Proof question.

Why would you do that?


That's fine. (in fact, this finishes the entire proof as TenofThem alludes to earlier in the thread [and is not what you said in any of your above posts!])


Not fine - you're not allowed to assume x>0 here. You need to show that if x^2>x then x>1 here without any further assumptions about x (including x>0).

[Unless]

You're starting with what you're trying to show, which you shouldn't do.

$x>1$ implies that $x-1>0$ so $(x-1)(x-1)>0$

You can do this last step because one of the axioms is that if $a>0, b>0$ then $ab >0$

$x^2 - 2x + 1 > 0$

$x^2 > 2x -1$

Now using the fact that $x>1$

$x^2 > x + x -1 > x$

There's probably a much quicker and neater way of doing it though.

You made mention of dividing by x in your first reply, and made no mention of showing that x>1 => x^2>x, hence my unease given that it could potentially confuse the OP into believing that showing x^2>x => x>1 is enough (regardless of whether that was what you wanted to show or not). I required a lot of extra information before what you were trying to do made any sense to me.

Anyway, I suppose this thread has been done to death.

Yes. My way. There's absolutely nothing wrong with my proof. The "you're starting with what you are trying to show" is nonsense. It doesn't matter whether you start with x>1 or x²>x.
Please don't criticize other peoples work if you don't have a strong argument to do so (rather than, "you shouldnt start with what you are trying to show". Thanks.

You made mention of dividing by x in your first reply, and made no mention of showing that x>1 => x^2>x, hence my unease given that it could potentially confuse the OP into believing that showing x^2>x => x>1 is enough (regardless of whether that was what you wanted to show or not). I required a lot of extra information before what you were trying to do made any sense to me.

Anyway, I suppose this thread has been done to death.

I never said your way is wrong, it's just bad practice and completely unnecessary to do it that way - that's all.

What is wrong with saying:

We know x > 1, so we shall check whether x^2 > x is consistent with this. As x >1 > 0 we can divide by x to get x > 1, which is true. Hence, x^2>x is true when x >1.

This was what I said originally.

Then please tell me how it is "bad practice"? And how is it unnecessary?

You're just throwing around words. How is your way less unnecessary then mine, if you do additional work which isn't needed to prove such a simple equation? Wouldn't your way be unnecessary then? Wouldn't rather that be considered bad practice?

Because, as I replied to someone else, when you start with what you're trying to show you're assuming it's true - you have nothing left to prove, the mere use of it to do your calculations means you're assuming it's true.

Because, as I replied to someone else, when you start with what you're trying to show you're assuming it's true - you have nothing left to prove, the mere use of it to do your calculations means you're assuming it's true.

It seems like you are confusing verification with proof. This discussion is frustrating.

I proved x²>x for x>1. Nothing left to say.

What are you on about? It's fine to do it for verification, not when you're trying to prove something.

Do you see what's wrong with this proof in showing $\sqrt{ab} \leq \frac{1}{2}(a+b)$

$\sqrt{ab} \leq \frac{1}{2}(a+b)$ gives $2\sqrt{ab} \leq a+b$, so $4ab \leq (a+b)^2$. Expanding gives $0 \leq a^2 - 2ab + b^2$ that is $0 \leq (a-b)^2$ then since squares are always positive, this is true. Hence $\sqrt{ab} \leq \frac{1}{2}(a+b)$

Firstly, you can't square both sides of an inequality. Another thing that's different is that the question asked:

prove x^2 > x given that x > 1 not just prove x^2>x

What are you on about? It's fine to do it for verification, not when you're trying to prove something.

Do you see what's wrong with this proof in showing $\sqrt{ab} \leq \frac{1}{2}(a+b)$

$\sqrt{ab} \leq \frac{1}{2}(a+b)$ gives $2\sqrt{ab} \leq a+b$, so $4ab \leq (a+b)^2$. Expanding gives $0 \leq a^2 - 2ab + b^2$ that is $0 \leq (a-b)^2$ then since squares are always positive, this is true. Hence $\sqrt{ab} \leq \frac{1}{2}(a+b)$

What I am "on about" is that my proof is a completely valid and legitimate proof. You don't make sense if you say that the proof that "x²>x for x>1" is incorrect if you start with "x²>x" to work your way to "x>1", but is completely fair and valid if you start with "x>1" then landing at "x²>x". We are doing exactly the same thing except for the fact that you start at a different place. Jesus. Stop treating others like kids especially if you have no basis to support your claim that my prove would not be considered one.

Sorry, I forgot to add you can assume $a,b \geq 0$

What I am "on about" is that my proof is a completely valid and legitimate proof. You don't make sense if you say that the proof that "x²>x for x>1" is incorrect if you start with "x²>x" to work your way to "x>1", but is completely fair and valid if you start with "x>1" then landing at "x²>x". We are doing exactly the same thing except for the fact that you start at a different place. Jesus. Stop treating others like kids especially if you have no basis to support your claim that my prove would not be considered one.

I'm late to the party and just being picky but what the OP posted was "x^2 > x for x>1".

That strikes me as an abridged version of a question to be honest - I'd expect a question to read something like "Prove that if x>1, then x^2 > x" or "Prove that x^2 > x for all x such that x > 1".

There'd be far less contention here if the direction of the implication had been made more explicit in the the OP's post

I was starting to give in before you came along!