Spivak Calculus - Chapter 1 Problems

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    Calculus Spivak
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Homework Help Overview

The discussion revolves around proving properties related to the equality of powers, specifically addressing the cases when n is odd and even in the context of Spivak's Calculus. Participants are exploring the implications of the statements x^n = y^n under these conditions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of contrapositive reasoning and factorization in their proofs. There are questions about the validity of certain approaches and the clarity of the assumptions being made, particularly regarding the use of exponent rules.

Discussion Status

The conversation is ongoing, with participants providing counterexamples and suggesting alternative methods for proving the statements. There is an acknowledgment of confusion regarding the rules and assumptions that can be applied, indicating a productive exploration of the topic.

Contextual Notes

Participants express concern about their understanding of the properties of real numbers and the rigor of their proofs, particularly in relation to the constraints of the homework context.

fedlibs
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I have begun in my spare time to work through Spivak's Calculus. Although I have been stuck on many problems, I am most troubled by my general clumsiness and nonelegance of answers (particularly proofs). It always seems that there likely is a much simpler route available, and yet I forgo this route for a long-winded tedious argument. (Perhaps of my lack of proof experience)

Homework Statement



6 c) Prove that if x^{n}=y^{n} and n is odd, then x=y.
6 c) Prove that if x^{n}=y^{n} and n is even, then x=y or x=-y.

Homework Equations



12 Properties of Real Numbers

The Attempt at a Solution



For 6c, I considered using the contrapositive and claiming that:
If y≠x, then either y>x or x>y, which implies x^{n}<y^{n} or x^{n}>y^{n}, which should complete the proof? (This was proven as the first part of the problem.)

However, I tried to use a more direct proof from considering the factorization of the terms. This required a long winded explanation that required several rewritings for different cases, and an absolute value claim that I believe to be non-rigorous.

Is there a simple way to do these proofs directly? (And is the contrapositive proof I provided sound?)
 
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Here is a counter example to your first proof. Let y = 2, x = -2. Now y > x. But x^2 = y^2.How about if x^n = y^n and n odd, then divide both sides by x^n, factor the n, and show x/y = 1 which leads to x = y. Then make a special case for when x is 0.

Similarly for n even.
 
Thank you for your quick reply.

Here is a counter example to your first proof. Let y = 2, x = -2. Now y > x. But x^2 = y^2.

Sorry about my inclarity. The proof I provided was only for problem 6c) where n is odd case. I realize this will not work for the even case.


How about if x^n = y^n and n odd, then divide both sides by x^n, factor the n, and show x/y = 1 which leads to x = y. Then make a special case for when x is 0.

Similarly for n even.

Interesting, thank you for this. I have just been somewhat confused about what facts I am and am not allowed to use. Thus, I was generally hesitant with the general exponent rules. (Should I think this way?)
 
So, is there a particularly clean way to show these through the factorization?

(I am uneasy about using the exponent rules at this point of the book.)
 

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