Induction proof inequality

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So I got this induction proof question but I can't seem to make a logical statement in one part of it:

The question is , $a_n + 1 = 5 - frac6a_n + 2$ with
$a_1 = 1$ . Prove by induction that $a_n < 4$ for $n geq 1$

I reached up to the proof where I need to prove $a_k+1 <4$

Proof

$a_k <4 implies a_k + 2<6 $

The next step I want to put is:

$frac6 a_k +2 >1$

However I can only justify this statement if $a_k > -2$ but I can't seem to prove that or find any info in the question to suggest that it.

Can anyone help me with the proof or my theory?

inequality induction self-learning proof-theory

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edited Aug 2 at 6:37

ben1806

33

asked Aug 2 at 5:48

user122343

445

add a comment | 

up vote
0
down vote

favorite

So I got this induction proof question but I can't seem to make a logical statement in one part of it:

The question is , $a_n + 1 = 5 - frac6a_n + 2$ with
$a_1 = 1$ . Prove by induction that $a_n < 4$ for $n geq 1$

I reached up to the proof where I need to prove $a_k+1 <4$

Proof

$a_k <4 implies a_k + 2<6 $

The next step I want to put is:

$frac6 a_k +2 >1$

However I can only justify this statement if $a_k > -2$ but I can't seem to prove that or find any info in the question to suggest that it.

Can anyone help me with the proof or my theory?

inequality induction self-learning proof-theory

share|cite|improve this question

edited Aug 2 at 6:37

ben1806

33

asked Aug 2 at 5:48

user122343

445

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

So I got this induction proof question but I can't seem to make a logical statement in one part of it:

The question is , $a_n + 1 = 5 - frac6a_n + 2$ with
$a_1 = 1$ . Prove by induction that $a_n < 4$ for $n geq 1$

I reached up to the proof where I need to prove $a_k+1 <4$

Proof

$a_k <4 implies a_k + 2<6 $

The next step I want to put is:

$frac6 a_k +2 >1$

However I can only justify this statement if $a_k > -2$ but I can't seem to prove that or find any info in the question to suggest that it.

Can anyone help me with the proof or my theory?

inequality induction self-learning proof-theory

share|cite|improve this question

edited Aug 2 at 6:37

ben1806

33

asked Aug 2 at 5:48

user122343

445

So I got this induction proof question but I can't seem to make a logical statement in one part of it:

The question is , $a_n + 1 = 5 - frac6a_n + 2$ with
$a_1 = 1$ . Prove by induction that $a_n < 4$ for $n geq 1$

I reached up to the proof where I need to prove $a_k+1 <4$

Proof

$a_k <4 implies a_k + 2<6 $

The next step I want to put is:

$frac6 a_k +2 >1$

However I can only justify this statement if $a_k > -2$ but I can't seem to prove that or find any info in the question to suggest that it.

Can anyone help me with the proof or my theory?

inequality induction self-learning proof-theory

share|cite|improve this question

edited Aug 2 at 6:37

ben1806

33

asked Aug 2 at 5:48

user122343

445

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 6:37

ben1806

33

edited Aug 2 at 6:37

ben1806

33

edited Aug 2 at 6:37

ben1806

33

33

asked Aug 2 at 5:48

user122343

445

asked Aug 2 at 5:48

user122343

445

asked Aug 2 at 5:48

user122343

445

445

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote

Hint:   write it as $;a_n + 1 -4 = dfraca_n-4a_n + 2,$, then (prove and) use that $,a_n+2 gt 0,$.

share|cite|improve this answer

answered Aug 2 at 6:46

dxiv

53.7k64796

add a comment | 

up vote
0
down vote

Assuming you have shown the base case, suppose $0< a_k < 4$ for some $k in mathbbN$. Then we have that $0< a_k + 2 < 6$, which implies
beginalign*
frac6a_k + 2 > 1
endalign*
(divide both sides of the inequality by $a_k + 2$). Also notice that $frac6a_k + 2 < 3$ since we have assumed $a_k > 0$ and hence $a_k + 2 > 2$.

beginalign*
0 < a_k+1 = 5 - frac6a_k + 2 < 4
endalign*
as desired.

share|cite|improve this answer

edited Aug 2 at 6:39

answered Aug 2 at 6:02

matt stokes

1047

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I appreciate your help but can you justify how the implication ( <1) is true? I can't seem to wrap my head around it
  – user122343
  Aug 2 at 6:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If ak was 2 for example , the inequality won't hold
  – user122343
  Aug 2 at 6:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry I made a mistake. It should be 6/(a_k + 2) >1.
  – matt stokes
  Aug 2 at 6:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If ak was -3 wouldn't the statement not hold also . I'm a bit confused on this since I don't know for sure if ak itself ( the terms of the sequence ) is greater than 0 . I assume it is from plugging in values but I feel as though I can't just write that worhout proving for all
  – user122343
  Aug 2 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Notice a_1 = 1 and the statement holds (this is the base case). Now for the induction part of the proof, if we assume a_n < 4 for arbitrary n, we have just shown that a_n+1<4. This implies that a_n can never be -3.
  – matt stokes
  Aug 2 at 6:19
  

  
  
  
  

 | 
show 1 more comment

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Hint:   write it as $;a_n + 1 -4 = dfraca_n-4a_n + 2,$, then (prove and) use that $,a_n+2 gt 0,$.

share|cite|improve this answer

answered Aug 2 at 6:46

dxiv

53.7k64796

add a comment | 

up vote
1
down vote

Hint:   write it as $;a_n + 1 -4 = dfraca_n-4a_n + 2,$, then (prove and) use that $,a_n+2 gt 0,$.

share|cite|improve this answer

answered Aug 2 at 6:46

dxiv

53.7k64796

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint:   write it as $;a_n + 1 -4 = dfraca_n-4a_n + 2,$, then (prove and) use that $,a_n+2 gt 0,$.

share|cite|improve this answer

answered Aug 2 at 6:46

dxiv

53.7k64796

Hint:   write it as $;a_n + 1 -4 = dfraca_n-4a_n + 2,$, then (prove and) use that $,a_n+2 gt 0,$.

share|cite|improve this answer

answered Aug 2 at 6:46

dxiv

53.7k64796

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 6:46

dxiv

53.7k64796

answered Aug 2 at 6:46

dxiv

53.7k64796

answered Aug 2 at 6:46

dxiv

53.7k64796

53.7k64796

add a comment | 

add a comment | 

up vote
0
down vote

Assuming you have shown the base case, suppose $0< a_k < 4$ for some $k in mathbbN$. Then we have that $0< a_k + 2 < 6$, which implies
beginalign*
frac6a_k + 2 > 1
endalign*
(divide both sides of the inequality by $a_k + 2$). Also notice that $frac6a_k + 2 < 3$ since we have assumed $a_k > 0$ and hence $a_k + 2 > 2$.

beginalign*
0 < a_k+1 = 5 - frac6a_k + 2 < 4
endalign*
as desired.

share|cite|improve this answer

edited Aug 2 at 6:39

answered Aug 2 at 6:02

matt stokes

1047

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I appreciate your help but can you justify how the implication ( <1) is true? I can't seem to wrap my head around it
  – user122343
  Aug 2 at 6:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If ak was 2 for example , the inequality won't hold
  – user122343
  Aug 2 at 6:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry I made a mistake. It should be 6/(a_k + 2) >1.
  – matt stokes
  Aug 2 at 6:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If ak was -3 wouldn't the statement not hold also . I'm a bit confused on this since I don't know for sure if ak itself ( the terms of the sequence ) is greater than 0 . I assume it is from plugging in values but I feel as though I can't just write that worhout proving for all
  – user122343
  Aug 2 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Notice a_1 = 1 and the statement holds (this is the base case). Now for the induction part of the proof, if we assume a_n < 4 for arbitrary n, we have just shown that a_n+1<4. This implies that a_n can never be -3.
  – matt stokes
  Aug 2 at 6:19
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

Assuming you have shown the base case, suppose $0< a_k < 4$ for some $k in mathbbN$. Then we have that $0< a_k + 2 < 6$, which implies
beginalign*
frac6a_k + 2 > 1
endalign*
(divide both sides of the inequality by $a_k + 2$). Also notice that $frac6a_k + 2 < 3$ since we have assumed $a_k > 0$ and hence $a_k + 2 > 2$.

beginalign*
0 < a_k+1 = 5 - frac6a_k + 2 < 4
endalign*
as desired.

share|cite|improve this answer

edited Aug 2 at 6:39

answered Aug 2 at 6:02

matt stokes

1047

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I appreciate your help but can you justify how the implication ( <1) is true? I can't seem to wrap my head around it
  – user122343
  Aug 2 at 6:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If ak was 2 for example , the inequality won't hold
  – user122343
  Aug 2 at 6:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry I made a mistake. It should be 6/(a_k + 2) >1.
  – matt stokes
  Aug 2 at 6:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If ak was -3 wouldn't the statement not hold also . I'm a bit confused on this since I don't know for sure if ak itself ( the terms of the sequence ) is greater than 0 . I assume it is from plugging in values but I feel as though I can't just write that worhout proving for all
  – user122343
  Aug 2 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Notice a_1 = 1 and the statement holds (this is the base case). Now for the induction part of the proof, if we assume a_n < 4 for arbitrary n, we have just shown that a_n+1<4. This implies that a_n can never be -3.
  – matt stokes
  Aug 2 at 6:19
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

up vote
0
down vote

Assuming you have shown the base case, suppose $0< a_k < 4$ for some $k in mathbbN$. Then we have that $0< a_k + 2 < 6$, which implies
beginalign*
frac6a_k + 2 > 1
endalign*
(divide both sides of the inequality by $a_k + 2$). Also notice that $frac6a_k + 2 < 3$ since we have assumed $a_k > 0$ and hence $a_k + 2 > 2$.

beginalign*
0 < a_k+1 = 5 - frac6a_k + 2 < 4
endalign*
as desired.

share|cite|improve this answer

edited Aug 2 at 6:39

answered Aug 2 at 6:02

matt stokes

1047

Assuming you have shown the base case, suppose $0< a_k < 4$ for some $k in mathbbN$. Then we have that $0< a_k + 2 < 6$, which implies
beginalign*
frac6a_k + 2 > 1
endalign*
(divide both sides of the inequality by $a_k + 2$). Also notice that $frac6a_k + 2 < 3$ since we have assumed $a_k > 0$ and hence $a_k + 2 > 2$.

beginalign*
0 < a_k+1 = 5 - frac6a_k + 2 < 4
endalign*
as desired.

share|cite|improve this answer

edited Aug 2 at 6:39

answered Aug 2 at 6:02

matt stokes

1047

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 6:39

edited Aug 2 at 6:39

edited Aug 2 at 6:39

answered Aug 2 at 6:02

matt stokes

1047

answered Aug 2 at 6:02

matt stokes

1047

answered Aug 2 at 6:02

matt stokes

1047

1047

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I appreciate your help but can you justify how the implication ( <1) is true? I can't seem to wrap my head around it
  – user122343
  Aug 2 at 6:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If ak was 2 for example , the inequality won't hold
  – user122343
  Aug 2 at 6:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry I made a mistake. It should be 6/(a_k + 2) >1.
  – matt stokes
  Aug 2 at 6:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If ak was -3 wouldn't the statement not hold also . I'm a bit confused on this since I don't know for sure if ak itself ( the terms of the sequence ) is greater than 0 . I assume it is from plugging in values but I feel as though I can't just write that worhout proving for all
  – user122343
  Aug 2 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Notice a_1 = 1 and the statement holds (this is the base case). Now for the induction part of the proof, if we assume a_n < 4 for arbitrary n, we have just shown that a_n+1<4. This implies that a_n can never be -3.
  – matt stokes
  Aug 2 at 6:19
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I appreciate your help but can you justify how the implication ( <1) is true? I can't seem to wrap my head around it
  – user122343
  Aug 2 at 6:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If ak was 2 for example , the inequality won't hold
  – user122343
  Aug 2 at 6:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry I made a mistake. It should be 6/(a_k + 2) >1.
  – matt stokes
  Aug 2 at 6:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If ak was -3 wouldn't the statement not hold also . I'm a bit confused on this since I don't know for sure if ak itself ( the terms of the sequence ) is greater than 0 . I assume it is from plugging in values but I feel as though I can't just write that worhout proving for all
  – user122343
  Aug 2 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Notice a_1 = 1 and the statement holds (this is the base case). Now for the induction part of the proof, if we assume a_n < 4 for arbitrary n, we have just shown that a_n+1<4. This implies that a_n can never be -3.
  – matt stokes
  Aug 2 at 6:19
  

  
  
  
  

I appreciate your help but can you justify how the implication ( <1) is true? I can't seem to wrap my head around it
– user122343
Aug 2 at 6:04

I appreciate your help but can you justify how the implication ( <1) is true? I can't seem to wrap my head around it
– user122343
Aug 2 at 6:04

If ak was 2 for example , the inequality won't hold
– user122343
Aug 2 at 6:09

If ak was 2 for example , the inequality won't hold
– user122343
Aug 2 at 6:09

Sorry I made a mistake. It should be 6/(a_k + 2) >1.
– matt stokes
Aug 2 at 6:11

Sorry I made a mistake. It should be 6/(a_k + 2) >1.
– matt stokes
Aug 2 at 6:11

If ak was -3 wouldn't the statement not hold also . I'm a bit confused on this since I don't know for sure if ak itself ( the terms of the sequence ) is greater than 0 . I assume it is from plugging in values but I feel as though I can't just write that worhout proving for all
– user122343
Aug 2 at 6:14

If ak was -3 wouldn't the statement not hold also . I'm a bit confused on this since I don't know for sure if ak itself ( the terms of the sequence ) is greater than 0 . I assume it is from plugging in values but I feel as though I can't just write that worhout proving for all
– user122343
Aug 2 at 6:14

Notice a_1 = 1 and the statement holds (this is the base case). Now for the induction part of the proof, if we assume a_n < 4 for arbitrary n, we have just shown that a_n+1<4. This implies that a_n can never be -3.
– matt stokes
Aug 2 at 6:19

Notice a_1 = 1 and the statement holds (this is the base case). Now for the induction part of the proof, if we assume a_n < 4 for arbitrary n, we have just shown that a_n+1<4. This implies that a_n can never be -3.
– matt stokes
Aug 2 at 6:19

 | 
show 1 more comment

 

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