# The contradiction method used to prove that the square root of a prime is irrational

The contradiction method given in certain books to prove that sqare root of a prime is irrational also shows that sqare root of $4$ is irrational, so how is it acceptable? e.g. Suppose $\sqrt{4}$ is rational,
\begin{align} \sqrt{4} &=p/q \qquad\text{where pand q are coprimes} \\ 4 &=p^2/q^2\\ 4q^2&=p^2 \tag{1} \\ 4&\mid p^2\\ 4&\mid p\\ \text {let }p&=4m \qquad\text{for some natural no. m} \\ p^2&=16m^2\\ 4q^2&=16m^2 \qquad\text{(from (1) )}\\ q^2&=4m^2\\ 4& \mid q^2\\ 4&\mid q \end{align} but this contradicts our assumption that $p$ and $q$ are coprime since they have a common factor $p$. Hence $\sqrt{4}$ is not rational. But we know that it is a rational. Why?

• 4 divides p^2 does not imply 4 divides p. – Vincent Pfenninger Aug 1 '13 at 14:48
• In any proof method, it's important to note exactly when the hypotheses come into play, so you can see if it can be used elsewhere. In particular, the $\text{prime}\mid n^2\implies \text{prime}\mid n$ steps in this proof method work because of properties of prime numbers. (Michael's and MJD's answers generalize that as far as it can be generalized.) – Cameron Buie Aug 1 '13 at 15:05

If $4q^2 = p^2$, $4$ is a factor of $p^2$, but it does not follow that $4$ is a factor of $p$ only that $2$ is a factor of $p$. For example, $4$ is a factor of $36$, but $4$ is not a factor of $6$ (but $2$ is).
In general, if $a$ is prime and $a$ is a factor of $pq$ then $a$ must be a factor of $p$ or a factor of $q$ (or both); in particular, if $a$ is a factor of $p^2$, then $a$ must be a factor of $p$. It then follows that if $m$ is a product of distinct primes, then if $m$ is a factor of $p^2$, $m$ must be a factor of $p$. So the method used to show that the square root of a prime number is irrational extends to numbers which are the product of distinct primes, but no further.
• Actually it extends to any nonsquare natural $>0.$ – Bill Dubuque Jan 8 '14 at 5:59
• How about proving that $\sqrt{12}$ is irrational without separately considering the prime factors. – marty cohen Jun 4 '15 at 6:05
The step from $4\mid p^2$ to $4\mid p$ is wrong. For example, take $p=6$. Then $4\mid 6^2$ but it is not true that $4\mid 6$.
In general, $q\mid p^2$ implies $q\mid p$ only for squarefree $q$. A number $q$ is "squarefree" if it is not divisible by any square larger than 1.
Let $p$ be prime and suppose that $\sqrt{p}$ is rational. Then $\sqrt{p}=\frac{r}{s}$ for some integers $r$ and $s$ with $s$ not zero. Then $s^2 p=r^2=q_1 \ldots q_n$ for some primes $q_i$ by the fundamental theorem of arithmetic. Note that $r^2$ and $s^2$ have an even number of prime factors by expanding squares. But then $s^2 p$ has an odd number of prime factors. This is a contradiction. Then $\sqrt{p}$ is irrational.