Determine if integral domain or field

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Homework Help Overview

The discussion revolves around determining whether a given set forms an integral domain or a field, focusing on the properties of real numbers and expressions involving square roots. Participants explore the definitions and criteria for both algebraic structures.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the properties of integral domains and fields, questioning the existence of zero divisors and multiplicative inverses. There are attempts to manipulate expressions involving square roots to find inverses and clarify definitions.

Discussion Status

The conversation is active, with participants providing hints and exploring various approaches to the problem. Some participants express uncertainty about the validity of their reasoning or the definitions involved, while others suggest methods for rationalizing expressions and finding inverses.

Contextual Notes

There is a focus on the implications of working within the rationals and the challenges posed by the presence of irrational numbers, particularly in relation to the definitions of fields and integral domains.

Shackleford
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I found the unity to be 1. I'm not sure how to determine 3) for each.

Integral Domain:

1) D has to be commutative ring (under multiplication)
2) D has to have a unity not equal to zero (also under multiplication)
3) D has to have no zero divisors (ab = 0)

Field:

1) F has to be commutative ring (under multiplication)
2) F has to have a unity not equal to zero (also under multiplication)
3) Each nonzero element in F has to have a multiplicative inverse

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110805_224110.jpg?t=1312602377

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110805_224012.jpg?t=1312602392
 
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The real numbers have no zero divisors, so it's a pretty safe bet that both are at least integral domains. To determine if there is an inverse, don't start from scratch. Multiply (a+sqrt(2)*b)*(a-sqrt(2)*b) and try to figure out where to go from there.
 
Dick said:
The real numbers have no zero divisors, so it's a pretty safe bet that both are at least integral domains. To determine if there is an inverse, don't start from scratch. Multiply (a+sqrt(2)*b)*(a-sqrt(2)*b) and try to figure out where to go from there.

Ah! Duh. I was focusing too much on the rationals a,b and forgetting that the term represents a real number.

So, there's no way to determine if there's an inverse if you start from scratch?

Why did you choose this? Just a good guess?

(a+sqrt(2)*b)*(a-sqrt(2)*b)
 
Shackleford said:
Ah! Duh. I was focusing too much on the rationals a,b and forgetting that the term represents a real number.

So, there's no way to determine if there's an inverse if you start from scratch?

Why did you choose this? Just a good guess?

(a+sqrt(2)*b)*(a-sqrt(2)*b)

Yes, good guess. It gets rid of the sqrt(2). Which you have to do if you are finally going to get just 1. Now go from there.
 
Dick said:
Yes, good guess. It gets rid of the sqrt(2). Which you have to do if you are finally going to get just 1. Now go from there.

(a+sqrt(2)*b)*(a-sqrt(2)*b) = 1

a2 - 2b2 = 1

a2 = 1 + 2b2

If b = 1, then a2 = 3

Then, a = sqrt(3), which is not a rational number. This is not a Field.
 
Shackleford said:
(a+sqrt(2)*b)*(a-sqrt(2)*b) = 1

a2 - 2b2 = 1

a2 = 1 + 2b2

If b = 1, then a2 = sqrt(3).

Then, a is not a rational number. This is not a Field.

Ok, my guess was really just a hint. a^2-2*b^2 is unlikely to be 1. Now think about (a+b*sqrt(2))*k*(a-b*sqrt(2))=1 where k is some constant in your system. Can you solve for k if you are working in the rationals?
 
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Dick said:
Ok, my guess was really just a hint. a^2-2*b^2 is unlikely to be 1. Now think about (a+b*sqrt(2))*k*(a-b*sqrt(2))=1 where k is some constant in your system. Can you solve for k if you are working in the rationals?

Is my counterexample correct? Did I show it's not a Field?

k = [a2 - 2b2]-1
 
Shackleford said:
Is my counterexample correct? Did I show it's not a Field?

k = [a2 - 2b2]-1

Well, no. Using that value of k, I'd say the inverse of a+sqrt(2) is (a-sqrt(2))/(a^2-2), isn't it? Your main worry over the rationals is whether a^2-2*b^2 could be zero.
 
Dick said:
Well, no. Using that value of k, I'd say the inverse of a+sqrt(2) is (a-sqrt(2))/(a^2-2), isn't it? Your main worry over the rationals is whether a^2-2*b^2 could be zero.

Okay, you're right. There is a valid inverse if b = 1.

a2 - 2b2 = 0

a = sqrt(2)b
 
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  • #10
\frac{1}{a+ b\sqrt{2}}

Rationalise the denominator.
 
  • #11
HallsofIvy said:
\frac{1}{a+ b\sqrt{2}}

Rationalise the denominator.

You can't?
 
  • #12
Shackleford said:
You can't?
I'm sure he can !

But the point is that you do it.
 
  • #13
SammyS said:
I'm sure he can !

But the point is that you do it.

No. I'm saying you can't rationalize the denominator because there will always be root two down there. Is that right?
 
  • #14
I thought we were getting towards being able to agree that the inverse of (a+b*sqrt(2)) is (a-b*sqrt(2))/(a^2-2*b^2)? In so far as that is defined. Is it defined over the rationals? Let's not go backwards here.
 
  • #15
Dick said:
I thought we were getting towards being able to agree that the inverse of (a+b*sqrt(2)) is (a-b*sqrt(2))/(a^2-2*b^2)? In so far as that is defined. Is it defined over the rationals? Let's not go backwards here.

Oh, sorry. I guess that's next logical conclusion.

Is the inverse defined over the rationals? Not if sqrt(2) is in there.
 
  • #16
You are still missing the whole point! (a+ b)(a- b)= a^2- b^2. I'm sure you learned that long ago. So what do you get if you multiply both numerator and denominator of
\frac{1}{a+ b\sqrt{2}}
by
a- b\sqrt{2}
 
  • #17
I don't see why people in this thread have an aversion to "starting from scratch". Linear algebra is a very useful technique in abstract algebra, and both of the questions
  • What, when multiplied by (a + b\sqrt{2}), gives 1?
  • What, when multiplied by (a + b\sqrt{2}), gives 0?
are linear systems of equations in two unknowns. (treating a and b as "knowns")

Rationalizing denominators is a good trick, but far from required for this problem.
 
  • #18
Shackleford said:
Oh, sorry. I guess that's next logical conclusion.

Is the inverse defined over the rationals? Not if sqrt(2) is in there.

What do you mean 'if sqrt(2) is in there'? Do you mean sqrt(2) is in the rationals? Can you elaborate a bit?
 
  • #19
Dick said:
What do you mean 'if sqrt(2) is in there'? Do you mean sqrt(2) is in the rationals? Can you elaborate a bit?

Rationalizing is when you get rid of the square root in the denominator? I was wrong. You can rationalize the denominator with the difference of two squares trick.

[1/sqrt(2)] = sqrt(2)/2
 
  • #20
Hurkyl said:
I don't see why people in this thread have an aversion to "starting from scratch". Linear algebra is a very useful technique in abstract algebra, and both of the questions
  • What, when multiplied by (a + b\sqrt{2}), gives 1?
  • What, when multiplied by (a + b\sqrt{2}), gives 0?
are linear systems of equations in two unknowns. (treating a and b as "knowns")

Rationalizing denominators is a good trick, but far from required for this problem.

As established earlier, nothing non-zero multiplied by a real number yields zero. Thus, that system is not possible. The inverse is [a - b*sqrt(2)] / [a2 - 2b2].
 
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  • #21
How can you be certain that a- 2b^2 is not 0 for some a and b?
 
  • #22
HallsofIvy said:
How can you be certain that a- 2b^2 is not 0 for some a and b?

I corrected the typo. It should be a2.Then a = sqrt(2)*b.
 

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