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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,458

Hi

Cany anyone of you, using mathematical induction or otherwise, give a flawless proof of (a^n-b^n) always being divisble by (a-b), if the statement is true?

I shall be glad even if someone gives a counter-proof.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Are there any restrictions on what a, b and n can be?

Presumably a ≠b, but do they all have to be integers as well, or all rationals, or something?

Why did the vector cross the road?

It wanted to be normal.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,458

The only condition is a,b belong to Natural numbers.

Thanks mathsyperson.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**espeon****Real Member**- Registered: 2006-02-05
- Posts: 2,586

hmm if a=2 and b=3 and n=2 then 4-9=-5 which isnt divisible by a-b=-1 and aaaaaaaaaaaaaaaaaaaaaa im getting confused! *Gets dizzy* it would of worked if i knew if -5 was divisible by -1 lol

Presenting the Prinny dance.

Take this dood! Huh doood!!! HUH DOOOOD!?!? DOOD HUH!!!!!! DOOOOOOOOOOOOOOOOOOOOOOOOOD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Should be a simple proof by induction.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Yes, but that doesn't need to be an induction.

will do

**X'(y-Xβ)=0**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

for instance,

a(aa+ab+bb)-b(aa+ab+bb)= (a-b)(aa+ab+bb)= a^3-b^3

**X'(y-Xβ)=0**

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,458

Thanks George and Ricky, I shall study the proof you have given.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

for instance,

a(aa+ab+bb)-b(aa+ab+bb)= (a-b)(aa+ab+bb)= a^3-b^3

You know better than anyone George that an instance does not make a proof.

I'm not sure how you would algebraically solve your formula, would you mind showing me the steps?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,458

I am looking forward to a clear proof by mathematical induction, a flawless elegant proof. Ricky, George and mathsyperon, help me. Help from any other source is most welcome.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Ricky wrote:

for instance,

a(aa+ab+bb)-b(aa+ab+bb)= (a-b)(aa+ab+bb)= a^3-b^3You know better than anyone George that an instance does not make a proof.

I'm not sure how you would algebraically solve your formula, would you mind showing me the steps?

Sure, I shall illustrate Post #6 in detail.

Typically, we wanna know if a[sup]n[/sup]-b[sup]n[/sup] could be expressed as (a-b)A, where A is some polynomial.

A could be find out-

Hence

and

Using notation, the proof would be:

Hence

t and k are indexes representing integars, so we can equate them when we do the following algebra.

**X'(y-Xβ)=0**

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