A funny proof, but what is wrong with it?

[jobsism]

Here's a question:-

Prove that for all positive integers n,

[(n+1)/2]^n >= n!

And here's a funny proof for it:-

Assume to the contrary that, for all positive integers n,

[(n+1)/2]^n < n!

However, for n=2,

(3/2)^2 > 2!

Therefore, our assumption must be false.

And hence, for all positive integers n, [(n+1)/2]^n >= n!

Now, I know that this proof can't be correct, because I've seen the real proof, and it's a marvel, making use of algebraic inequalities, and the proof above simply seems too simple compared to it. But I wonder, what's the mistake with the above proof?

 

[micromass]

You've made a logic mistake.

Consider the sentence

"All men are mortal"

This is true. But what's the negation of this sentence?? Is it

"All men are immortal"?

or is it

"There is a man that is immortal"??

It's the second one.

So, if your assertion is

"For all natural numbers n holds the inequality blabla"

then the negation of that is

"There is a natural number n such that the inequality does not hold"

Your proof shows that the inequality holds for n=2. But this does not contradict the negation, as there could still be an n such that it doesn't hold.

 

[bmxicle]

The assumption that for all possible integers n, n!>[(n+1)/2]^n is violated by your counter example, so that demonstrates that your assumption is false. You have not checked all possible values of n so it says nothing about any of those values. Your assumption being false says nothing about the case of only some n's satisfying the inequality.

 

[jobsism]

Oh...now I understand. Thanks, micromass and bmxicle! :D