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Real Analysis

  1. add Prove that \(\sqrt{3}\) is irrational. Does a similar argument work to show \(\sqrt{6}\) is irrational?

    Solution: Assume, for contradiction, that there exist integers \(p\) and \(q\) such that \(\gcd(p, q) = 1\) and: \[ \left( \frac{p}{q} \right)^2 = 3 \tag{1}\] Next, equation (1) implies: \[ p^2 = 3q^2 \tag{2} \] This shows that \(p^2\) is a multiple of 3. The only way for a perfect square to be a multiple of 3 is if its factors also include 3. Therefore, \(p\) must also be a multiple of 3. Let \(p = 3k\), where \(k\) is an integer. Substituting this into (2) gives: \[ (3k)^2 = 3q^2 \] \[ 9k^2 = 3q^2 \] \[ 3k^2 = q^2 \tag{3} \] This shows that \(q^2\) is a multiple of 3, and hence, \(q\) is a multiple of 3. This contradicts our hypothesis that \(\gcd(p, q) = 1\). From this reasoning, we can only assume that equation (1) cannot hold for any integers \(p\) and \(q\). Thus, \(\sqrt{3}\) is irrational.

    Q.E.D.

    Yes, a similar argument works to show that \(\sqrt{6}\) is irrational. We can replace equation (2) with: \[ p^2 = 6q^2 \tag{4} \] This shows that \(p\) is a multiple of 6. Let \(p = 6k\) and substitute into (4): \[ (6k)^2 = 6q^2 \] \[ 36k^2 = 6q^2 \] \[ 6k^2 = q^2 \] This tells us that \(q\) is a multiple of 6. By the same reasoning as above, \(\sqrt{6}\) is irrational.

  2. add Show that there is no rational number \(r\) satisfying \(2^r = 3\)

    Solution: Suppose that, for contradiction, there exists a rational number such that \(2^r = 3\). Let \(r = p/q\) where \(\gcd(p,q) = 1\) for \(p,q \in \mathbb{Z}\). We will consider three cases:
    1. \(p/q > 0\)
    2. \(p/q < 0\)
    3. \(p/q = 0\)

    Case 1:

    Suppose that \(p/q > 0 \). Then \[ 2^{p/q}=3 \implies 2^p=3^q \] Since \(2^p\) is even and \(3^q\) is odd for all \(p,q \in \mathbb{N}\), this is a contradiction and no such \(r > 0\) exists.

    Case 2:

    Suppose that \(p/q < 0\). Without loss of generality, assume that \(p<0\) and \(q>0\). Then \[ 2^{-{p/q}}=\frac{1}{2^{p/q}}=3 \implies \frac{1}{2^p}=3^q\] Then \(\frac{1}{2^p} < 1\) while \(3^q> 1\). This is a contradiction and no such \(r < 0 \) exists.

    Case 3:

    Suppose that \(p/q=0\). Then \[2^0=3 \implies 1=3\] This is a contradiction and no such \(r=0\) exists.

    Hence, there is no rational number \(r\) satisfying \(2^r = 3\).

    Q.E.D.