# Numbers that are not rational

• $$\sqrt{2}$$ 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.