Question

prove thatv√3is irrational
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Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

$⇒ \sqrt{3} = p/q$

$⇒ 3 = \frac{{p}^{2} }{{q}^{2}} \: \: (Squaring \: on \: both \: the \: sides)$

$⇒ 3 {q}^{2}  = {p}^{2}$

.......….......…………(1)

It means that 3 divides p^2and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

$⇒ {p}^{2} = 9 {r}^{2}$

………………………………..(2)

from equation (1) and (2)

$⇒ 3q^2 = 9r^2$

$⇒ q^2 = 3r^2$

Where q^2 is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

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