Irrationality of $\sqrt{24}$

Question

Please, help me show that $\sqrt{24}$ is irrational by contradiction. I tried to prove using the fact that $24=2×2×2×3$ and that $\sqrt2$ and $\sqrt3$ are irrational, but the product of irrationals isn't always irrational.

Answer 1

One can adapt any of the proofs for the squart root of 2 to this problem. I provide one here:

By definition $\sqrt{24}$ is a root to $x^2 - 24 = 0$. By the Rational root test the only candidates for rational roots of this polynomial are $\pm 1, 2, 3, 4, 6, 8, 12, 24$. It is easy to check none of these square to $24$, so $x^2 - 24$ has no rational roots.

Answer 2

Always approach by definitions. What you're doing is a proof by example which isn't rigorous.

To show that $\sqrt{24}$ is irrational, suppose, by method of contraction, that $\sqrt{24}$ is rational.

We define a rational number $R$ such that $R = \frac{a}{b}, a\in\mathbb Z , b \in\mathbb Z\text{ and }b\ne0$ , and $a$ and $b$ are in lowest terms.

We have that $\sqrt{24}$ = $\frac{a}{b}$.

Can you do some algebra to reach a contradiction?

Answer 3

Assume $\sqrt{24}$ is rational and let $N$ to be the smallest natural number such that $N\sqrt{24}$ is natural. Then:

$N(\sqrt{24}-4)\sqrt{24}=24N-4N\sqrt{24}$

is also a natural number, and $N(\sqrt{24}-4)$ is a natural number with the property that $N(\sqrt{24}-4)<N$. This contradicts the minimality of $N$, thus $\sqrt{24}$ is irrational. Done.

Answer 4

Notice that $\sqrt{24} = 2\sqrt 6$. Hence, if we show that $\sqrt 6$ is irrational, we're done. Suppose to the contrary that $\sqrt 6$ is rational, hence $\sqrt{6} = a/b$ for nonzero, coprime integers $a$ and $b$. Then, $a^2 = 6b^2$. We will contradict the uniqueness of prime factorization. That is, if $2^{c_1}\dotsb p_k^{c_k}$ and $2^{d_1}\dotsb q_\ell^{d_\ell}$ are two prime factorizations of a positive integer with the $p_i$'s and $q_i$'s arranged in increasing order, then $k=\ell$, $p_i = q_i$ and $c_i = d_i$ for each $i=1,\ldots, k$. Factor $a^2$ in increasing order as a product of primes $2^{2c_1}3^{2c_2}\dotsb p_k^{2c_k}$ and $6b^2 = 2^{2d_1+1}3^{2d_2+1}\dotsb p_k^{2c_k}$. The exponents of $2$ and $3$ cannot simultaneously be even and odd, so $\sqrt 6$ is irrational. Hence, $2\sqrt 6 = \sqrt{24}$ is irrational.

Answer 5

If $\sqrt{24}=\dfrac{a}{b}$, then $a^2=24b^2$. Now consider the powers of $2$ on both sides. On the left you have an even exponent. On the right you have an odd exponent. Contradiction.

This is exactly the same proof for the irrationality of $\sqrt{2}$. The common point is that the exponent of $2$ in both $2$ and $24$ is odd.

The same argument works for $\sqrt{n}$, with $n$ not a square: just consider a prime whose exponent in the factorization of $n$ is odd.