proof

**nmatthies1** · October 4th 2008, 08:12 AM

I am facing the following problem. I have proven by induction that if x, y are positive then x < y => x^n < y^n but now I have to prove the converse (not specified using which kind of proof) ie. that if both x, y are positive then x^n < y^n => x<y and I'm not really sure how I should start. I have thought of proving that it is true for n-1 and therefore it must be true for n=1. But then there would be no base step. Could I do it like that? Or is there another way?

**Sean12345** · October 4th 2008, 09:36 AM

Hi nmatthies1,

You can prove by induction as before. We have;

$x^n<y^n \implies x<y~~~P(n)$

First we shall prove that $P(1)$ is true then that $P(k)\implies P(k+1)$ is true. Let $n=1$ ;

$LHS=~x^1<y^1 \implies x<y~=RHS$ and thus $P(1)$ is true. Now assume $P(k)$ is indded true and hence

$<br /> x^k<y^k \implies x<y$

Since $x<y$ and $x,y >0$ we can multiply the $LHS$ side of the inequality by $x$ and the $RHS$ by $y$ and not change the inequality hence,

$x^{k+1}<y^{k+1}$

Thus $P(k)\implies P(k+1)$ is true. Since this is the case and $P(1)$ is true then we can conclude that $P(n)$ is true $\forall n \in \mathbb {Z}$ and $n\geq 1$ .

Alternatively we could prove this directly. We know that $x^n<y^n$ and thus it follows that $x^n-y^n<0$ and hence,

$<br /> (x-y)\left(\sum^{n-1}_{k=0}{x^ky^{n-1-k}}\right)<0$

which is negative iff $x-y<0 \implies x<y$ provided that $x,y >0$ which is given in this case.

Hope this helps.

**nmatthies1** · October 5th 2008, 02:13 PM

Thanks that really helped! I've still got to get a good grasp of proof by induction, but I think I'm getting there

The second proof looks really interesting too, but how did you get $(x-y)\left(\sum^{n-1}_{k=0}{x^ky^{n-1-k}}\right)<0$ from $x^n-y^n<0$ ?

**Showcase_22** · October 6th 2008, 08:28 AM

DUDE! You totally go to Warwick university!!!!

What course are you doing?

Someone managed to solve this? I thought they didn't imply each other:

$x^n<y^n \Rightarrow x<y$

For n=1:

$x<y \Rightarrow x<y$

That's true for n=1!

Assume true for n=k:

$x^k<y^k \Rightarrow x<y$

For n=k+1

$x^{k+1}<y^{k+1} \Rightarrow x<y$

$x^k x^1<y^k y^1 \Rightarrow x<y$

$\frac{x^k}{y^k}<\frac{y}{x} \Rightarrow \frac{x}{y}<1$

Now the above line doesn't make sense so I thought that the converse couldn't be proven (it states "try" in the question). I must have been a bit wrong!

Anyway, pm me with the course that you do. We might already know each other!!

**nmatthies1** · October 6th 2008, 10:02 AM

I do the Maths Bsc, you?

**Sean12345** · October 6th 2008, 10:33 AM

Oops in my original post I forgot to mention that

$x^n-y^n=(x-y)\left(\sum^{n-1}_{k=0}{x^ky^{n-1-k}}\right)<0$

only holds if $n \in \mathbb {Z}$ and $n\geq 2$ . Doesn't make much of a difference since the statement is clearly true for $n=1$ .

Originally Posted by nmatthies1

The second proof looks really interesting too, but how did you get $(x-y)\left(\sum^{n-1}_{k=0}{x^ky^{n-1-k}}\right)<0$ from $x^n-y^n<0$ ?

Without going into a thorough proof...... Say we wanted $x^2-y^2$ . Then we would want $(x-y)(x+y)$ . This is because we want the terms with $x$ and $y$ as a product to cancel. Similarly for $x^3-y^3$ we would want $(x-y)(x^2+xy+y^2)$ and for $x^4-y^4$ we would want $(x-y)(x^3+x^2y+xy^2+y^3)$ . Notice that we always want $(x-y)$ as the first bracket then all terms positive in the next bracket since then the negative terms will cancel out the positive terms because the terms are symmetrical. Hence as a generalisation we can say;

$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+.....+xy^{n-2}+y^{n-1})$ and thus more formally,

$x^n-y^n=(x-y)\left(\sum^{n-1}_{k=0}{x^ky^{n-1-k}}\right)$

Originally Posted by Showcase_22

$\frac{x^k}{y^k}<\frac{y}{x} \Rightarrow \frac{x}{y}<1$

Now the above line doesn't make sense so I thought that the converse couldn't be proven (it states "try" in the question). I must have been a bit wrong!

I can't see why that line doesn't make sense...... If $\frac{x^k}{y^k}<\frac{y}{x}$ and $x,y>0$ then it must imply that $x<y$ . I think here the direct proof is better.

As an added note I might be joining you next year as I have applied to Warwick for a maths degree of course.

**Showcase_22** · October 6th 2008, 10:55 AM

I like the sum of the series you posted. It's really handy! I'll remember it for a later date.

I can't see why that line doesn't make sense...... If and then it must imply that . I think here the direct proof is better.

What I was thinking was that since $\frac{y}{x}>1$ then it does not necessarily imply $<br /> \frac{x^k}{y^k}<\frac{y}{x} \Rightarrow \frac{x}{y}<1<br />$ .

My reasoning for this was that $\frac{y}{x}>1$ . For example, let's say $\frac{y}{x}=1.2$ . Therefore $\frac{x^k}{y^k}<1.2$ so this does not imply $\frac{x}{y}<1$ .

I would try and prove this using algebra, except when I started to do so it became increasingly confusing to convey my thought processes. I hope using the example of 1.2 made more sense.

Anyway, I could be completely wrong so let me know what you think.

As an added note I might be joining you next year as I have applied to Warwick for a maths degree of course.

You're making us look bad!!! Stop it!!!

Seriously, you seem pretty good at maths. What A -levels are you taking and have you applied at Oxford or Cambridge??

Don't think that Warwick is filled with silly people like me. I reckon i'm about average. Some people are (miles) better, some people aren''t so good. I think you'll enjoy it here.

**Sean12345** · October 7th 2008, 11:15 AM

Originally Posted by Showcase_22

My reasoning for this was that $\frac{y}{x}>1$ . For example, let's say $\frac{y}{x}=1.2$ . Therefore $\frac{x^k}{y^k}<1.2$ so this does not imply $\frac{x}{y}<1$ .

Look at it this way........ say $\frac{y}{x}=1.2 \implies \frac{y^k}{x^k}=1.2^k \implies \frac{x^k}{y^k}=\frac{1}{1.2^k}<1$ and therefore $\frac{x^k}{y^k}<1 \implies \frac{x}{y}<1$ .

Does this help?

Originally Posted by Showcase_22

Seriously, you seem pretty good at maths. What A -levels are you taking and have you applied at Oxford or Cambridge??

Thank you....

I am taking further maths, economics and physics and yes I have applied to Oxford. However Warwick is a university that is up there right at the top and isn't off the Oxbridge par by much.

**Showcase_22** · October 7th 2008, 01:33 PM

hmm, I see that.....

That's one question I got wrong lol =S

Thank you....

I am taking further maths, economics and physics and yes I have applied to Oxford. However Warwick is a university that is up there right at the top and isn't off the Oxbridge par by much.

Are you taking STEP and AEA? Some people here have taken both and have S's and merits in the two.

If you want an indication of what the Warwick maths homework is like, look for some of my other posts. Most of the stuff I post on here are homework questions I get stuck on.

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