Proving that $\sqrt { 15 } $ is irrational
Prove that $\sqrt { 15 } $ is irrational.
Steps I took:
I want to do a proof by contradiction.
Let us assume $\sqrt { 15 }$ is rational.
This means that: $\sqrt { 15 } =\frac { a }{ b } ;$ such that $a$ and $b$ are coprime.
$$15=\frac { a^{ 2 } }{ b^{ 2 } } $$
$$\Rightarrow 15b^{ 2 }=a^{ 2 }$$
Where do I go from here?
$\endgroup$ 46 Answers
$\begingroup$You have shown that $3$ and $5$ divide $a^2$, so $3$ and $5$ also divide $a$ (remember that if a prime divides a product, then it divides at least one of the factors). Now $a=15k$, so you get $b^2 = 15k^2$. Do the same thing to show that $15|b$ and get a contradiction.
$\endgroup$ 7 $\begingroup$Use the rational root theorem: If the polynomial $a_n x^n + \dotsb + a_0$ with integer coefficients, such that $a_n \ne 0$ and $a_0 \ne 0$ has a rational zero $u / v$, then $u$ divides $a_0$ and $v$ divides $a_n$.
Proof: Substitute the zero, and multiply through by $v^n$ to get:
$\begin{align} a_n u^n + a_{n - 1} u^{n - 1} v + \dotsb + a_1 u v^{n - 1} + a_0 v^n = 0 \end{align}$
The right hand side is divisible by $u$ and $v$, and so is the left. But on the left all terms are divisible by $u$, except possibly the last one, so $u$ must divide $a_0$. In the same way, $v$ divides $a_n$.
Now consider $x^2 - 15 = 0$. By the above, any rational root must be an integer ($u$ divides 15, while $v$ divides $1$). But $\sqrt{15}$ isn't an integer, so it is irrational.
$\endgroup$ $\begingroup$This is getting really old.
Copy and paste from another answer of mine.
If $n$ is a positive integer that is not a square of an integer, then $\sqrt{n}$ is irrational.
Let $k$ be such that $k^2 < n < (k+1)^2$. Suppose $\sqrt{n}$ is rational. Then there is a smallest positive integer $q$ such that $\sqrt{n} = p/q$.
Then $\sqrt{n} = \sqrt{n}\frac{\sqrt{n}-k}{\sqrt{n}-k} = \frac{n-k\sqrt{n}}{\sqrt{n}-k} = \frac{n-kp/q}{p/q-k} = \frac{nq-kp}{p-kq} $.
Since $k < \sqrt{n} < k+1$, $k < p/q < k+1$, or $kq < p < (k+1)q$, so $0 < p-kq < q$. We have thus found a representation of $\sqrt{n}$ with a smaller denominator, which contradicts the specification of $q$.
$\endgroup$ $\begingroup$Suppose that $ a, b $ are coprime numers. Note that $3 |b^2 $ then $3|b|$. If $3|b|$ then $3|a $, absurd!
$\endgroup$ 0 $\begingroup$I'll give a proof for more general statement: Let $p$ be prime and $a$ a positive integer coprime with $p$. Then $\sqrt{ap}$ is irrational.
Assume that $ap = \frac{x^2}{y^2}$ where $x,y\in\mathbb Z$ are coprime. Then we have $apy^2 = x^2\implies p|x^2\implies p|x$, because $p$ is prime. Thus, $x = px'$ and $apy^2 = p^2x'^2\implies ay^2 = px'^2\implies p|ay^2$. Since $p$ and $a$ are coprime, we have that $p|y^2\implies p|y$. Contradiction, as $x,y$ are coprime by our assumption. Hence, $\sqrt{ap}$ is irrational.
Specially, $\sqrt p$ is irrational for any prime, just like $\sqrt{p_1p_2\cdots p_k}$ for distinct primes $p_1,\ldots,p_k$. Thus, $\sqrt{15} = \sqrt{3\cdot 5}$ is irrational.
Also, because for positive reals we have $\sqrt{xy} = \sqrt{x}\sqrt y$, for any positive integer $n$, we can express $\sqrt n = m\sqrt{p_1\cdots p_k}$ for some distinct primes $p_1,\ldots,p_k$. If $k = 0$, $n$ is a perfect square, so $\sqrt n$ is integer, otherwise, $\sqrt n$ is irrational, by our previous argument that $\sqrt{p_1\cdots p_k}$ is irrational. Now, since $15$ is not a perfect square, $\sqrt {15}$ must be irrational.
$\endgroup$ 2 $\begingroup$Using: Square root of integer is integer if rational:
Suppose $\sqrt {15}$ is rational, then we'd have a positive integer $x$ such that $x^2=15$, by the lemma below, we know that $1 \leq x=\sqrt{15} \leq 15$.
Therefore we have a systematic way to check if $\sqrt {15}$ is rational: is any of the numbers $1^2,2^2,...,15^2$ equal to $15$? If the answer is negative, then our number is irrational.
$\bf Lemma$: If $z\geq 1 \rightarrow z\geq \sqrt z$.
Proof: Let $z\geq 1 \rightarrow z^2\geq z$ take square roots and we get $|z|=z\geq \sqrt z ._\square$
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