Question

Q.Prove that 3√2 is irrational.
Q. Prove that 3+2 √5 is irrational.
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Answer 1

Answer:

means our assumption is wrong. Hence √3 is irrational. Question 3 : Prove that 3 √2 is a irrational. Solution : Let us assume 3 √2 as rational. 3 √2 = a/b √2 = a/3b. Since √2 is irrational Since 3, a and b are integers a/3b be a irrational number. So it contradicts. Hence 3 √2 is irrational number.

Answer 2

$\bf{( 1 ) \:Prove \:that \:3√2 \:is \:irrational.}$

$\bf{Solution:-}$

Let,

$\bf{3√2 \:is \: a \:rational \:number}$

$\bf{Therefore \:3√2 \:= \:p/q}$

$\bf{and \:q \:≠ \:0}$

$\bf{⇒\:√2 \:= \:p/3q}$

Therefore, = p/3q (rational) √2 (therefore, √2 will be a rational number) (But it is impossible). √2 is a irrational number.

$\bf{So, \:3√2 \:is \:irrational}$

$\bf{( 2 ) \:Prove \:that \:3+2√5 \:is \:irrational.}$

$\bf{Solution:-}$

Let,

3+2√5 be a rational number.

Therefore, 3+2√5 = p/q [where p and q are integers, q ≠ 0, and p and q are co-prime]

$\bf{⇒2√5 = p/q – 3}$

$\bf{⇒2√5 = p– 3q / q}$

$\bf{⇒√5 = p–3q /2q}$

We know that,

p/q is a rational number.

therefore, p–2q / 2q is also a rational number.

Then, √5 is also a rational number.

But, √5 is an irrational number.

This contradicts our assumption.

$\bf{Therefore, 3+2√5 \:is \:an \:irrational \:number}$

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