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

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CStudent

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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.
Thanks.
 
CStudent said:
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.
Thanks.
Click to expand...

A common trick for this sort of thing is to bring in a*b, which can act as a bridge between a*a and b*b. For example, try adding that to each side of the first equation.

For the second problem, I suppose you have noticed that 1 = -1 in this field, which is a special case of the goal. Can you use that?

When you don't know how to start, just do something; even if your first step doesn't take you closer to the goal, it will take you to a different place from which you might see the goal better.

Then show us what you find, even if it seems like nothing. Once you're moving, we can steer you better!
 
Dr.Peterson said:
A common trick for this sort of thing is to bring in a*b, which can act as a bridge between a*a and b*b. For example, try adding that to each side of the first equation.

For the second problem, I suppose you have noticed that 1 = -1 in this field, which is a special case of the goal. Can you use that?

When you don't know how to start, just do something; even if your first step doesn't take you closer to the goal, it will take you to a different place from which you might see the goal better.

Then show us what you find, even if it seems like nothing. Once you're moving, we can steer you better!
Click to expand...
Great explanation! Thanks.

So the first one is solved by distributivity and commutativity - we get that if a!=-b then a=b and if the opposite is true then a=-b

For the second one, given that 1=-1 we multiply this equation by a and get a=-a
 
CStudent said:
Great explanation! Thanks.

So the first one is solved by distributivity and commutativity - we get that if a!=-b then a=b and if the opposite is true then a=-b

For the second one, given that 1=-1 we multiply this equation by a and get a=-a
Click to expand...

Good; I'll trust that you have filled in the details using only field axioms or theorems you have proved.

What you say about the first is essentially what I first saw; you can simplify the logic somewhat by putting everything on one side and factoring further: (aa + ab) - (ab + ab) = 0...

Have you worked out the second half of each problem?
 
Dr.Peterson said:
Good; I'll trust that you have filled in the details using only field axioms or theorems you have proved.

What you say about the first is essentially what I first saw; you can simplify the logic somewhat by putting everything on one side and factoring further: (aa + ab) - (ab + ab) = 0...

Have you worked out the second half of each problem?
Click to expand...
The properties are just as I mentioned no? Commutativity and distributivity.

Thanks!
 
CStudent said:
The properties are just as I mentioned no? Commutativity and distributivity.

Thanks!
Click to expand...

Those are some of the field axioms. The question is, have you written out a proof that explicitly justifies every step? I can't say you are correct without seeing those details (not that I have to).
 
CStudent said:
I think that I have done it, yes.
Thank you!
Click to expand...
Excellent!! Now can you show us your work? If it is correct, then others can learn from it. If it is wrong/incomplete we can help you finish the proof.
 
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