Question

Show that $\sqrt{p}$is irrational for every positive prime integer p. [Hint: What are the roots of ${\mathbf{x}}^{\mathbf{2}}-{\mathbf{p}}^{\mathbf{2}}$? Do you prefer this proof to the one in Exercises 30 and 31 of Section 1.37].

Short answer

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

Step-by-step solution

Rational Root test  

Consider that $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$as a polynomial with integer coefficients. When $r\ne 0$and the rational number$\frac{r}{s}$ (In lowest terms) is a root of $f\left(x\right)$then$r/a_0$ and $s/a_0$.

Show that pis irrational for every positive prime integer p

Assume that $\sqrt{p}$is rational and that $\sqrt{p}=\frac{r}{s}$. Then, $\frac{r}{s}$would be a root of $x^2-p$. This implies that$\frac{r}{p}$ and $\frac{s}{1}$according to the rational root test. As a result,$\frac{r}{s}\in \left\{1,-1,p,-p\right\}$ .

However, neither of these are roots of $x^2-p$, which is a contradiction.

As a result, it concludes that $\sqrt{p}$is irrational.

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