Question

prove that√2+√3 is irrational

Answer 1

Answer:

√2+√3 is a irrational number

Step-by-step explanation:

Assume that√2+√3 is an irrational number

√2+√3=a/b (where a and b are integers)

squaring on both the sides

(√2+√3)^=(a/b)^

2+3+2 √2*√3=a^/b^

5+2√6=a^/b^

2√6=a^/b^- 5

2√6=a^-5b^/b^

√6=a^- 5b^/2b^

we know that√6 is an irrational number

irrational number cannot be equal to a rational number

√2+√3 not equal to a/b √2+√3 is a irrational number

Answer 2

Answer:

$\sqrt{2}+\sqrt{3}$ is irrational.

Step-by-step explanation:

Given:

$\sqrt{2}+\sqrt{3}$

To prove that $\sqrt{2}+\sqrt{3}$ is irrational.

Step 1

Let us assume that $\sqrt{2}+\sqrt{3}$ is a rational number

Then. there exist coprime integers $p, q, q \equiv 0$ such that

So it can be written in the form p/q

$& \sqrt{2}+\sqrt{3}=\frac{p}{q}$

$\Rightarrow & \frac{p}{q}-\sqrt{3}=\sqrt{2}$

Squaring on both sides, we get

$&\Rightarrow\left(\frac{p}{q}-\sqrt{3}\right)^{2}=(\sqrt{2})^{2}$

Step 2

We know that

$(a - b)^{2} = a^{2} + b^{2} - 2ab$

So the equation $\left(\frac{p}{q}-\sqrt{3}\right)^{2}$can be written as

Substitute in the above equation we get

$&\Rightarrow \frac{p^{2}}{q^{2}}-2 \frac{p}{q} \sqrt{3}+(\sqrt{3})^{2}=2$

$&\Rightarrow \frac{p^{2}}{q^{2}}-2 \frac{p}{q} \sqrt{3}+3=2$

$&\Rightarrow \frac{p^{2}}{q^{2}}+1=2 \frac{p}{q} \sqrt{3}$

$&\Rightarrow \frac{p^{2}+q^{2}}{q^{2}} \times \frac{q}{2 p}=\sqrt{3}$

simplifying the above equation, we get

$&\Rightarrow \frac{p^{2}+q^{2}}{2 p q}=\sqrt{3}$

Step 3

Since, $p, q$ are integers, $\frac{p^{2}+q^{2}}{2 p q}$ is a rational numbers.

$\Rightarrow \sqrt{3}$ is a rational number.

This contradicts the fact that $\sqrt{3}$ is irrational.

Hence, our assumption is wrong.

Therefore, $\sqrt{2}+\sqrt{3}$ is irrational.

#SPJ2