Let x be a real number. Prove if x>1 then x^2>x.

- anonymous

Let x be a real number. Prove if x>1 then x^2>x.

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- anonymous

Induction might work here.

- OakTree

Do you know what induction is, @zonazoo?

- anonymous

yes, but I don't see how you would use that here.

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## More answers

- OakTree

Okay, so let's try a different approach. Let's look at the behavior of x^2 for all x>1. Do you see anything useful you could use there?

- anonymous

ummm, well, im not sure what you are looking for. if you square any number that is greater than 1, the answer will always be greater than the number you squared... the y value of the graph will always be greater on the quadratic than the linear function past 1. I would of thought you need to use contradiction, like suppose x

- OakTree

Contradiction would be a great way to prove this. Again, induction would be best, but contradiction would only be a line or two longer.

- anonymous

im not sure how you would do either.

- OakTree

Okay. So we have \[x^2 - x < 0\]So \[x(x-1) < 0\]But this means that exactly one of x or x-1 must be negative, since two negatives is positive and two positives is positive. Following so far?

- anonymous

yes... and this method would be by contradiction correct.

- RadEn

x > 1
multiply by x on both sides, we get
x^2 > x
:)

- OakTree

Right. Because we're assuming that something is wrong at the beginning.
Anyway, so can you think of what the next step might be?

- anonymous

idk, just stating that either x<0 or x<1, which would be a contradiction since x>1???

- OakTree

Kind of. Since exactly one of x, x-1 must be negative, we know that x-1 is -1 and x is 0. But that means that x is not greater than 1. Contradiction. QED. Get it?

- anonymous

yes I do. What is QED?

- OakTree

Quod erat demonstrandum. It's Latin for "what was required to be proved" - it's basically a formal way of saying that the proof is finished.

- anonymous

oh okay. Thank you for all the help.

- OakTree

No problem.

- anonymous

I suppose induction may not be as easy as I had thought... the base case, namely, is giving me trouble. But if you restrict the domain here to be the natural numbers, it's easier to work out.
Base case: \(x=2\). You have \(2^2=4>2\), so the inequality holds.
Assume it holds for \(x=k\), so that \(k^2>k\).
Next, show it holds for \(x=k+1\), i.e. \((k+1)^2>k+1\).
The trick here is to write this inequality in terms of that given by the previous assumption:
\[(k+1)^2=\color{red}{k^2}+\color{blue}{2k+1}>\color{red}{k}+\color{blue}{1}\]
You know the red part on the left is greater than the red on the right, so now it's a matter of showing \(2k+1>1\), which is true for all \(k>0\). So the inequality holds for \(x=k+1\).
But again, I'm not sure how one would adjust this for the real number system.

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