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Proof question.

The following question has me quite puzzled.
x^2>x for x>1

I don't know how I can prove this to be right or wrong.

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Original post by ShredAlert

The following question has me quite puzzled.
x^2>x for x>1

I don't know how I can prove this to be right or wrong.

You can safely multiply through inequalities by positive quantities (like x here, for example) and not have to worry about the sign changing direction.

You need to start from what you know and show what you want; what is it that you know about x here?

Original post by ShredAlert

The following question has me quite puzzled.
x^2>x for x>1

I don't know how I can prove this to be right or wrong.

Perhaps:

Spoiler

OP

So this is what I got so far.
x²>x
x²-x>0
x(x-1)>0

kind of hit a brick wall after this.

Original post by ShredAlert

So this is what I got so far.
x²>x
x²-x>0
x(x-1)>0

kind of hit a brick wall after this.

If x>1 is x positive or negative
If x>1 is (x-1) positive or negative

Original post by ShredAlert

So this is what I got so far.
x²>x
x²-x>0
x(x-1)>0

kind of hit a brick wall after this.

If x(x-1)>0 both factors in the product must be either greater than 0 or less than 0.
(Negative x Negative), (Positive x Positive) both make positive.
What does this tell you about x?

Original post by ShredAlert

So this is what I got so far.
x²>x
x²-x>0
x(x-1)>0

kind of hit a brick wall after this.

Since x is positive you can just divide by x.

Original post by ShredAlert

So this is what I got so far.
x²>x
x²-x>0
x(x-1)>0

kind of hit a brick wall after this.

Generally speaking, you shouldn't start from what you're trying to show - usually it's bad logic (although it's not so much a problem here given that the steps are reversible, but still).

What you know is that x>1. Can you start from this and get x^2>x? (hint: it's literally a one step proof using the first line of my previous post).

Original post by joostan

If x(x-1)>0 both factors in the product must be either greater than 0 or less than 0.
(Negative x Negative), (Positive x Positive) both make positive.
What does this tell you about x?

Original post by metaltron

Since x is positive you can just divide by x.

Without further explanation/comments, all that you've done here is shown that x^2>x => x>1; which is not what the OP is trying to show.

Original post by Farhan.Hanif93

Without further explanation/comments, all that you've done here is shown that x^2>x => x>1; which is not what the OP is trying to show.

Having worked through surely one could then work back and show that this is true such that x>1 <=> x² > x (or x<0)and hence the question.
Or have I misunderstood. . . :s-smilie:

:s-smilie:

(edited 11 years ago)

Wouldn't a sketch be of any help? :/

Original post by Farhan.Hanif93

Without further explanation/comments, all that you've done here is shown that x^2>x => x>1; which is not what the OP is trying to show.

The further explanation was that x is positive, so you can have double implication arrows.

Original post by joostan

Having worked through surely one could then work back and show that this is true such that x>1 <=> x² > x
Or have I misunderstood. . . :s-smilie:

:s-smilie:

That wasn't evident from your post.

Original post by metaltron

The further explanation was that x is positive, so you can have double implication arrows.

Your double implications would require even further explanation - when showing that x^2 > x => x>1 (which is not necessarily true), you're not allowed to assume x>0.

(edited 11 years ago)

Original post by Farhan.Hanif93

Of course, but that wasn't evident from either post.

As for Metaltron's; your double implications would require even further explanation - when showing that x^2 > x => x>1, you're not allowed to assume x>0.

well it says x>1 .

OP

Original post by Farhan.Hanif93

Generally speaking, you shouldn't start from what you're trying to show - usually it's bad logic (although it's not so much a problem here given that the steps are reversible, but still).

What you know is that x>1. Can you start from this and get x^2>x? (hint: it's literally a one step proof using the first line of my previous post).

According to your suggestion I have done the following.
If x>1
multiplying both sides by x gives us x²>x
therefore the statement is true?
Please tell me if this is correct and if so sufficient in detail.
Thanks for all the help.

Original post by metaltron

well it says x>1 .

Which is exactly where the flaw in your logic lies. You claim to be able to use double implications here but, given that the converse is false, this isn't possible. (i.e. x^2>x ≠> x>1) So far, you've claimed that x^2>x, divided by x (assuming that x>1) and shown that x>1.

Original post by ShredAlert

According to your suggestion I have done the following.
If x>1
multiplying both sides by x gives us x²>x
therefore the statement is true?
Please tell me if this is correct and if so sufficient in detail.
Thanks for all the help.

Yes.

Original post by Farhan.Hanif93

Which is exactly where the flaw in your logic lies. You claim to be able to use double implications here but, given that the converse is false, this isn't possible. (i.e. x^2>x ≠> x>1) So far, you've claimed that x^2>x, divided by x (assuming that x>1) and shown that x>1.

Yes.

I assumed x>0 not x>1. So the end argument says:

Since x>1, x >0 so x^2>x. The other way round is if x is positive x^2 > x goes to x > 1 which is consistent with x being positive. Also, this is so messy that I have to make clear that I was just finishing off what the OP wrote.

It's straight forward.

x²>x and we know that x>1.
So you can divide both sides by x since x is not 0, and also the sign wont change since x is positive.

we get
x²/x>x/x which is equivalent to x>1: this is true. thus proven.

If you want a more complicated solution just prove by induction.

Original post by Giveme45

It's straight forward.

x²>x and we know that x>1.
So you can divide both sides by x since x is not 0, and also the sign wont change since x is positive.

we get
x²/x>x/x which is equivalent to x>1: this is true. thus proven.

If you want a more complicated solution just prove by induction.

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.

Original post by metaltron

I assumed x>0 not x>1.

Why would you do that?

Since x>1, x >0 so x^2>x.

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!])

The other way round is if x is positive x^2 > x goes to x > 1 which is consistent with x being positive.

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]

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