Question

Prove 6+3√5 is irrational

Answer 1

Step-by-step explanation:

Let us assume that 6 + 3√5 is rational.

So, we can write it as:

6 + 3√5 = (a/b). {a and b are integers, b≠0, a and b are co-primes}

⇒ 3√5 = (a/b) - 6

⇒ 3√5 = (a - 6b)/b

⇒ √5 = (a - 6b)/3b

∴ (a - 6b)/3b is a rational number. So, √5 should also be rational number.

But √5 is irrational.

Which contradicts our assumption is wrong.

∴ Therefore, 6+ 3√5 is irrational number.

Hope it helps!

Answer 2

Answer:

Let us assume that is 6+3

5

rational .

So, we can write it as:

6+3

5

=(

b

a

). [a and b are integers, b



=0,a and b are co- primes]

⇒3

5

=

b

a

−6

⇒3

5

=

b

(a−6b)

⇒

5

=

3b

(a−6b)

∴

3b

(a−6b)

is a ration number.

So,

5

should also be rational number.

But

5

is irrational.

Which contradicts our assumption is wrong.

Therefore,

∴6+3

5

is irrational number.