\(\sqrt{2}\) is not rational:

Explain why the following two statements are equivalent:
\(\sqrt{2}\) is a rational number;
There is an integer \(n>0\) such that \(n\sqrt{2}\) is an integer

The two statements are equivalent because both are false.
\(\sqrt{2}\) is not a rational number.

The product of two integers will always be an integer. The product of a integer and an irrational is not an integer. Since n is an integer and \(\sqrt{2}\) is irrational, \(n\sqrt{2}\) is not an integer.
Both statements are false.

Assuming (2) above, show that \(n^*=n(\sqrt{2}-1)\) satisfies:

\(n^*\) is an integer;

\[n^*=n(\sqrt{2}-1)\] \[n^* = n\sqrt{2} - n\]
The difference of two integers will be an integer. \(\therefore n^*\) is an integer.

\(n^*>0\);

\[n^*=n(\sqrt{2}-1)\] \[n^* +n = n\sqrt{2}\] \[n^* = n\sqrt{2} - n\]

Since n is an integer and \(n>0\), \(n^*\) will have to be greater than 0 as well.

\(n^*\sqrt{2}\) is an integer; \[n^*=n(\sqrt{2}-1)\] \[n^* = n\sqrt{2} - n\] \[(n^*)\sqrt{2} = (n\sqrt{2} - n)\sqrt{2}\] \[n^*\sqrt{2} = 2n - n\sqrt{2}\]

We know \(2n\) and \(n\sqrt{2}\) are integers. The difference of two integers is an integer.
\(\therefore n^*\sqrt{2}\) is an integer.

\(n^*<n.\) \[n^*=n(\sqrt{2}-1)\] \[\frac{n^*}{\sqrt{2}-1} = n\] \(\sqrt{2}-1\) is a number greater than zero.
\(\therefore n^*<n.\)

Obtain a contradiction by taking n to be the smallest positive integer for which (2) is true, and so conclude that \(\sqrt{2}\) is not a rational number.
Smallest positive integer \(n> 0\) is when n = 1. Then \(1\sqrt{2}\) = \(\sqrt{2}\)
There are no two integers for p,q where \(\sqrt{2} = \frac{p}{q}\).
\(\therefore \sqrt{2}\) is irrational.

Convert the above into a structured proof that \(\sqrt{2}\) is not a rational number.
There is no rational number whose square is 2.
\(\sqrt{2}\) is not rational.
Suppose there is a rational number r such that \(r^2 =2\) then \(r = \frac{p}{q}\) where \(p, q > 0\) are integers. We can assume gcd(p,q) = 1. \[(\frac{p}{q})^2 = 2\] \[\frac{p^2}{q^2} = 2\] \[p^2 = 2q^2\] where p^2, 2, q^2 are integers
\(2|(2q^2)\) so \(2|p^2\) will equal \(2|p * p\) Since 2 is prime \(2|p\).
Thus, \(p = 2p_1\), for some integer p_1. \[(2p_1)^2 = 2q^2\] \[4p_1^2 = 2q^2\] \[2p_1^2 = q^2\] \[2|(2p_1^2) = 2|q^2\] Since \(2p_1 = p\) \[2|p^2 = 2|q^2\] \[2|p = 2|q\] \(\therefore\) By \(2|p\) and \(2|q\) \(gcd(p,q) \ne 1\) Proof by contradiction.

Modify your structured proof to show that \(\sqrt{5}\) is not a rational number.
There is no rational number whose square is 5.
\(\sqrt{5}\) is not rational.
Suppose there is a rational number r such that \(r^2 =5\) then \(r = \frac{p}{q}\) where \(p, q > 0\) are integers. We can assume gcd(p,q) = 1. \[(\frac{p}{q})^2 = 5\] \[\frac{p^2}{q^2} = 5\] \[p^2 = 5q^2\] where p^2, 5, q^2 are integers
\(5|(5q^2)\) so \(5|p^2\) will equal \(5|p * p\) Since 5 is prime \(5|p\).
Thus, \(p = 5p_1\), for some integer p_1. \[(5p_1)^2 = 5q^2\] \[25p_1^2 = 5q^2\] \[5p_1^2 = q^2\] \[5|(5p_1^2) = 5|q^2\] Since \(5p_1 = p\) \[5|p^2 = 5|q^2\] \[5|p = 5|q\] \(\therefore\) By \(5|p\) and \(5|q\) \(gcd(p,q) \ne 1\) Proof by contradiction.

Modify the structured proof to show that if k is not a perfect square then \(\sqrt{k}\) is not a rational number.

There is no rational number whose square is k.
\(\sqrt{k}\) is not rational.
Suppose there is a rational number r such that \(r^2 =k\) then \(r = \frac{p}{q}\) where \(p, q > 0\) are integers. We can assume gcd(p,q) = 1. \[(\frac{p}{q})^2 = k\] \[\frac{p^2}{q^2} = k\] \[p^2 = kq^2\] where p^2, k, q^2 are integers
\(k|(kq^2)\) so \(k|p^2\) will equal \(k|p * p\) k has to be prime so \(k|p\).
Thus, \(p = kp_1\), for some integer p_1. \[(kp_1)^2 = kq^2\] \[k^{2}p_1^2 = kq^2\] \[kp_1^2 = q^2\] \[k|(kp_1^2) = k|q^2\] Since \(kp_1 = p\) \[k|p^2 = k|q^2\] \[k|p = k|q\] \(\therefore\) By \(k|p\) and \(k|q\) \(gcd(p,q) \ne 1\) Proof by contradiction.

Why does the proof break down for perfect squares?

This proof breaks down for perfect squares because the square root of a perfect square lets say \(\sqrt{n^2}\) will always be an integer n. Therefore n is rational.

Show that \(\log _2(3)\) is not a rational number.
Assume that \(\log_2 (3) = \frac{p}{q}\) where p, q are integers.

\[2^{\log_{2} (3)} = 3\] \[2^\frac{p}{q} = 3\] \[2^{p-q} = 3\] \(\nexists\) two integers for p,q.
Also \[(2^\frac{p}{q})^q = (3)^q\] \[2^p = 3^q\] \(2^p\) will always be even and \(3^q\) will always be odd.

\(\therefore log_2(3)\) is not a rational number. Proof by contradiction.

Show that \(\log_{\sqrt{2}}(3)\) is not a rational number.
Assume that \(\log_\sqrt{2} (3) = \frac{p}{q}\) where p, q are integers.

\[\sqrt{2}^{\log_{\sqrt{2}} (3)} = 3\] \[\sqrt{2}^\frac{p}{q} = 3\] \[(\sqrt{2}^\frac{p}{q})^q = (3)^q\] \[\sqrt{2}^p = 3^q\] \[(\sqrt{2}^p)^2 = (3^q)^2\] \[2^p = 3^{2q}\] \( 2^p\) will always be even and \(3^2q\) will always be odd.

Show that \(\log_{10}(2)\) is not a rational number. \[\log_{10}(2) = \frac{p}{q}\] where p,q are integers \[10^{\frac{p}{q}} = 2\] \[10^{p-q} = 2\] \( \nexists\) two integers for p,q.
Also \[(10^{\frac{p}{q}})^q = 2^q\] \[10^p = 2^q\] \[2^p5^p \ne 2^q\] because 2^q is not divisible by 5.

Investigate for which positive integers m,n the number \(\log_m(n)\) is rational.
Assume \[m^\frac{p}{q} = n\] \[(m^\frac{p}{q})^q = (n)^q\] \[m^p = n^q\]

\(m = a_{1}^{\mu_{1}} , a_{2}^{\mu_{2}}, a_{3}^{\mu_{3}}, a_{i}^{\mu_{i}}\) where \(\mu_{i}\) are integers \(\ge 1\) and \(a_{i}\) are primes.
\(n = b_{1}^{v_{1}} , b_{2}^{v_{2}}, b_{3}^{v_{3}}, b_{i}^{v_{i}}\) where \(v_{i}\) are integers \(\ge 1\) and \(b_{i}\) are primes.

\[m^{p-q} = n\]

The difference between two integers is an integer. Then m raise to a certain power, z \(\in \mathbb{I}\) set of integers, will have to equal n.

For a positive integer n, when can \(n^{1/n}\) be a rational number (other than n=1)?
Assume \(n^\frac{1}{n} = \frac{p}{q}\)

\[n = \frac{p^n}{q^n}\] \[nq^n = p^n\]

Plugging in positive integers z > 1 into n we see that all numbers are irrational.
\(\therefore 1\) is the only positive integer to make \(n^{1/n}\) a rational number.