Question

prove that √3 is an irrational

Answer 1

Let us assume the contrary that ✓3 is an rational number..
✓3=a/b(co-primes)
✓3=a/b(Squaring both side)
3=a square / b square
3b^2=a^2
3 | a^2
3 | a......1
a=3c for some integer
a^2=9c^2
3b^2=9c^2
b^2=3c^2
3 | b^2
=)3 | b.....2
From 1 and 2 we observed that A and B have at least 3 AS Common Factor but this contradicts the fact that A and B are coprimes this means that our assumption is not correct...
Hence ✓3 is an irrational number







Hope it helps you



#MAXER#

Answer 2

Hii dear here is your answer

This answer is verified ✅✅✅

If possible ,let $\sqrt{3}$be rational and let its simplest form be
$\frac{a}{b} .$

Then,a and b are integers having no common factor other than 1 , and b does not equal 0.

Now,
$\sqrt{3} = \frac{a}{b} \\ 3 = \frac{ {a}^{2} }{ {b}^{2} }$
$3 {b}^{2} = {a}^{2} ......(1)\\ 3 \: divides \: {a}^{2} \\ 3 \: divides \: a$

Let a = 3c for some integer c.

Putting a = 3c in (1), we get

$3 {b}^{2} = 9 {c}^{2} \\ {b}^{2} = 3 {c}^{2} \\ 3 \: divides \: {b}^{2} \\ 3 \: divides \: b$

Thus,3 is a common factor of a and b.

But, this contradicts the fact that a and b have no common factor other than 1.

The contradiction arises by assuming that
$\sqrt{3}$
is rational.

Hence,
$\sqrt{3 } \: is \: irrational...$