypothesis $(1)$ to prove thesis $(2)$. It's in here when I hesitate if the next one that I am going to write is correct:



First way:



We multiply hypothesis $(1)$ by $left(x+1right)$ and we have:



$x^{k}left(x+1right)<left[left(x+1right)^{k}-1right]left(x+1right)$



$x^{k}left(x+1right)<left(x+1right)^{k+1}-left(x+1right)$



Last expression divided by $left(x+1right)$ we have again the expression $(1)$:



$displaystyle frac{x^{k}left(x+1right)<left(x+1right)^{k+1}-left(x+1right)}{left(x+1right)}$



$x^{k}<left(x+1right)^{k}-1$



Second way:



If we multiply $(2)$ by $x$ we have:



$xx^{k}<xleft[left(x+1right)^{k}-1right]$



$x^{k+1}<xleft(x+1right)^{k}-x$



And if we again divided last expression by $x$, we arrive at the same result



$displaystyle frac{x^{k+1}<xleft(x+1right)^{k}-x}{x}$



$x^{k}<left(x+1right)^{k}-1$



I do not find another way to prove this demonstration, another way to solve the problem is using Newton's theorem binomial coeficients, but the prove lies in the technical using of mathematical induction. If someone can help me, I will be very grateful with him/her!
Thanks
-Víctor Hugo-






algebra-precalculus induction







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that both $x^n$ and $1$ are coefficients in the binomial expansion of $(x+1)^n$, and any other terms are positive since $x>0$. The inequality thus follows.
    $endgroup$
    – Math1000
    Dec 20 '18 at 23:41










  • $begingroup$
    You have to prove that if $k$ holds, then $k+1$ also holds. You haven't actually done that in either case, you've just arrived at the original statement where $n=k$.
    $endgroup$
    – Alex S
    Dec 20 '18 at 23:56










  • $begingroup$
    Just a general recommendation: you'll never get anywhere in proofs with the "one step forward, one step back" method.
    $endgroup$
    – Cameron Buie
    Dec 21 '18 at 1:07
















4












$begingroup$


I have to prove this preposition by mathematical induction:



$$left(x^n+1right)<left(x+1right)^n quad forall ngeq 2 quad text{and}quad x>0,,, n in mathbb{N}$$



I started the prove with $n=2$:



$left(x^{2}+1right)<left(x+1right)^{2}$



$x^{2}+1<x^{2}+2x+1$



We see that;



$x^{2}+1-x^{2}-1<2x$



$0<2x$



Then



$x>0$



And this one carries out for $n=2$



Now for $quad n=k quad$ (Hypothesis)



$left(x^{k}+1right)<left(x+1right)^{k}$



We have



$displaystyle x^{k}<left(x+1right)^{k}-1ldots quad (1)$



Then, we must prove for $quad n= k+1 quad$ (Thesis):



$x^{k+1}+1<left(x+1right)^{k+1}$



We develop before expression as:



$x^{k+1}<left(x+1right)^{k+1}-1ldots quad (2)$



According to the steps of mathematical induction, the next stpe would be use the hypothesis $(1)$ to prove thesis $(2)$. It's in here when I hesitate if the next one that I am going to write is correct:



First way:



We multiply hypothesis $(1)$ by $left(x+1right)$ and we have:



$x^{k}left(x+1right)<left[left(x+1right)^{k}-1right]left(x+1right)$



$x^{k}left(x+1right)<left(x+1right)^{k+1}-left(x+1right)$



Last expression divided by $left(x+1right)$ we have again the expression $(1)$:



$displaystyle frac{x^{k}left(x+1right)<left(x+1right)^{k+1}-left(x+1right)}{left(x+1right)}$



$x^{k}<left(x+1right)^{k}-1$



Second way:



If we multiply $(2)$ by $x$ we have:



$xx^{k}<xleft[left(x+1right)^{k}-1right]$



$x^{k+1}<xleft(x+1right)^{k}-x$



And if we again divided last expression by $x$, we arrive at the same result



$displaystyle frac{x^{k+1}<xleft(x+1right)^{k}-x}{x}$



$x^{k}<left(x+1right)^{k}-1$



I do not find another way to prove this demonstration, another way to solve the problem is using Newton's theorem binomial coeficients, but the prove lies in the technical using of mathematical induction. If someone can help me, I will be very grateful with him/her!
Thanks
-Víctor Hugo-






algebra-precalculus induction







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that both $x^n$ and $1$ are coefficients in the binomial expansion of $(x+1)^n$, and any other terms are positive since $x>0$. The inequality thus follows.
    $endgroup$
    – Math1000
    Dec 20 '18 at 23:41










  • $begingroup$
    You have to prove that if $k$ holds, then $k+1$ also holds. You haven't actually done that in either case, you've just arrived at the original statement where $n=k$.
    $endgroup$
    – Alex S
    Dec 20 '18 at 23:56










  • $begingroup$
    Just a general recommendation: you'll never get anywhere in proofs with the "one step forward, one step back" method.
    $endgroup$
    – Cameron Buie
    Dec 21 '18 at 1:07














4












4








4


1



$begingroup$


I have to prove this preposition by mathematical induction:



$$left(x^n+1right)<left(x+1right)^n quad forall ngeq 2 quad text{and}quad x>0,,, n in mathbb{N}$$



I started the prove with $n=2$:



$left(x^{2}+1right)<left(x+1right)^{2}$



$x^{2}+1<x^{2}+2x+1$



We see that;



$x^{2}+1-x^{2}-1<2x$



$0<2x$



Then



$x>0$



And this one carries out for $n=2$



Now for $quad n=k quad$ (Hypothesis)



$left(x^{k}+1right)<left(x+1right)^{k}$



We have



$displaystyle x^{k}<left(x+1right)^{k}-1ldots quad (1)$



Then, we must prove for $quad n= k+1 quad$ (Thesis):



$x^{k+1}+1<left(x+1right)^{k+1}$



We develop before expression as:



$x^{k+1}<left(x+1right)^{k+1}-1ldots quad (2)$



According to the steps of mathematical induction, the next stpe would be use the hypothesis $(1)$ to prove thesis $(2)$. It's in here when I hesitate if the next one that I am going to write is correct:



First way:



We multiply hypothesis $(1)$ by $left(x+1right)$ and we have:



$x^{k}left(x+1right)<left[left(x+1right)^{k}-1right]left(x+1right)$



$x^{k}left(x+1right)<left(x+1right)^{k+1}-left(x+1right)$



Last expression divided by $left(x+1right)$ we have again the expression $(1)$:



$displaystyle frac{x^{k}left(x+1right)<left(x+1right)^{k+1}-left(x+1right)}{left(x+1right)}$



$x^{k}<left(x+1right)^{k}-1$



Second way:



If we multiply $(2)$ by $x$ we have:



$xx^{k}<xleft[left(x+1right)^{k}-1right]$



$x^{k+1}<xleft(x+1right)^{k}-x$



And if we again divided last expression by $x$, we arrive at the same result



$displaystyle frac{x^{k+1}<xleft(x+1right)^{k}-x}{x}$



$x^{k}<left(x+1right)^{k}-1$



I do not find another way to prove this demonstration, another way to solve the problem is using Newton's theorem binomial coeficients, but the prove lies in the technical using of mathematical induction. If someone can help me, I will be very grateful with him/her!
Thanks
-Víctor Hugo-






algebra-precalculus induction







share|cite|improve this question











$endgroup$




I have to prove this preposition by mathematical induction:



$$left(x^n+1right)<left(x+1right)^n quad forall ngeq 2 quad text{and}quad x>0,,, n in mathbb{N}$$



I started the prove with $n=2$:



$left(x^{2}+1right)<left(x+1right)^{2}$



$x^{2}+1<x^{2}+2x+1$



We see that;



$x^{2}+1-x^{2}-1<2x$



$0<2x$



Then



$x>0$



And this one carries out for $n=2$



Now for $quad n=k quad$ (Hypothesis)



$left(x^{k}+1right)<left(x+1right)^{k}$



We have



$displaystyle x^{k}<left(x+1right)^{k}-1ldots quad (1)$



Then, we must prove for $quad n= k+1 quad$ (Thesis):



$x^{k+1}+1<left(x+1right)^{k+1}$



We develop before expression as:



$x^{k+1}<left(x+1right)^{k+1}-1ldots quad (2)$



According to the steps of mathematical induction, the next stpe would be use the hypothesis $(1)$ to prove thesis $(2)$. It's in here when I hesitate if the next one that I am going to write is correct:



First way:



We multiply hypothesis $(1)$ by $left(x+1right)$ and we have:



$x^{k}left(x+1right)<left[left(x+1right)^{k}-1right]left(x+1right)$



$x^{k}left(x+1right)<left(x+1right)^{k+1}-left(x+1right)$



Last expression divided by $left(x+1right)$ we have again the expression $(1)$:



$displaystyle frac{x^{k}left(x+1right)<left(x+1right)^{k+1}-left(x+1right)}{left(x+1right)}$



$x^{k}<left(x+1right)^{k}-1$



Second way:



If we multiply $(2)$ by $x$ we have:



$xx^{k}<xleft[left(x+1right)^{k}-1right]$



$x^{k+1}<xleft(x+1right)^{k}-x$



And if we again divided last expression by $x$, we arrive at the same result



$displaystyle frac{x^{k+1}<xleft(x+1right)^{k}-x}{x}$



$x^{k}<left(x+1right)^{k}-1$



I do not find another way to prove this demonstration, another way to solve the problem is using Newton's theorem binomial coeficients, but the prove lies in the technical using of mathematical induction. If someone can help me, I will be very grateful with him/her!
Thanks
-Víctor Hugo-






algebra-precalculus induction




algebra-precalculus induction






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 23:46









Cameron Williams

22.5k43680




22.5k43680










asked Dec 20 '18 at 23:34









Víctor VázquezVíctor Vázquez

232




232












  • $begingroup$
    Note that both $x^n$ and $1$ are coefficients in the binomial expansion of $(x+1)^n$, and any other terms are positive since $x>0$. The inequality thus follows.
    $endgroup$
    – Math1000
    Dec 20 '18 at 23:41










  • $begingroup$
    You have to prove that if $k$ holds, then $k+1$ also holds. You haven't actually done that in either case, you've just arrived at the original statement where $n=k$.
    $endgroup$
    – Alex S
    Dec 20 '18 at 23:56










  • $begingroup$
    Just a general recommendation: you'll never get anywhere in proofs with the "one step forward, one step back" method.
    $endgroup$
    – Cameron Buie
    Dec 21 '18 at 1:07


















  • $begingroup$
    Note that both $x^n$ and $1$ are coefficients in the binomial expansion of $(x+1)^n$, and any other terms are positive since $x>0$. The inequality thus follows.
    $endgroup$
    – Math1000
    Dec 20 '18 at 23:41










  • $begingroup$
    You have to prove that if $k$ holds, then $k+1$ also holds. You haven't actually done that in either case, you've just arrived at the original statement where $n=k$.
    $endgroup$
    – Alex S
    Dec 20 '18 at 23:56










  • $begingroup$
    Just a general recommendation: you'll never get anywhere in proofs with the "one step forward, one step back" method.
    $endgroup$
    – Cameron Buie
    Dec 21 '18 at 1:07
















$begingroup$
Note that both $x^n$ and $1$ are coefficients in the binomial expansion of $(x+1)^n$, and any other terms are positive since $x>0$. The inequality thus follows.
$endgroup$
– Math1000
Dec 20 '18 at 23:41




$begingroup$
Note that both $x^n$ and $1$ are coefficients in the binomial expansion of $(x+1)^n$, and any other terms are positive since $x>0$. The inequality thus follows.
$endgroup$
– Math1000
Dec 20 '18 at 23:41












$begingroup$
You have to prove that if $k$ holds, then $k+1$ also holds. You haven't actually done that in either case, you've just arrived at the original statement where $n=k$.
$endgroup$
– Alex S
Dec 20 '18 at 23:56




$begingroup$
You have to prove that if $k$ holds, then $k+1$ also holds. You haven't actually done that in either case, you've just arrived at the original statement where $n=k$.
$endgroup$
– Alex S
Dec 20 '18 at 23:56












$begingroup$
Just a general recommendation: you'll never get anywhere in proofs with the "one step forward, one step back" method.
$endgroup$
– Cameron Buie
Dec 21 '18 at 1:07




$begingroup$
Just a general recommendation: you'll never get anywhere in proofs with the "one step forward, one step back" method.
$endgroup$
– Cameron Buie
Dec 21 '18 at 1:07










2 Answers
2






active

oldest

votes


















3












$begingroup$

Suppose that $(1+x)^n>1+x^n$ for some $nge 2$. Then
$$
(1+x)^{n+1}=(1+x)^n(1+x)>(1+x^n)(1+x)=1+x+x^n+x^{n+1}>1+x^{n+1}
$$
since $x>0$ where in the first inequality we used the induction hypothesis.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    We will prove it is true with $n=k$. Indeed we need to prove it is true with $n=k+1$.



    Or we need to prove $x^{k+1}+1<left(x+1right)^{k+1}$



    We have: $RHS=(x+1)cdot (x+1)^k>(x+1)cdot (x^k+1)$ (Hypothesis)



    $=x(x^k+1)+(x^k+1)=x^{k+1}+x^k+x+1$



    $>x^{k+1}+1=RHS$ $(x>0)$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes









      3












      $begingroup$

      Suppose that $(1+x)^n>1+x^n$ for some $nge 2$. Then
      $$
      (1+x)^{n+1}=(1+x)^n(1+x)>(1+x^n)(1+x)=1+x+x^n+x^{n+1}>1+x^{n+1}
      $$
      since $x>0$ where in the first inequality we used the induction hypothesis.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Suppose that $(1+x)^n>1+x^n$ for some $nge 2$. Then
        $$
        (1+x)^{n+1}=(1+x)^n(1+x)>(1+x^n)(1+x)=1+x+x^n+x^{n+1}>1+x^{n+1}
        $$
        since $x>0$ where in the first inequality we used the induction hypothesis.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Suppose that $(1+x)^n>1+x^n$ for some $nge 2$. Then
          $$
          (1+x)^{n+1}=(1+x)^n(1+x)>(1+x^n)(1+x)=1+x+x^n+x^{n+1}>1+x^{n+1}
          $$
          since $x>0$ where in the first inequality we used the induction hypothesis.






          share|cite|improve this answer









          $endgroup$



          Suppose that $(1+x)^n>1+x^n$ for some $nge 2$. Then
          $$
          (1+x)^{n+1}=(1+x)^n(1+x)>(1+x^n)(1+x)=1+x+x^n+x^{n+1}>1+x^{n+1}
          $$
          since $x>0$ where in the first inequality we used the induction hypothesis.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 23:40









          Foobaz JohnFoobaz John

          22.8k41452




          22.8k41452























              2












              $begingroup$

              We will prove it is true with $n=k$. Indeed we need to prove it is true with $n=k+1$.



              Or we need to prove $x^{k+1}+1<left(x+1right)^{k+1}$



              We have: $RHS=(x+1)cdot (x+1)^k>(x+1)cdot (x^k+1)$ (Hypothesis)



              $=x(x^k+1)+(x^k+1)=x^{k+1}+x^k+x+1$



              $>x^{k+1}+1=RHS$ $(x>0)$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                We will prove it is true with $n=k$. Indeed we need to prove it is true with $n=k+1$.



                Or we need to prove $x^{k+1}+1<left(x+1right)^{k+1}$



                We have: $RHS=(x+1)cdot (x+1)^k>(x+1)cdot (x^k+1)$ (Hypothesis)



                $=x(x^k+1)+(x^k+1)=x^{k+1}+x^k+x+1$



                $>x^{k+1}+1=RHS$ $(x>0)$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  We will prove it is true with $n=k$. Indeed we need to prove it is true with $n=k+1$.



                  Or we need to prove $x^{k+1}+1<left(x+1right)^{k+1}$



                  We have: $RHS=(x+1)cdot (x+1)^k>(x+1)cdot (x^k+1)$ (Hypothesis)



                  $=x(x^k+1)+(x^k+1)=x^{k+1}+x^k+x+1$



                  $>x^{k+1}+1=RHS$ $(x>0)$






                  share|cite|improve this answer









                  $endgroup$



                  We will prove it is true with $n=k$. Indeed we need to prove it is true with $n=k+1$.



                  Or we need to prove $x^{k+1}+1<left(x+1right)^{k+1}$



                  We have: $RHS=(x+1)cdot (x+1)^k>(x+1)cdot (x^k+1)$ (Hypothesis)



                  $=x(x^k+1)+(x^k+1)=x^{k+1}+x^k+x+1$



                  $>x^{k+1}+1=RHS$ $(x>0)$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 20 '18 at 23:44









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