Prove the theorems using mathematical induction.

[yy205001]

Homework Statement

Prove the theorems using mathematical induction.

$\forall n$ $\in N$, n$\geq 4$ $\rightarrow$n²$\leq n!$

Thanks in advance!

Homework Equations

The Attempt at a Solution

First, check the base case which is n=4.
$\Rightarrow$n=4$\geq$4-True

$\Rightarrow$4²$\leq$4*3*2*1

$\Rightarrow$16$\leq$24-True

Therefore, P(1) is true.

Now, check for the case n=k+1, to prove P(k)$\rightarrow$P(k+1).
Assume P(k) is true.
$\Rightarrow$k$\geq$4
$\Rightarrow$k²$\leq$k!

$\Rightarrow$k+1$\geq$4+1 -Add 1 on both side
$\Rightarrow$k+1$\geq$5
$\Rightarrow$(k+1)²=k²+2k+1
$\Rightarrow$(k+1)!=(k+1)k!
...

And I stop here can't get further to prove P(k+1) is true.
Any help is appreciated.

 

[HallsofIvy]

Homework Statement

Prove the theorems using mathematical induction.

$\forall n$ $\in N$, n$\geq 4$ $\rightarrow$n²$\leq n!$

Thanks in advance!

Homework Equations

The Attempt at a Solution

First, check the base case which is n=4.
$\Rightarrow$n=4$\geq$4-True

$\Rightarrow$4²$\leq$4*3*2*1

$\Rightarrow$16$\leq$24-True

Therefore, P(1) is true.

Now, check for the case n=k+1, to prove P(k)$\rightarrow$P(k+1).
Assume P(k) is true.
$\Rightarrow$k$\geq$4
$\Rightarrow$k²$\leq$k!

$\Rightarrow$k+1$\geq$4+1 -Add 1 on both side
$\Rightarrow$k+1$\geq$5
$\Rightarrow$(k+1)²=k²+2k+1
$\Rightarrow$(k+1)!=(k+1)k!

Rather than multiply the square note that what you want to prove is that
$(k+1)^2= (k+1)(k+1)\le (k+1)k!$ which, because k+1 is positive, is
the same as proving that $k+1\le k!$. You might want to do another induction to prove that is true first.

...

And I stop here can't get further to prove P(k+1) is true.
Any help is appreciated.

 

[yy205001]

Rather than multiply the square note that what you want to prove is that
$(k+1)^2= (k+1)(k+1)\le (k+1)k!$ which, because k+1 is positive, is
the same as proving that $k+1\le k!$. You might want to do another induction to prove that is true first.

Thanks HallsofIvy,
So I should start at proving (n+1)(n+1)≤(n+1)n! $\Rightarrow$ n+1≤n!
First, Check the base case n=4:
$\Rightarrow$(4)+1≤4!
$\Rightarrow$5≤24
So, the base case is true.

Now, Assume P(k) is true, then prove prove P(k+1):
$\Rightarrow$k+1≤k! -is true (Premise)
Also, let n=k+1:
$\Rightarrow$k+1+1≤(k+1)!
$\Rightarrow$(k+1)+1≤(k+1)k!
$\Rightarrow$1+1/(k+1)≤k! [Since k+1 is positive, so the inequality sign won't change]
Since 1+k≤k! is true and k>1/(k+1)
Therefore, P(k+1) is true
P(k)$\rightarrow$P(k+1)
And prove $\forall n$$\in N$n≥4$\rightarrow$n²≤n!

Does this look alright??

 

[HallsofIvy]

Thanks HallsofIvy,
So I should start at proving (n+1)(n+1)≤(n+1)n! $\Rightarrow$ n+1≤n!

No, that follows trivially by dividing both sides by n+1. What I said was that you should show that, for all $n\ge 4$, $n+1\le n!$ (which is, in fact, what you are doing).

First, Check the base case n=4:
$\Rightarrow$(4)+1≤4!
$\Rightarrow$5≤24
So, the base case is true.

Now, Assume P(k) is true, then prove prove P(k+1):
$\Rightarrow$k+1≤k! -is true (Premise)
Also, let n=k+1:
$\Rightarrow$k+1+1≤(k+1)!

You cannot assert this because you do not know if it is true yet. This is what you want to prove.

$\Rightarrow$(k+1)+1≤(k+1)k!
$\Rightarrow$1+1/(k+1)≤k! [Since k+1 is positive, so the inequality sign won't change]
Since 1+k≤k! is true and k>1/(k+1)
Therefore, P(k+1) is true

Strictly speaking what you have done is prove "if P(k+1) is true then P(k) is true". Do it the other way around: if [itex]k+1\le k![/tex] then $(k+1)(k+1)\le k!(k+1)= (k+1)!$<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> P(k)$\rightarrow$P(k+1)<br /> And prove $\forall n$$\in N$n≥4$\rightarrow$n²≤n!<br /> <br /> Does this look alright?? </div> </div> </blockquote>[/itex]