Question

Prove that √12 is irrational

Answer 1

Answer:

A rational number is what we can express as a fraction of two integers.

like $\frac{p}{q}$. If we could express $\sqrt{12}$ as a fraction then it would be a rational number. But we can't do it so it is irrational.

Step-by-step explanation:

Let's say . $\sqrt{12}$ is rational.

so we could write it as a fraction of two rational numbers $\frac{p}{q}$.

$\sqrt{12} = \frac{p}{q}\\ (\sqrt{12})^2 = (\frac{p}{q})^2\\12 = \frac{p^2}{q^2}\\12q = \frac{p^2}{q}$

look, p and q both are integers. so 12q would be also an integer. but $\frac{p^2}{q}$ can't be an integer as p is integer and q is also integer. so it would be a fraction .

so,$12q \neq \frac {p^2}{q}$

so $\sqrt{12}$ can't be a rational number.

So it is obviously an irrational value.

Answer 2

We have to prove that √12 is irrational.

So, let us assume √12 is a rational number. This implies that √12 can be expressed in the form of p/q, where q ≠ 0 and p and q are co-primes.

$\Longrightarrow \sf \sqrt{12} = \dfrac{p}{q}$

Squaring on both sides we get:

$\Longrightarrow \sf \Big(\sqrt{12}\Big)^2 = \Bigg(\dfrac{p}{q}\Bigg)^2$

$\Longrightarrow \sf 12 = \dfrac{p^2}{q^2}$

$\Longrightarrow \sf 12q^2 = p^2$

⇒ 12 divides p²

∴ 12 divides p as well. → Relation (1)

(Theorem Used: If p is a prime number and divides a², then p divides 'a' as well where 'a' is a positive integer)

Let us take p = 12c, for any positive integer c.

⇒ 12q² = p²

⇒ 12q² = (12c)²

⇒ 12q² = 144c²

⇒ q² = (144c²)/12

⇒ q² = 12c²

⇒ 12 divides q²

∴ 12 divides q as well. → Relation (2)

From relation 1 & relation 2, we can say that p & q have other factors other than themselves and 1. This contradicts my statement that p & q are co-primes. This is due to my incorrect assumption that √12 is a rational number.

∴ √12 is an irrational number.