Question

prove that root 3 is irrational ​

Answer 1

Step-by-step explanation:

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number

Answer 2

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Let √3 be rational and let it's simple form be a/b

Then, a and b are integers having no common factor other than 1, and b ≠ 0.

Now,

√3 = a/b

Squaring both side

(√3)² = (a/b)²

$\rm\longrightarrow\:$3 = a²/b²

$\rm\longrightarrow\:$3b² = a²...............(i)

$\rm\longrightarrow\:$3 divides a² [∴ 3 divides 3b²]

$\rm\longrightarrow\:$3 divides a

(∴ 3 is prime and 3 divides a² $\rm\longrightarrow\:$ 3 divides a )

let a = 3c for some integer c

putting a = 3c in eq (i) ,we get

3b² = 9c² $\rm\longrightarrow\:$b² = 3c²

$\rm\longrightarrow\:$ 3 divides b² [∴3 divides 3c²]

$\rm\longrightarrow\:$ 3 divides b

(∴ 3 is prime and 3 divides b²$\rm\longrightarrow\:$ 3 divides b)

Thus,

3 is a common factor of a and b

but this contradiction the fact that a and b have no factor other than 1

This contradiction arises by assuming √3 is rational.

Hence, √3 is irrational.