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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 appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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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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The only condition is a,b belong to Natural numbers.
Thanks mathsyperson.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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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
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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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Yes, but that doesn't need to be an induction.
will do
X'(y-Xβ)=0
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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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Thanks George and Ricky, I shall study the proof you have given.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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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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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.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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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
Using notation, the proof would be:
Hence
X'(y-Xβ)=0
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