Subgroup Proof: H In R* With Rational Squares

Hey everyone! Today, we're diving into an interesting problem in abstract algebra, specifically dealing with subgroups of the multiplicative group of non-zero real numbers. We're going to prove that a particular set H is indeed a subgroup and then explore if we can generalize a certain condition. So, let's get started!

Defining the Group and the Subset

First, let's lay down the groundwork. We're given ${ R^* }$, which represents the set of all non-zero real numbers. This set forms a group under the usual multiplication operation. Remember, for a set to be a group, it needs to satisfy four key properties: closure, associativity, the existence of an identity element, and the existence of inverses.

Now, we have a subset H of ${ R^* }$. This is where things get interesting. H is defined as the set of all elements x in ${ R^* }$ such that ${ x^2 }$ is a rational number. In mathematical notation:

${ H = \{ x \in R^* \mid x^2 \text{ is rational} \} }$

Our mission, should we choose to accept it (and we do!), is to prove that H is a subgroup of ${ R^* }$. But what exactly does it mean to be a subgroup? Well, a subgroup is a subset of a group that itself forms a group under the same operation. To prove that H is a subgroup, we need to show that it satisfies the subgroup criterion.

The Subgroup Criterion

The subgroup criterion is a handy shortcut. Instead of checking all four group axioms, we only need to verify three conditions:

1. Closure: If a and b are in H, then their product ${ ab }$ must also be in H.
2. Identity: The identity element of the parent group (in this case, ${ R^* }$) must be in H.
3. Inverses: If a is in H, then its inverse ${ a^{-1} }$ must also be in H.

Let's break down each of these conditions in the context of our problem.

Proving Closure in H

Closure is the first hurdle. We need to show that if we take any two elements from H and multiply them, the result is still an element of H. In other words, if ${ x }$ and ${ y }$ are in H, then ${ xy }$ must also be in H. Remember, an element is in H if its square is a rational number.

So, let's assume ${ x }$ and ${ y }$ are in H. This means that ${ x^2 }$ and ${ y^2 }$ are both rational numbers. Now, consider the square of their product, ${ (xy)^2 }$. Using the properties of exponents, we can rewrite this as:

${ (xy)^2 = x^2y^2 }$

Since ${ x^2 }$ and ${ y^2 }$ are rational, their product, ${ x^2y^2 }$, is also rational (because the product of two rational numbers is always rational). Therefore, ${ (xy)^2 }$ is rational. This tells us that ${ xy }$ satisfies the condition to be in H, and we've successfully proven closure!

Identity Element in H

Next up is the identity element. The identity element in ${ R^* }$ (under multiplication) is 1. We need to show that 1 is also an element of H. This is pretty straightforward. Let's square 1:

${ 1^2 = 1 }$

Since 1 is a rational number, 1 satisfies the condition to be in H. So, the identity element is indeed in H. Halfway there, guys!

Proving Inverses in H

Finally, we need to tackle inverses. For every element x in H, we need to show that its inverse, ${ x^{-1} }$ (which is just ${ 1/x }$), is also in H. So, let's assume x is in H. This means, once again, that ${ x^2 }$ is a rational number.

Now, let's consider the square of the inverse, ${ (x^{-1})^2 }$. We can rewrite this as:

${ (x^{-1})^2 = (1/x)^2 = 1/x^2 }$

Since ${ x^2 }$ is rational, its reciprocal, ${ 1/x^2 }$, is also rational (as long as ${ x^2 }$ is not zero, which it isn't since x is a non-zero real number). Therefore, ${ (x^{-1})^2 }$ is rational. This means that ${ x^{-1} }$ satisfies the condition to be in H. We've shown that inverses exist within H.

Conclusion: H is a Subgroup

We've successfully checked all three conditions of the subgroup criterion: closure, identity, and inverses. Therefore, we can confidently conclude that H is indeed a subgroup of ${ R^* }$. Awesome job, team!

Generalizing the Exponent: Can We Replace '2' with Any Positive Integer?

Now, for the second part of our adventure: Can we replace the exponent '2' in the definition of H with any positive integer and still have a subgroup? This is a fantastic question that pushes us to think more deeply about the properties we've just explored.

Let's consider a more general set, let's call it ${ H_n }$, defined as:

${ H_n = \{ x \in R^* \mid x^n \text{ is rational} \} }$

where n is a positive integer. We need to determine if ${ H_n }$ is a subgroup of ${ R^* }$ for all positive integers n.

Case n = 1: A Trivial Subgroup

Let's start with the simplest case: n = 1. In this case, ${ H_1 }$ is the set of all non-zero real numbers x such that ${ x^1 = x }$ is rational. This is simply the set of all non-zero rational numbers, which we know forms a subgroup of ${ R^* }$. So, for n = 1, ${ H_n }$ is a subgroup.

Case n > 1: The Challenge of Closure

Now, let's consider the case where n is greater than 1. The identity and inverse properties are fairly straightforward to verify, similar to what we did for the case n = 2. The real challenge lies in the closure property. We need to determine if, for any two elements x and y in ${ H_n }$, their product ${ xy }$ is also in ${ H_n }$.

In other words, if ${ x^n }$ and ${ y^n }$ are rational, is ${ (xy)^n = x^ny^n }$ necessarily rational? The answer, as it turns out, depends on the value of n.

When n is Even

If n is an even integer, the same logic we used for n = 2 applies. If ${ x^n }$ and ${ y^n }$ are rational, then their product ${ x^ny^n }$ is also rational. So, closure holds, and ${ H_n }$ is a subgroup when n is even.

When n is Odd

However, things get trickier when n is an odd integer. While the product of two rational numbers is always rational, we need to be careful about the roots involved. Let's consider a counterexample to illustrate this.

Suppose n = 3. Let's take ${ x = \sqrt[3]{2} }$ and ${ y = \sqrt[3]{4} }$. Notice that:

${ x^3 = (\sqrt[3]{2})^3 = 2 }$

${ y^3 = (\sqrt[3]{4})^3 = 4 }$

Both ${ x^3 }$ and ${ y^3 }$ are rational numbers, so x and y are elements of ${ H_3 }$. However, let's look at their product:

${ xy = \sqrt[3]{2} \cdot \sqrt[3]{4} = \sqrt[3]{8} = 2 }$

And now, let's cube the product:

${ (xy)^3 = 2^3 = 8 }$

It seems like the product is still in H3. Let's consider another example: Let's take x = 2^(1/3) and y = 2^(4/3). Notice that:

${ x^3 = (2^(1/3))^3 = 2 }$

${ y^3 = (2^(4/3))^3 = 16 }$

Both ${ x^3 }$ and ${ y^3 }$ are rational numbers, so x and y are elements of ${ H_3 }$. However, let's look at their product:

${ xy = 2^(1/3) \cdot 2^(4/3) = 2^(5/3) }$

${ (xy)^3 = (2^(5/3))^3 = 32 }$

It seems like the product is still in H3. Let's try to come up with a counterexample.

Let's take ${ x = \sqrt[3]{2} }$ and ${ y = -\sqrt[3]{4} }$. Notice that:

${ x^3 = (\sqrt[3]{2})^3 = 2 }$

${ y^3 = (-\sqrt[3]{4})^3 = -4 }$

Both ${ x^3 }$ and ${ y^3 }$ are rational numbers, so x and y are elements of ${ H_3 }$. However, let's look at their product:

${ xy = \sqrt[3]{2} \cdot (-\sqrt[3]{4}) = -\sqrt[3]{8} = -2 }$

Let's try a different approach, suppose ${ xy = \sqrt[3]{2} \cdot \sqrt[5]{2} }$, then ${ x^3 = 2 }$ and ${ y^5 = 2 }$, however, ${ (xy)^{15} = 2^8 }$, and ${ (xy)^n }$ is not rational. However, this doesn't prove that ${ H_n }$ is not a subgroup when n is odd.

Let's consider ${ x = 2^{1/3} }$ and ${ y = 3^{1/3} }$, then ${ x^3 = 2 }$ and ${ y^3 = 3 }$, thus x and y are in ${ H_3 }$. However, ${ xy = 2^{1/3}3^{1/3} }$, thus ${ (xy)^3 = 6 }$, so it seems like xy is in ${ H_3 }$.

This approach is not bearing fruit, we need to find a general proof whether ${ H_n }$ is a subgroup.

Let's check closure. If ${ x, y \in H_n }$, then ${ x^n }$ and ${ y^n }$ are rational. ${ (xy)^n = x^ny^n }$, since both are rational, the product is rational, so it is closed.

Identity: 1 is still the identity, and ${ 1^n = 1 }$, so the identity is still in ${ H_n }$.

Inverses: If x is in ${ H_n }$, then ${ x^n }$ is rational. We need to check if ${ x^{-1} }$ is in ${ H_n }$, thus, we need to check if ${ (x^{-1})^n }$ is rational. ${ (x^{-1})^n = (x^n)^{-1} }$, since ${ x^n }$ is rational, ${ (x^n)^{-1} }$ is also rational. Thus, inverses exist.

Thus, ${ H_n }$ is a subgroup for all positive integers n.

Final Answer

So, the answer to our question is: Yes, the exponent '2' can be replaced by any positive integer, and H will still be a subgroup of ${ R^* }$. We've shown that the subgroup criterion holds for all positive integers n. High five!

Wrapping Up

We've taken a deep dive into the world of subgroups and explored how changing a seemingly small condition can lead to interesting generalizations. We've proven that the set of real numbers whose nth power is rational forms a subgroup of ${ R^* }$ for any positive integer n. This is a testament to the beauty and power of abstract algebra! Keep exploring, guys, and never stop questioning!