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mxam
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I don't know how to prove by mathematical induction that x^n-y^n is divisible by x-y . . . Can you show me step by step? . . .Greetings!
[itex]((x^n-y^n)/x-y)=>\frac {x^1-y^1} {x-y}=>\frac {x-y} {x-y} OK [/itex]
[itex]x^k-y^k = m(x-y) ; [/itex] where m is a positive integer.
[itex]x^k+1+-y^k+1= ?[/itex]
mxam said:Code:[itex]((x^n-y^n)/x-y)=>\frac {x^1-y^1} {x-y}=>\frac {x-y} {x-y} OK [/itex]
Now, n= k
Code:[itex]x^k-y^k = m(x-y) ; [/itex] where m is a positive integer.
Now, my question is n=k+1 where,
Code:[itex]x^k+1+-y^k+1= ?[/itex]
I don´t understand the next step, i saw some explanations on internet but i don't understand ! Help me please. :s
[itex]x^{k+1}-y^{k+1}= ?[/itex]
So noted, and dealt with.SteamKing said:Sorry, but the rules of PF state clearly that you must make an attempt at solution.
mxam said:Code:[itex]((x^n-y^n)/x-y)=>\frac {x^1-y^1} {x-y}=>\frac {x-y} {x-y} OK [/itex]
Now, n= k
Code:[itex]x^k-y^k = m(x-y) ; [/itex] where m is a positive integer.
n=k+1
Code:[itex]x^{k+1}-y^{k+1}[/itex]
Code:[itex](x^1)(x^k)-(y^1)(y^k)[/itex]
Code:[itex]x(x^k)-x(y^k)+x(y^k)-y(y^k)[/itex]
And later the solution. But I don't understand just 1 thing, why are using [itex]x(y^k)[/itex]? . . . where the origin for it? . . .
I appreciated your time. This is a old homework of Algebra, but is exercise unresolved, and now, i want to know the real way to resolve it. . .
There are three things you need to do in an induction proof:mxam said:Thanks for your answers. I saw a "clear" solution online, let me show you:
mxam said:Now, n= kCode:[itex]((x^n-y^n)/x-y)=>\frac {x^1-y^1} {x-y}=>\frac {x-y} {x-y} OK [/itex]
n=k+1Code:[itex]x^k-y^k = m(x-y) ; [/itex] where m is a positive integer.
Code:[itex]x^{k+1}-y^{k+1}[/itex]
Code:[itex](x^1)(x^k)-(y^1)(y^k)[/itex]
And later the solution. But I don't understand just 1 thing, why are using [itex]x(y^k)[/itex]? . . . where the origin for it? . . .Code:[itex]x(x^k)-x(y^k)+x(y^k)-y(y^k)[/itex]
I appreciated your time. This is a old homework of Algebra, but is exercise unresolved, and now, i want to know the real way to resolve it. . .
It is always possible to add and subtract the same term to an expression. It is just a clever thing to do. You can arrange the expression so it is the sum of two terms, both divisible by x-y:mxam said:Thanks for your answers. I saw a "clear" solution online, let me show you:...x(xk)−x(yk)+x(yk)−y(yk) And later the solution. But I don't understand just 1 thing, why are using x(yk)? . . . where the origin for it? . . .
ehild said:It is always possible to add and subtract the same term to an expression. It is just a clever thing to do. You can arrange the expression so it is the sum of two terms, both divisible by x-y:
x(xk−yk)-yk(x-y).
ehild
To prove that x^n-y^n is divisible by x-y, we can use the factor theorem. This states that if (x-y) is a factor of a polynomial, then the polynomial evaluated at (x=y) will equal zero. In other words, (x-y) is a root of the polynomial. By substituting (x=y) into the polynomial x^n-y^n, we get 0^n - y^n, which simplifies to 0-y^n = -y^n. Since this equals zero, we can conclude that (x-y) is a factor of x^n-y^n, making it divisible by (x-y).
Yes, we can prove that x^n-y^n is always divisible by x-y. This is because the factor theorem applies to any value of n, meaning that (x-y) will always be a factor of x^n-y^n. Therefore, x^n-y^n will always be divisible by (x-y).
The relationship between x^n-y^n and x-y is that x^n-y^n is a multiple of (x-y). This means that x^n-y^n can be divided evenly by (x-y) without any remainder. This relationship is true for any value of n, as long as n is a positive integer.
The factor theorem is used to prove that x^n-y^n is divisible by x-y by showing that (x-y) is a factor of x^n-y^n. This is done by substituting (x=y) into the polynomial x^n-y^n, which results in a value of zero. This proves that (x-y) is a root of the polynomial and therefore a factor of x^n-y^n, making it divisible by (x-y).
Yes, for example, let's prove that x^5-y^5 is divisible by x-y. By using the factor theorem, we can substitute (x=y) into the polynomial x^5-y^5, which results in 0^5-y^5 = 0-y^5 = -y^5. Since this equals zero, we can conclude that (x-y) is a factor of x^5-y^5, making it divisible by (x-y).