Question

If p is a positive prime, prove that$\sqrt{p}$ is irrational.

Short answer

It is proved that $\sqrt{p}$ is an irrational number.

Step-by-step solution

Definition of Prime

An integer p is known as prime. When $p\ne 0,\pm 1$ and $\pm 1$and role="math" localid="1646227164627" $\pm p$would be the only divisors of p .

Show that p is irrational

Theorem 1.7 states that any integer n except $0,\pm 1$ would be the product of primes.

Let p be a positive prime number.

Let $\sqrt{p}$ be a rational number. Then there is an integer a, b in which $\sqrt{p}=\frac{a}{b}$.

Square on both sides of the equation $\sqrt{p}=\frac{a}{b}$ as follows:

$$\begin{gathered} {\left(\sqrt{p}\right)}^2=\frac{a^2}{b^2} \\ p=\frac{a^2}{b^2} \\ {a}^2=p{b}^2 \end{gathered}$$

It implies that $\frac{p}{a^2}$ where $a^2$can be expressed as $a·a$, therefore, $\frac{p}{a·a}$.

However, if a prime number divides the product of two integers, it should divide at least one of them.

Now, let$a=pm$, with m as an integer.

Square on both sides of the above equation as follows:

$\begin{array}{rcl}{a}^2& =& {\left(pm\right)}^2\left(\because a^2=p{b}^2\right)\\ p{b}^2& =& {\left(pm\right)}^2\\ p{b}^2& =& p^2{m}^2\\ b^2& =& p{m}^2\end{array}$

It implies that p divides $b^2$. However, $b^2=b·b$. It implies that $\frac{p}{b}$because .

Therefore, a and b contain at least one common multiple p , which is a contradiction to the assumption that a and b are coprime.

Hence, it is proved that$\sqrt{p}$is an irrational number.