Algebra Problem with Rationals and Proofs

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    Algebra Proofs
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Homework Help Overview

The discussion revolves around proving properties related to rational numbers and expressions involving square roots, specifically focusing on equations of the form a + b√2 = c + d√2 and the non-zero condition of a² - 2b².

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore whether the differences (a-c) and (b-d) can be rational or irrational, and how this affects the proof. There are attempts to factor the second part of the problem and relate it to the first part. Questions arise about the implications of assuming certain values are non-zero and how that leads to contradictions.

Discussion Status

Participants are actively engaging with the problem, offering insights and clarifications. Some express newfound understanding and confidence in approaching the proofs, while others encourage continued exploration of the second part of the problem.

Contextual Notes

There is an emphasis on the rationality of certain variables and the implications of their relationships, with participants questioning the setup and assumptions of the problem.

silvermane
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Hello fellow forum buddies :)

Homework Statement


a.) Prove that if a+b[tex]\sqrt{2}[/tex] = c+d[tex]\sqrt{2}[/tex] with a,b,c,d all in Q, then
a = c and b = d.
b.) Prove that a^2 - 2b^2 with a, b in Q is nonzero unless a=b=0

The Attempt at a Solution


I really don't know where to start. Any tips would be nice. I just need to see something that I'm not seeing at the moment. :(
I don't need an answer, just a little jolt lol. It will be greatly appreciated! :)
 
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For part a, is (a-c) in Q? Is (b-d)[tex]\sqrt{2}[/tex]?

For part b, factor the equation and try to use the result from part a.
 
I guess he ^ already took care of you ;-).

Post again if you need more clearification.
 
fzero said:
For part a, is (a-c) in Q? Is (b-d)[tex]\sqrt{2}[/tex]?

For part b, factor the equation and try to use the result from part a.

AHHH... wow I'm glad I see it now. So I just need to solve a and c with respect to the fact that they are rational, then b and d with the fact that they're not rational?
 
╔(σ_σ)╝ said:
I guess he ^ already took care of you ;-).

Post again if you need more clearification.

Awe phanku :)
 
silvermane said:
AHHH... wow I'm glad I see it now. So I just need to solve a and c with respect to the fact that they are rational, then b and d with the fact that they're not rational?

b and d are rational, however, the product with sqrt(2) is not. I believe this is what you meant.

I believe you can see that if b-d was non-zero a contradiction ensues. :-)

EDIT

lmao.
U r velcome :-).
 
silvermane said:
AHHH... wow I'm glad I see it now. So I just need to solve a and c with respect to the fact that they are rational, then b and d with the fact that they're not rational?

Well if x = a - c is nonzero, can we find y such that x = y [tex]\sqrt{2}[/tex]? Is such a y compatible with y = b-d?
 
Ahhhh that should do it lol. I wrote that down and when I saw each of your replies, it was confirmed. I think I can finish it from here. Thanks so much! :)
 
Okay. Post if you get stuck in part b or something.
 

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