Question

prove that 3/root5 is irrational​

Answer 1

$\huge\green{Answer}$

Δ To prove:-

3√5 is irrational.

Δ Proof:-

Let us assume that 3√5 is a rational number.

So,

3√5 = p/q, where p and q are integers, q≠0 and p and q are co-primes. i.e. HCF(p,q) = 1.

⇒ √5 = p×3/q.

⇒ √5 = 3p/q = integer/integer.

so,

3p/q which is equal to √5 is rational number.

Therefore,

√5 is also a rational number.

But it contradicts the fact that √5 is irrational.

This contradiction has arisen because of our wrong assumption.

So our assumption that 3√5 is rational was wrong.

And hence we can conclude that 3√5 is an irrational number.

$<marquee>HENCE PROVED</marquee>$

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Answer 2

Question: To prove $\sf 3\sqrt{5}$ is an irrational number.

Solution:

Let us assume that $\sf 3\sqrt{5}$ is a rational number. This means that

$\Longrightarrow \sf 3\sqrt{5} = \dfrac{p}{q}\\ \\ \\\sf Transposing \ 3 \ to \ the \ other \ side \ we \ get,\\ \\ \\\sf \Longrightarrow \sqrt{5} = \dfrac{p}{3q} \ \ \ \ \longmapsto \textcircled{\scriptsize {\sf 1}}$

Now let's assume √5 to be a rational number. Therefore it can be expressed of the form p/q where 'p' and 'q' are positive co-prime integers and q ≠ 0.

$\Longrightarrow \sf \sqrt{5} = \dfrac{p}{q}$

Squaring on both sides we get,

$\Longrightarrow \sf \left(\sqrt{5}\right)^2 = \left(\dfrac{p}{q}\right)^2\\ \\ \\\Longrightarrow \sf \ 5 = \dfrac{p^{2}}{q^{2}}$

$\Longrightarrow \sf 5q^2 = p^2\\ \\ \\\sf If \ 5 \ divides \ p^2, \ then \ 5 \ divides \ p \ as \ well.\ \ \ \longmapsto \textcircled{\scriptsize 2}$

(If 'a', a positive prime integer divides p², another integer, then 'a' divides 'p' as well.)

$\sf Let \ p = \ 5c \ for \ a \ positive \ integer \ 'c \ '.\\ \\ \\\sf Substitute \ the \ value \ of \ 'p' \ in \ 5q^2 = p^2\\ \\ \\\Longrightarrow \sf 5q^2 = p^2\\ \\\Longrightarrow \sf 5q^2 = (5c)^2\\ \\\Longrightarrow \sf {5 \! \! \! {/}}q^2 = {25 \! \! \! \! {/}} \ c^2\\ \\\Longrightarrow \sf q^2 = 5c^2\\ \\\sf If \ 5 \ divides \ 'q^2 \ ',\\ \\ \sf 5 \ divides \ 'q' \ as \ well.\ \ \ \longmapsto \textcircled{\scriptsize \sf {3}}$

From 3 and 4, we can say that 'p' and 'q' are not co-primes as they have another factor '5' despite us saying they're co-primes. This contradiction has arisen due to our wrong assumption that 'p' and 'q' are co-primes. This is because we've assumed √5 to be a rational number.

$\therefore \sf \underline{\underline{{\sqrt{5} \ is \ a \ Irrational \ number.}}}$

Now, we know that √5 is irrational. Now lets go back to Equation 1.

$\displaystyle \Longrightarrow \sf \ \sqrt{5} = \dfrac{p}{3q}$

Here, p/3q is rational, but √5 is irrational. (Proved).

But Irrational numbers are not equal to rational numbers.

This contradicts my statement that 3√5 is rational, therefore my statement is wrong.

$\boxed{\boxed{\implies \sf 3 + \sqrt{5} \ is \ Irrational. \ }}$