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dchilow

Member

Registered: 2007-03-05

Posts: 27

Need help with abstract algebra!

Let G = Z x Z, be the group consisting of all ordered pairs with entries in Z.
Let H be the set of all elements of R of the form a + b√2 where a and b are integers.
R denotes the real numbers and Z denotes the integers and both R and Z are groups under addition.

Assume a + b√2 and c + d√2 are elements of H.  Prove that if a + b√2 = c +d√2, then a = c and
b = d.  To do this, you may use that √2 is not a rational number without proving it.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Need help with abstract algebra!

Writing

you immediately see that the RHS is a rational number since the LHS is a rational number. If b ≠ d, then 1⁄(d−b) would be a rational number. And then

would be a product of two rational numbers and so would be itself rational. This is contradiction. Hence we must have b = d, whence a = c.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Need help with abstract algebra!

Alternate solution:

if a + b√2 = c +d√2, then (a - c) + (b-d)√2 = 0.  Note that an integer plus an irrational is irrational and that a non-zero integer times an irrational is also irrational (proofs are trivial).  Thus, if b - d is not 0, then (a - c) + (b-d)√2 is irrational.  This is a contradiction, and the conclusion follows.

I like this version of the proof because it sort of describes {1, √2} as the basis of the two dimensional Z-module Z[√2].  If we replace a-c with lambda_1 and b-d with lambda_2, then this proof is the equivalent of proving that 1 and √2 are linearly independent.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Need help with abstract algebra!

if a + b√2 = c +d√2, then (a - c) + (b-d)√2 = 0.  Note that an integer plus an irrational is irrational and that a non-zero integer times an irrational is also irrational (proofs are trivial).  Thus, if b - d is not 0, then (a - c) + (b-d)√2 is irrational.  This is a contradiction, and the conclusion follows.

That’s exactly what my proof is, only stated in different words. It is not a different proof.

Last edited by JaneFairfax (2007-12-03 12:17:31)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Need help with abstract algebra!

Sure it is.  First off, the way I'm stating it, I'm emphasizing that linear independence aspect.  Your version hides it.  Second, I used more general facts about irrational numbers.  Rather than showing that square root of 2 is not the product of two rationals, I showed that 0 can not equal an irrational.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."