Prove that is an irrational number.

Question

KnowledgeBoat · Accepted Answer

We will use the proof by contradiction.

Assume $\sqrt{5}$ is a rational number. Then it can be written as $\dfrac{p}{q}$ , where p and q are integers with q ≠ 0 and gcd(p, q) = 1 (lowest form).

⇒ $\sqrt{5} = \dfrac{p}{q}$

Squaring both sides :

$\Rightarrow 5 = \dfrac{p^2}{q^2} \\[1em] \Rightarrow 5q^2 = p^2.$

So, p² is divisible by 5, which means p is also divisible by 5.

Let p = 5k for some integer k. Substituting :

$\Rightarrow 5q^2 = (5k)^2 \\[1em] \Rightarrow 5q^2 = 25k^2 \\[1em] \Rightarrow q^2 = 5k^2.$

So, q² is divisible by 5, which means q is also divisible by 5.

But this means both p and q are divisible by 5, contradicting our assumption that gcd(p, q) = 1.

Hence, our initial assumption is wrong.

Hence, $\sqrt{5}$ is an irrational number.

Mathematics

Prove that $\sqrt{5}$ is an irrational number.

Whole Numbers

Answer

Related Questions

Consider this puzzle: What is the square root of –1? We know that $1 \times 1 = 1$ . We also know that $(-1) \times (-1) = 1$ . There is no Real Number that, when multiplied by itself, results in a negative number. Thus, $\sqrt{-1}$ cannot exist on number line.

Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:
(i) $\dfrac{3}{50}$
(ii) $\dfrac{2}{9}$

Convert the following decimal numbers in the form of $\dfrac{p}{q}$ .
(i) 12.6
(ii) 0.0120
(iii) $3.0\overline{52}$
(iv) $1.2\overline{35}$
(v) $0.\overline{23}$
(vi) $2.0\overline{5}$
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(viii) $3.12\overline{5}$
(ix) $2.\overline{1625}$

Locate the following rational numbers on the number line.
(i) 0.532
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