10. Let \( p \in \mathbb{N} \) be a prime and define \( Q_p \) to be the set of all rational numbers which have a denominator that is a power of \( p \) when written in reduced form. Specifically, if \( \frac{a}{b} \in Q_p \) and \(\gcd(a,b) = 1\), then \( b = p^i \) for some \( i \in \mathbb{N} \).

Show that \( Q_p \leq Q_2 \).

To show that Qp is a subset of Q2, we need to prove that any rational number in Qp can be expressed in the form of Q2. Let's consider a rational number ba in Qp, where a and b are coprime integers and b is a power of p (b = pi).

Since p is a prime number, its prime factorization is p = 2^k, where k is a non-negative integer. Therefore, b = pi = (2^k)^i = 2^(ki).

Now, let's rewrite the rational number ba in terms of Q2. We can express b as 2^(ki) and a as 2^m, where m is a non-negative integer. Therefore, ba = (2^m)/(2^(ki)) = 2^(m-ki).

Since m and ki are both non-negative integers, m-ki is also a non-negative integer. Hence, the rational number ba can be expressed in the form of Q2, which means Qp is a subset of Q2.

Learn more about inclusion of sets of rational numbers here:

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