Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


I am attempting to replicate a proof Hardy provided in his book - A Course of Pure Mathematics, yet am having trouble with one of the steps. I was wondering if someone can explain how he made a step in the proof.


If $M$ and $N$ are integers which have no common factor, and neither of which is a perfect square, $\sqrt{M}$ and $\sqrt{N}$ are dissimilar surds.

His Proof up to the Point of Confusion

Suppose that $\sqrt{M}$ and $\sqrt{N}$ are similar surds. Then we can instead write them as: $\sqrt{M}=\dfrac{p}{q} \sqrt{\dfrac{t}{u}}$ and $\sqrt{N}=\dfrac{r}{s} \sqrt{\dfrac{t}{u}}$

Then $\sqrt{MN}$ is evidently rational, and therefore (from a previous example) integral.

The example he is referring to

An algebraic equation,

$x^n+p_1 x^{n-1} +p_2 x^{n-2}+...+p_n=0$

with integral coefficients, cannot have a rational but non-integral root.

My Question

How was Hardy able to determine that $\sqrt{MN}$ was integral from that example he was referring to?

share|cite|improve this question
up vote 4 down vote accepted

We want to show that if $C$ is an integer, and $\sqrt{C}$ is rational, then $\sqrt{C}$ is an integer.

So we want to show that any rational solution of the equation $x^2-C=0$ is actually an integer.

Let $\frac{a}{b}$ be a rational solution of the equation. We may without lloss of generality assume that $\frac{a}{b}$ is in lowest terms, that is, that no integer $\gt 1$ is a common divisor of $a$ and $b$. We may also without loss of generality assume that $b$ is positive.

Substituting in the equation, we find that $\left(\frac{a}{b}\right)^2-C=0$.

Multiply through by $b$. We find that $$a^2=b^2C.$$ Now let $p$ be any prime divisor of $a^2$. Since $a^2=b^2 C$, the prime $p$ must divide $a^2$, so it must divide $a$. This is impossible, since $\frac{a}{b}$ is in lowest terms.

We conclude that $b$ has no prime divisors, meaning that $b=1$. It follows that $C=a^2$, a perfect square, so $\sqrt{C}=a$, an integer.

Remark: The argument we used for the polynomial equation $x^2-C=0$ can, without significant modification, be used to prove the theorem about $$x^n+p_1 x^{n-1} +p_2 x^{n-2}+\cdots+p_n=0$$ that you quoted. Once it has been proved, it can be applied to our particular equation $x^2-C=0$.

share|cite|improve this answer

Because $\:x = \sqrt{MN}\:$ is a root of $\:x^2 - MN = 0\:$ so, being rational, it is integral, by said theorem (which is known as the rational root test).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.