Question

23. Prove that 4+ 3√2 is irrational number.​

Answer 1

Answer:

Let 4+3√2 be a rational number. Then both 4+3√2 and 4 are rational. ⇒ 4+3√2 – 4 = 3√2 = rational [∵Difference of two rational numbers is rational] ⇒ 3√2 is rational. ⇒ 1/3 3√2 is rational. [∵ Product of two rational numbers is rational] ⇒ √2 is rational. This contradicts the fact that √2 is irrational when 2 is prime √2 is irrational Hence 4 + 3√2 is irrational.

Answer 2

Step-by-step explanation:

If possible, let 4+3√2 be rational.

Then,4+3√2 is rational and 4 is rational.

[(4+3√2)+4] is rational.

[Sum of two rationals is rational]

Then, 3√2is rational.

Let the simplest form of √2 be a/b.

Then, a and b are integers having no common factor other than 1, and b not equal to 0.

Now,

√2 = a/b

2b² = a²

2 divides a² [2 divides 2b²]

2 divides a

Let a=2c for some integer c.

Therefore,

2b² = 4c²

b² = 2c²

2 divides b² [2 divides 2c²]

2 divides b

Thus, 2 is a common factor of a and b.

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

So, √2 is irrational.

Hence, 4+3√2 is irrational.