Question

prove that 3 + 5 square root 2 is an irrational number​

Answer 1

Answer:

Let us assume that 3+5√2 is Rational

this, it can be expressed in the form of p /q where p and q are rational numbers and q not equal to zero.

Step-by-step explanation:

$3 + 5 \sqrt{2} = \frac{x}{y} \\ 5 \sqrt{2} = \frac{x}{y} - 3 \\ 5\sqrt{2} = \frac{x - 3y}{y} \\ \sqrt{2} = \frac{x - 3y}{y} \times \frac{1}{5} \\ \sqrt{2} = \frac{x - 3y}{5y}$

Since, √2 is irrational

irrational is not equal to Reational

This contradicting factor has arisen due to our Wrong assumption

therefore, 3+5√2 is irrational.

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Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let us assume the contrary that 3 + 5 √2 is a rational number.

Then there exists co-prime positive integers a and b such that

⇒ 3 + 5 √2 = a/b

⇒ 5 √5 = a/b

⇒ √2 = a - 3b/5b

Therefore, √2 is a rational number.

(∵ a, b are integers ∵ a - 3b/5b is a rational number.)

This contradicts the fact that √2 is an irrational number.

So, our supposition is wrong.

Hence, 3 + 5 √2 is an irrational number.