Question

prove that (3+5√2) is an irrational number, given that √2 is an irrational number.​

Answer 1

Given :

• √2 is irrational number.

To prove :

• 3+5√2 is irrational number as well .

Solution:

• Let us consider that 3 + 5√2 be a rational number .

Then 3+5√2 can be written in the form p/q where q≠ 0 .

$\tt \: 3 + 5 \sqrt{2} = \dfrac{p}{q}$

$\tt \implies 5 \sqrt{2} = \dfrac{p}{q} - 3$

$\implies \tt 5 \sqrt{2} = \dfrac{p - 3q}{q}$

$\implies \tt \sqrt{2} = \dfrac{ \dfrac{p - 3q}{q} }{5}$

$\implies \tt \sqrt{2} = \dfrac{p - 3q}{5q}$

Given that √2 is irrational .

Since LHS is irrational , so RHS should also be irrational . But RHS is rational . This contradicts the fact that √2 is irrational number .

This contradiction arises due to our consideration that 3+ 5√2 is a rational number . Hence , our consideration is wrong and 3+5√2 is an irrational number .

Hence proved .

Answer 2

first,

need to prove √2 is a irrational number.

so,

Lets first assume √2 to be rational.

√2 = a/b, where b is not equal to 0.

Here, a and b are co-primes whose HCF is 1.

√2 = a/b ( squaring both sides )… 2 = a^2/b^2

2b^2 = a^2 ….. Eq.1

Here, 2 divides a^2 also a ( bcz, If a prime number divides the square of a positive integer, then it divides the integer itself )

Now, let a = 2c ( squaring both sides )…

a^2 = 4c^2…Eq.2

Substituting Eq. 1 in Eq. 2,

2 b^2 = 4c^2

b^2 = 2c^2

2c^2 = b^2

2 divides b^2 as well as b.

Conclusion:

Here, a & b are divisible by 2 also. But our assumption that their HCF is 1 is being contradicted.

Therefore, our assumption that √2 is rational is wrong. Thus, it is irrational.

Now,

let us assume that 3+5√2 is a rational number ( a and b are co prime numbers)

3+5√2=a/b

5√2a/b-3

√2=a-3b/5b

a-3b/5b is a rational number.

we know that,

√2 is a irrational number.

so, our assumption is wrong

3+5√2 is a irrational number.