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

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Homework Help Overview

The discussion revolves around proving that if \( x^n = y^n \) for an odd integer \( n \), then it follows that \( x = y \). The participants are exploring the implications of this equation within the context of real numbers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to conclude the proof after establishing the equation \( x^{2k+1} = y^{2k+1} \). Some participants suggest breaking the problem into cases based on the relationship between \( x \) and \( y \), while others propose using induction or exploring the contrapositive of the statement.

Discussion Status

The discussion is active, with various approaches being suggested, including case analysis, induction, and the use of contrapositives. Participants are engaging with different methods to tackle the proof, indicating a productive exploration of the topic.

Contextual Notes

There is an emphasis on the necessity of assuming \( x \) and \( y \) are real numbers for the proof to hold. Additionally, some participants mention the relevance of properties introduced in related mathematical texts.

Antonio94
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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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