Proving H is a Subgroup of R^*

[MikeDietrich]

Homework Statement

Let R^* be a group of all nonzero real numbers under multiplication and let H = {x in R^*: sqrt (x) is rational. Prove or disprove H is a subgroup of R^*.

Homework Equations

all axioms must be satisfied

The Attempt at a Solution

associative: satisfied. Both R^* and H have same operation so if R^* is assoc. so is H
Identity: satisfied. sqrt(x) (1) = (1) sqrt(x) = sqrt(x)
Inverse: satisfied. (1/sqrt(x))(sqrt(x)/1) = e = 1
Closed: satisfied. the product of two rationals is a rational (therefore, in H)

I would say that H is a subgroup of R^*. Did I do anything wrong or am I missing something?

 

[micromass]

I think it is correct...

 

[fzero]

Homework Statement

Let R^* be a group of all nonzero real numbers under multiplication and let H = {x in H: sqrt (x) is rational.

I believe the set H should be H = {x in G: sqrt (x) is rational}.

The Attempt at a Solution

associative: satisfied. Both R^* and H have same operation so if R^* is assoc. so is H
Identity: satisfied. sqrt(x) (1) = (1) sqrt(x) = sqrt(x)
Inverse: satisfied. (1/sqrt(x))(sqrt(x)/1) = e = 1
Closed: satisfied. the product of two rationals is a rational (therefore, in H)

You have all of the right steps, but you seem to have misinterpreted the nature of the elements x. They are x such that $$\sqrt{x}$$ is rational, so you need to check that $$\sqrt{x^{-1}}$$ is rational and that $$\sqrt{xy}$$ is rational if $$x,y\in H$$.

 

[MikeDietrich]

Edit: H ={x in R^*:sqrt(x) is rational}
Thanks fzero!

 

[MikeDietrich]

fzero, I guess you are correct. Based on what you are saying I am misinterpreting the set. Is it not restricting x to real numbers (excluding zero) and that the sqrt(x) must be rational? Or, is it saying if I can show sqrt(x) is not rational then the set is not closed?

 

[JonF]

note your set is of x, not square root x. You're proofs are for square root x.

For example the product of two rationals being rational is irrelevant. You need to show if square root x and y are rational, (xy)^1/2 is rational.

 

[fzero]

Just to restate things, the set H is a set of all real numbers (excluding 0) that satisfy a certain condition. Closure requires that the products of these elements satisfy the same condition. Additionally the inverse of any of these elements must also satisfy the condition. Don't let the particular mathematical form of the condition obscure the idea involved.

 

[MikeDietrich]

I liked math so much better when numbers were involved :P

OK... I'm not the sharpest knife in the drawer so please bear with me. The axioms:

1) assoc - satisfied (same as above)
2) ident - satisfied (same as above)
3) inverse - satisfied (sqrt (x))(sqrt (x)^-1) = 1
4) closure - satisfied (sqrt (x))(sqrt (y)) = sqrt (xy) are in the set since sqrt (x) and sqrt (y) must be perfect squares in order for them to be rational and when multiplied the product is a rational number.

I get the impression that the set is not closed (based on the responses) but I am just not seeing it. Argh!

 

[fzero]

You have only repeated the same things that you checked in your first post. The point is not that those statements are wrong, it's that they aren't the logical things that you need to check to verify if H is a subgroup of $$\mathbb{R}^*$$.

Let's look at closure in detail. We take $$x,y\in H$$. By definition, $$\sqrt{x}$$ is rational, as is $$\sqrt{y}$$. It is not necessarily true that $$\sqrt{x}$$ or $$\sqrt{y}$$ are in $$H$$. For instance $$4\in H$$, since $$\sqrt{4} =2$$ is rational, but $$2\notin H$$, because $$\sqrt{2}$$ is irrational.

In order for $$xy$$ to be in $$H$$, $$\sqrt{xy}$$ must be rational. As above, it is not necessary that $$\sqrt{xy}\in H$$, in general it won't.