Question

Prove that root under5 is an irrational number​

Answer 1

Let us assume that √5 is a rational number.

we know that the rational numbers are in the form of p/q form where p,q are intezers.

so, √5 = p/q

     p = √5q

we know that 'p' is a rational number. so √5 q must be rational since it equals to p

but it doesnt occurs with √5 since its not an intezer

therefore, p =/= √5q

this contradicts the fact that √5 is an irrational number

hence our assumption is wrong and √5 is an irrational number.

hope it helped u :)

Answer 2

Answer:

Step-by-step explanation:

Suppose that $\sqrt{5}$ is rational

$\sqrt{5}$ = $\frac{a}{b}$ where a, b are co-prime integers and b ≠ 0

By cross multiplication, we get;

b $\sqrt{5}$ = a

squaring both the sides, we get;

$5b^{2}$ = $a^{2$                                                                      ⇒ equation 1

⇒5 divides  $a^{2$

⇒ 5 divides $a$ also                                                  ⇒ equation 2

a = $(5c)^{2}$ for some integer c                                     ⇒ equation 3

Substituting eq 3 in eq , we get;

$5b^{2}$ = $(5c)^{2}$

$5b^{2}$ = $25c^{2}$

$b^{2}$ = $5c^{2}$

⇒5 divides $b^{2}$

⇒ 5 divides b also                                                 ⇒ equation 4

From eq 2 and eq 4, we can say that;

5 is a factor common to both a and b. This is a contradiction since both a and b are co-prime integers.

∴ $\sqrt{5}$ is an irrational number

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