Question
Show that 5 + √7 is an irrational number.
✓
Answer
Let us assume that 5 + √7 is a rational number. So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that
$\begin{array}{r}5+\sqrt{7}=\frac{a}{b} \\ \therefore \quad \sqrt{7}=\frac{a}{b}-5\end{array}$
Since, ' $a$ ' and 'b' are integers, $\sqrt[a]{b}-5$ is a rational number and so $\sqrt{ } 7$ is a rational number. $\therefore$ But this contradicts the fact that $\sqrt{ } 7$ is an irrational number. Our assumption that $5+\sqrt{ } 7$ is a rational number is wrong.
$\therefore 5+\sqrt{7}$ is an irrational number.
$\begin{array}{r}5+\sqrt{7}=\frac{a}{b} \\ \therefore \quad \sqrt{7}=\frac{a}{b}-5\end{array}$
Since, ' $a$ ' and 'b' are integers, $\sqrt[a]{b}-5$ is a rational number and so $\sqrt{ } 7$ is a rational number. $\therefore$ But this contradicts the fact that $\sqrt{ } 7$ is an irrational number. Our assumption that $5+\sqrt{ } 7$ is a rational number is wrong.
$\therefore 5+\sqrt{7}$ is an irrational number.
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