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November 17th, 2018, 11:52 AM   #1
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Prove elementary rules in a field

Hey.

Given the field F, Prove:
* For any a, b in F, if a*a=b*b, then a=b or a=-b
* For any a, b in F, if a*a*a = b*b*b, then a=b

Given the field F where 1+1=0, Prove:
* For any a in F, -a=a
* For any a, b in F, a+b=a-b

The proofs have to be supported by the axioms of fields rigorously.

I don't know actually how even start proving it.
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November 17th, 2018, 01:46 PM   #2
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What have you tried on the first one, if if a*a=b*b, then a=b or a=-b? Can you see any obvious algebraic manipulations that will solve the problem?
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November 17th, 2018, 01:48 PM   #3
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If you can't make a good start, consider what you can prove by use of the axioms.
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November 17th, 2018, 08:10 PM   #4
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Maybe I can start by assuming a=b? And by transitivity we can get that a*a=b*b?
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November 20th, 2018, 06:32 AM   #5
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Quote:
Originally Posted by CStudent View Post
Maybe I can start by assuming a=b? And by transitivity we can get that a*a=b*b?
You're being asked to show that a*a=b*b implies something, but your argument shows that something implies a*a=b*b.

More precisely, fix elements a and b in your field F. Let A be the statement "a*a=b*b", B be the statement "a=b" and C be the statement "a=-b". You've been asked to show A $\implies$ B or C, but you've instead shown B $\implies$ A.

What you should instead do is assume A is true (that is, assume a*a=b*b), and then deduce that B or C is true (that is, that a=b or a=-b). (Or, equivalently, you could assume that the statement "B or C" is false, and then deduce that A is false. This works as P $\implies$ Q is equivalent to not(Q) $\implies$ not(P).)

____________________________________

Here are some slightly less involved "prove from the axioms" questions, to help you get used to this style of problem. These results will also be useful in answering the first question you've posted.

(1) A field axiom tells you that F has an element 0 (an "additive identity") such that a + 0 = a for all a in F. Prove that 0 is the only additive identity in F. In other words, show: if x is an element of F such that a + x = a for all a in F, then x = 0.

Similarly, an axiom says F has a "multiplicative identity" 1. (That is, a*1 = a for all a in F). Prove 1 is the only multiplicative identity of F.

(2) Another field axiom says, for any c in F, there exists an element d in F (an "additive inverse" of c) such that c + d = 0. Prove that d is the only additive inverse of c in F. In other words, show: if e is an element of F such that c + e = 0, then e = d.

Due to this exercise, we can talk of "the" additive inverse of c, rather than just "an" additive inverse of c. From now on, we will denote the additive inverse of c by -c.

(3) Prove that x*0 = 0 for all x in F.

(4) Suppose x and y are elements of F. Prove that, if x*y = 0, then x = 0 or y = 0.

(5) Suppose x and y are elements of F. Prove that -(x*y) = (-x)*y = x*(-y) = (-1)*x*y.

_______________________________

For what it's worth, the second implication you've posted (that a*a*a = b*b*b implies a=b) isn't true in general. For example, take F to be the field of complex numbers. Then if $a = e^{\frac{2\pi i}{3}}$ and $b = 1$, we have $a * a * a = e^{2\pi i} = 1 = b * b * b$, but $a \neq b$.

In a general field F, we only have the implication: a*a*a = b*b*b implies a = w*b, where w is a cuberoot of unity in F. (When I say w is a cuberoot of unity, I mean w*w*w = 1.)
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