induction problem

[rocketman421]

Are you smart enough to prove that for every natural number n,

 

(x[math]^{n}[/math] - y[math]^{n}[/math]) is divisible by (x - y), where x and y are two nonidentical real numbers?

 

I already solved its just it took me forever to get lol. If nobody solves it by the next time I visit, I'll post the solution.

[the tree]

I'm assuming that what's meant is to prove that

 

(xⁿ-yⁿ) is divisible by (x-y)

 

Seriously, preview your posts or at least them before your edit time is out.

 

Although in the real realm, the statement is trivial - so I think you're leaving something out as well.

[rocketman421]

well i solved it before but i lost my solution so here i coiped it from some site:

 

Prove that for every natural number n, (x - y) divides (x^n - y^n).

 

Our first step in mathematical induction is the base case; that is, for n = 1.

 

1) Base Case: Let n = 1. Then

(x^n - y^2) = (x - y), and (x - y) is obviously divisible by (x - y).

Therefore, the formula holds true for n = 1.

 

2) Induction Hypothesis: Assume the formula holds true for n = k. That is, assume that

 

(x - y) divides (x^k - y^k).

 

( We want to prove that (x - y) divides x^(k + 1) - y^(k + 1) )

 

But what does x^(k + 1) - y^(k + 1) equal?

 

x^(k + 1) - y^(k + 1)

 

I'm going to re-express these two terms.

 

(x^1)(x^k) - (y^1)(y^k)

x*(x^k) - y*(y^k)

 

I'm going to use a little trick, by "adding zero" in the middle.

 

x*(x^k) + 0 - y*(y^k)

 

I'm going to subtract x*(y^k) and add x*(y^k). After all, subtracting and then adding the same term is the same as adding zero.

 

x*(x^k) - x*(y^k) + x*(y^k) - y*(y^k)

 

Now I'm going to factor the first two terms and the last two terms.

 

x(x^k - y^k) + (y^k)(x - y)

 

Look closely at this;

(y^k)(x - y) is obviously divisible by (x - y).

By our induction hypothesis, we assumed that (x^k - y^k) is divisible by (x - y). Therefore, x(x^k - y^k) is divisible by (x - y).

 

The sum of two terms both divisible by (x - y) is also divisible by (x - y). Therefore, what we started with,

 

x^(k + 1) - y^(k + 1) is divisible by (x - y)

 

Therefore, by the principle of mathematical induction,

x^n - y^n is divisible by (x - y) for all natural numbers n.

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Merged post follows:

Consecutive posts merged

Thread Closed.

Edited July 20, 2009 by rocketman421
Thread closed due to uninterstingness.

[insane_alien]

the mods decide when a thread gets closed

[the tree]

Okay, you just left out that we were talking about real polynomials - leaving information out makes for fairly useless 'proofs'.

 

Although it should go without saying, imitating mod-voices and making decisions on their behalf is not cool.

[rocketman421]

tree go take another look at the problem:

 

"for every natural number n"

"where x and y are two nonidentical real numbers"

Edited July 21, 2009 by rocketman421

[the tree]

Yeah, you still need to specify that you're talking about polynomials, or that x and y are variables. Otherwise you're just stating that some real number is divisible by some other non-zero real number which goes without saying.

[Malteaser]

All

 

Given this statement, I've been wondering for some time now whether you can prove that (z-x)^2 is definitely not a divisor of z^n - x^n.

 

I think I can prove it iff (if an only if) (z-x) is divisible by z^(n-1). But I'm not sure I can prove it for all z. Can anyone help?

 

The proof for n=3 would be something like:

 

x^3 = z^3 -3(z-x)x^2 -3(z-x)x - (z-x)^3.

 

(you could prove this by drawing a cube of x^3 and working out which cuboids are needed to get z^3)

 

Which means that

 

z^3 - x^3 = 3(z-x)x^2 +3 (z-x)x + (z-x)^3.

 

Evidently this is divisible a (z-x), but is only divisble by (z-x)^2 if x^2 is divisble by (z-x).

 

Any help/pointers would be great.

 

Thanks.

 

Malteaser

 

As an aside, what does non-incidental mean please? I'm guessing it means mutually exclusive?