Proving x=y for odd n in the equation x^n=y^n | Help & Explanation

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For the equation x^n = y^n with odd n, it can be concluded that x = y under the condition that x and y are real numbers. The discussion suggests breaking the problem into cases based on the values of x and y, such as x < y or |x| < |y|. An alternative proof approach involves using the contrapositive of the statement, which maintains the truth value of the original assertion. Additionally, dividing both sides by y and letting t = x/y leads to the conclusion that t^n = 1 has only one real solution, t = 1. This confirms that x must equal y when n is odd.
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If x^n=y^n , n= odd
Then x=y

I know that an odd is n=2k+1

So x^(2k+1)=y^(2k+1)

I don't know how to finish it. Please help.
 
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You are necessarily dealing with reals for this result to be true, so you can break it into cases according to x < y, |x| < |y| etc. Try induction.
 
If this is from Spivak's Calculus, there is an alternative approach that avoids induction (as that property is first introduced in the next chapter). In particular, I used the fact that the contrapositive of a conditional statement has the same truth value as the original statement. Therefore, if you can prove that the contrapositive of the statement is true, you have proven the truth of the original statement. The contrapositive of this statement is particularly simple, as long as you write it properly.
 
You have too assume that x, y are reals first...

divide y both side, and let \frac{x^n}{y^n}=t deduct to proof t^n=1 has only one real solution which is one. which is equivalent to (t-1)\left(\sum_{i=0}^{n}t^i\right)=0 which is relative to the root of unity.
 
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