Question

Prove that the following are irrational.
(i)
1
2 (ii) 3 + 5 (iii) 6 + 2 (iv) 5 (v) 3 + 2 5

Answer 1

Answer:

(i) 5√2 Let 5√2 is a rational number. 5√2 = a/b, b ≠ 0, (where a, b are co-prime numbers) or √2 = a/5b ....(i) Since, a, b are integer. So, a/5b is a rational number. It is clear that from equation (i) √2 is a rational number which is contradict statement, since we know that √2 is irrational number. So, our hypothesis is wrong. Hence, 5√2 is a irrational number (ii) 2/√7 Let 2/√7 is a rational number. we find two integer such as 2/√7 = a/b(b ≠ 0) where a and be are co-primes 1/√7 = a/2b ∴ a and b is integers ∴ a/2b is a rational number ⇒ 1/7 also a rational number But 1/7 is not rational number. This is contradict. So, our hypothesis is wrong, Hence, 2/√7 is a irrational number. (iii) 3/2√5 Let 3/2√5 is a rational number. we find two integers such as 3/2√5 = a/b(b ≠ 0) Where a and b are co-prime. 1/√5 = 2a/3b ∴ a and b integer r. ∴ 2a/3b is a rational number ⇒ 1/√5 will be also a rational number. But 1/√5 is not rational number. It is a irrational number. This is contradict. So, our hypothesis is wrong. Hence, 3/2√5 is a irrational number. (iv) 4 + √2 Let 4 + √2 is rational number. Now we find two integers a and b such as 4 + √2 = a/b ⇒ √2 = a/b - 4 ⇒ √2 = (a - 4b)/b a, b, ab and 4 all are integers. (a - 4b)/b is a rational number. √2 will be a rational number This contradicts. our hypothesis is wrong. So, 4 + √2 is an irrational number.

Answer 2

$\huge\fbox\pink{✯Answer✯}$

let

$1) \frac{1}{ \sqrt{2} }$

is an rational number

$\frac{1}{ \sqrt{2} } \: be \: \frac{p}{q}$

where p and q are Integers and q is not equal to 0 where p and q are co-prime

$\frac{1}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } = \frac{p}{q}$

$\frac{ \sqrt{2} }{2} = \frac{p}{q}$

$\sqrt{2 = } \frac{2p}{q}$

where p and q are Integers

$\frac{2p}{q}$

is rational but √2 is irrational

our assumption was wrong .

$\huge\fbox\pink{✯Answer✯}$

$\frac{1}{ \sqrt{2} }$

Is rational number

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$3 + \sqrt{5}$

similarly as we solve the first problem we have to solve this then we get the answer that

$\huge\fbox\pink{✯Answer✯}$

$3 + \sqrt{5}$

is irrational number

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$\huge\boxed{\dag\sf\red{ANSWER}\dag}$

$5$

is rational number

$\huge\boxed{\dag\sf\red{ANSWER}\dag}$

$\sqrt{5}$

is irrational number

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$\huge\fbox\pink{✯Answer✯}$

$3 + \sqrt{5}$

is irrational number

$\huge\boxed{\dag\sf\red{Thanks}\dag}$