Question

prove that /3 is an irrational no
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Answer 1

★ Question.

→ Prove that √3 is an irrational number.

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→ fact.

★we know that a rational number is always written in the form of a/b

★ Solution.

If possible , let √3 be a rational number and its simplest form be

$\frac{a}{b}$ then, a and b are integers having no common factor .

other than 1 and b ≠ 0.

now, √3= $\longrightarrow\large\frac{a}{b} \longrightarrow 3= \frac{a^2}{b^2}$(squaring on both side )

$\longrightartow 3b^{2}=a^{2}........eq(i) \\ \longrightarrow 3\: divides \:a^{2} ( 3 divides b^{2})\\ \longrightarrow 3 \:divide \:a$

let a=3c for some integer c

putting a=3c in eq......(I)

or $3b^{2}=9c^{2} \longrightarrow b^{2}=3c \\ \longrightarrow 3\: divides\: b^{2} (∵3 divides 3c^{2} ) \\ \longrightarrow 3 divides a$

thus,

Thus 3 is a common factor of a and b

This contradicts the fact that a and b have no common factor other than 1.

The contradiction arises by assuming √3 is a rational.

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Answer 2

$\rule{300}2$

$\huge\tt{\underline{Solution:}}$

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

$<marquee>hope this helps you ✌ </marquee>$

$\rule{300}2$