Question

prove that root 2 + 1 is a irrational number​

Answer 1

let we assume that root 2 +1 is rational therefore there exist a unique pair of integer a and b such that..

$\sqrt{2 } + 1 = a \div b$

$\sqrt{2} = a - b \div b$

here a-b/b is a rational number but root 2 is irrational a rational cant be an irrational

therefore this contradiction arrises due to our wrong assumption that root2 -1 is rational is false hence it is an irrational...

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Answer 2

Answer:

To prove that √2 + 1 is an irrational number, we need to show that it cannot be expressed as a ratio of two integers.

Step-by-step explanation:

In other words, we need to prove that there are no integers p and q (with q ≠ 0) such that:

√2 + 1 = p/q

We will prove this by contradiction. Suppose, for the sake of contradiction, that √2 + 1 is a rational number, i.e., there exist integers p and q (with q ≠ 0) such that:

√2 + 1 = p/q

Squaring both sides of this equation, we get:

2 + 2√2 + 1 = p^2/q^2

Simplifying this equation, we get:

√2 = (p^2 - q^2)/(2q^2)

Since p^2 and q^2 are both integers, it follows that (p^2 - q^2) is also an integer. Therefore, √2 is a ratio of two integers, which contradicts the fact that √2 is an irrational number (which has been proven previously).

Therefore, our assumption that √2 + 1 is a rational number is false, and we have proved that √2 + 1 is an irrational number.

In conclusion, we have shown that √2 + 1 cannot be expressed as a ratio of two integers, which means that it is an irrational number.

To learn more about irrational numbers, click on the given link.

https://brainly.in/question/3537626

To learn more about integers, click on the given link.

https://brainly.in/question/140247

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