# Determine if integral domain or field

1. Aug 5, 2011

### Shackleford

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

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

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110805_224012.jpg?t=1312602392 [Broken]

Last edited by a moderator: May 5, 2017
2. Aug 5, 2011

### Dick

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.

3. Aug 5, 2011

### Shackleford

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)

4. Aug 5, 2011

### Dick

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.

5. Aug 5, 2011

### Shackleford

(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.

6. Aug 5, 2011

### Dick

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?

Last edited: Aug 5, 2011
7. Aug 5, 2011

### Shackleford

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

k = [a2 - 2b2]-1

8. Aug 6, 2011

### Dick

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.

9. Aug 6, 2011

### Shackleford

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

a2 - 2b2 = 0

a = sqrt(2)b

Last edited: Aug 6, 2011
10. Aug 6, 2011

### HallsofIvy

$\frac{1}{a+ b\sqrt{2}}$

Rationalise the denominator.

11. Aug 6, 2011

### Shackleford

You can't?

12. Aug 6, 2011

### SammyS

Staff Emeritus
I'm sure he can !!!

But the point is that you do it.

13. Aug 6, 2011

### Shackleford

No. I'm saying you can't rationalize the denominator because there will always be root two down there. Is that right?

14. Aug 6, 2011

### Dick

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. Aug 7, 2011

### Shackleford

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

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

16. Aug 7, 2011

### HallsofIvy

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. Aug 7, 2011

### Hurkyl

Staff Emeritus
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. Aug 7, 2011

### Dick

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. Aug 7, 2011

### Shackleford

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. Aug 7, 2011

### Shackleford

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].

Last edited: Aug 7, 2011
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