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Ordinary Induction

Mathematical induction proof; $g_k=3g_k-1 - 2g_k-2$Naive question about ordinary vs. transfinite inductionTips on constructing a proof by induction.Use induction to prove that $n! leq n^n-1$Induction well ordering principleProof By Induction for arbitrary integersProve by inductionOrdinary Induction vs. Strong Induction (on the same example)2 variable induction proof divisibilityIs this description of “ordinary induction” from Velleman's *How to Prove It* correct?

0

1

$begingroup$

Define a function $f:mathbbZtomathbbZ$ by $f(x)=2x^2-8x+1$. Use ordinary induction to prove that $f(x)geq 0$ for all integers $xgeq 4$.

A bit perplexed on the IS. We solve for P(K+1) right? Can someone help me prove this?

induction

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asked Mar 21 at 23:40

RobinRobin

525

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried doing the usual steps for an inductive proof? What was your working out so far?
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 23:50
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Define a function $f:mathbbZtomathbbZ$ by $f(x)=2x^2-8x+1$. Use ordinary induction to prove that $f(x)geq 0$ for all integers $xgeq 4$.

A bit perplexed on the IS. We solve for P(K+1) right? Can someone help me prove this?

induction

share|cite|improve this question

asked Mar 21 at 23:40

RobinRobin

525

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried doing the usual steps for an inductive proof? What was your working out so far?
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 23:50
  

  
  
  
  

add a comment | 

$begingroup$

Define a function $f:mathbbZtomathbbZ$ by $f(x)=2x^2-8x+1$. Use ordinary induction to prove that $f(x)geq 0$ for all integers $xgeq 4$.

A bit perplexed on the IS. We solve for P(K+1) right? Can someone help me prove this?

induction

share|cite|improve this question

asked Mar 21 at 23:40

RobinRobin

525

$endgroup$

share|cite|improve this question

asked Mar 21 at 23:40

RobinRobin

525

share|cite|improve this question

asked Mar 21 at 23:40

RobinRobin

525

asked Mar 21 at 23:40

RobinRobin

525

asked Mar 21 at 23:40

RobinRobin

525

2 Answers
2

active

oldest

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1

$begingroup$

All you need to do is show that:

1) $f(4) geq 0$ (base case)

2) $f(x+1) geq f(x)$ for $x geq 4$ (inductive argument)

For the latter, simply compute the expression $g(x)=f(x+1)-f(x)$ and show that $g(x) geq 0$ for $x geq 4$.

share|cite|improve this answer

edited Mar 21 at 23:58

answered Mar 21 at 23:51

DeepakDeepak

17.9k11640

$endgroup$

add a comment | 

1

$begingroup$

Let $f(x)=2x^2-8x+1=2(x-2)^2-7.$ To show $f(x)geq 0$ for all $xgeq4$, $xinmathbbZ$, we must show it is at least true for the base case $x=4$: $$f(4)=2(4)-7=1geq0$$ Now we assume the statement holds for some $x=n$ where $n>4,$ that is, assume $f(n)geq0, $ we shall show $f(n+1)geq 0$:$$f(n+1)=2(n-1)^2-7=2n^2-4n+2-7=2n^2-4n +(4n-4n)+1+1-7=(2n^2-8n+1)+(4n-6)$$ Thus $f(n+1)=(2n^2-8n+1)+(4n-6)=f(n)+(4n-6)$

Since $n>4$, $4n-6inmathbbZ^+$ so we have that since $f(n)geq0$ by our induction hypothesis, $$f(n+1)=f(n)+(textelement of $mathbbZ^+$)=$textelement of $mathbbZ^+$$$$

Thus since the statement is true for $f(n+1)$, it is true for all $ngeq4.$ Conclude by induction that this statement is true.

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answered Mar 21 at 23:59

coreyman317coreyman317

1,256522

$endgroup$

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1

$begingroup$

All you need to do is show that:

1) $f(4) geq 0$ (base case)

2) $f(x+1) geq f(x)$ for $x geq 4$ (inductive argument)

For the latter, simply compute the expression $g(x)=f(x+1)-f(x)$ and show that $g(x) geq 0$ for $x geq 4$.

share|cite|improve this answer

edited Mar 21 at 23:58

answered Mar 21 at 23:51

DeepakDeepak

17.9k11640

$endgroup$

add a comment | 

1

$begingroup$

All you need to do is show that:

1) $f(4) geq 0$ (base case)

2) $f(x+1) geq f(x)$ for $x geq 4$ (inductive argument)

For the latter, simply compute the expression $g(x)=f(x+1)-f(x)$ and show that $g(x) geq 0$ for $x geq 4$.

share|cite|improve this answer

edited Mar 21 at 23:58

answered Mar 21 at 23:51

DeepakDeepak

17.9k11640

$endgroup$

add a comment | 

$begingroup$

All you need to do is show that:

1) $f(4) geq 0$ (base case)

2) $f(x+1) geq f(x)$ for $x geq 4$ (inductive argument)

For the latter, simply compute the expression $g(x)=f(x+1)-f(x)$ and show that $g(x) geq 0$ for $x geq 4$.

share|cite|improve this answer

edited Mar 21 at 23:58

answered Mar 21 at 23:51

DeepakDeepak

17.9k11640

$endgroup$

share|cite|improve this answer

edited Mar 21 at 23:58

answered Mar 21 at 23:51

DeepakDeepak

17.9k11640

answered Mar 21 at 23:51

DeepakDeepak

17.9k11640

answered Mar 21 at 23:51

DeepakDeepak

17.9k11640

1

$begingroup$

Let $f(x)=2x^2-8x+1=2(x-2)^2-7.$ To show $f(x)geq 0$ for all $xgeq4$, $xinmathbbZ$, we must show it is at least true for the base case $x=4$: $$f(4)=2(4)-7=1geq0$$ Now we assume the statement holds for some $x=n$ where $n>4,$ that is, assume $f(n)geq0, $ we shall show $f(n+1)geq 0$:$$f(n+1)=2(n-1)^2-7=2n^2-4n+2-7=2n^2-4n +(4n-4n)+1+1-7=(2n^2-8n+1)+(4n-6)$$ Thus $f(n+1)=(2n^2-8n+1)+(4n-6)=f(n)+(4n-6)$

Since $n>4$, $4n-6inmathbbZ^+$ so we have that since $f(n)geq0$ by our induction hypothesis, $$f(n+1)=f(n)+(textelement of $mathbbZ^+$)=$textelement of $mathbbZ^+$$$$

Thus since the statement is true for $f(n+1)$, it is true for all $ngeq4.$ Conclude by induction that this statement is true.

share|cite|improve this answer

answered Mar 21 at 23:59

coreyman317coreyman317

1,256522

$endgroup$

add a comment | 

1

$begingroup$

Let $f(x)=2x^2-8x+1=2(x-2)^2-7.$ To show $f(x)geq 0$ for all $xgeq4$, $xinmathbbZ$, we must show it is at least true for the base case $x=4$: $$f(4)=2(4)-7=1geq0$$ Now we assume the statement holds for some $x=n$ where $n>4,$ that is, assume $f(n)geq0, $ we shall show $f(n+1)geq 0$:$$f(n+1)=2(n-1)^2-7=2n^2-4n+2-7=2n^2-4n +(4n-4n)+1+1-7=(2n^2-8n+1)+(4n-6)$$ Thus $f(n+1)=(2n^2-8n+1)+(4n-6)=f(n)+(4n-6)$

Since $n>4$, $4n-6inmathbbZ^+$ so we have that since $f(n)geq0$ by our induction hypothesis, $$f(n+1)=f(n)+(textelement of $mathbbZ^+$)=$textelement of $mathbbZ^+$$$$

Thus since the statement is true for $f(n+1)$, it is true for all $ngeq4.$ Conclude by induction that this statement is true.

share|cite|improve this answer

answered Mar 21 at 23:59

coreyman317coreyman317

1,256522

$endgroup$

add a comment | 

$begingroup$

Let $f(x)=2x^2-8x+1=2(x-2)^2-7.$ To show $f(x)geq 0$ for all $xgeq4$, $xinmathbbZ$, we must show it is at least true for the base case $x=4$: $$f(4)=2(4)-7=1geq0$$ Now we assume the statement holds for some $x=n$ where $n>4,$ that is, assume $f(n)geq0, $ we shall show $f(n+1)geq 0$:$$f(n+1)=2(n-1)^2-7=2n^2-4n+2-7=2n^2-4n +(4n-4n)+1+1-7=(2n^2-8n+1)+(4n-6)$$ Thus $f(n+1)=(2n^2-8n+1)+(4n-6)=f(n)+(4n-6)$

Since $n>4$, $4n-6inmathbbZ^+$ so we have that since $f(n)geq0$ by our induction hypothesis, $$f(n+1)=f(n)+(textelement of $mathbbZ^+$)=$textelement of $mathbbZ^+$$$$

Thus since the statement is true for $f(n+1)$, it is true for all $ngeq4.$ Conclude by induction that this statement is true.

share|cite|improve this answer

answered Mar 21 at 23:59

coreyman317coreyman317

1,256522

$endgroup$

share|cite|improve this answer

answered Mar 21 at 23:59

coreyman317coreyman317

1,256522

answered Mar 21 at 23:59

coreyman317coreyman317

1,256522

answered Mar 21 at 23:59

coreyman317coreyman317

1,256522