Question

prove that ✓2 is irrational​

Answer 1

To prove that the square root of 2 is irrational is to first assume that its negation is true. Therefore, we assume that the opposite is true, that is, the square root of 2 is rational. ... If 2 is a rational number, then we can express it as a ratio of two integers.

hope it helps you

please mark it brainliest

Answer 2

Question:

Prove that $\small{\tt{\sqrt{2}}}$ is irrational.

Answer:

Here, we have to prove that √2 is irrational.

We know that if a number is not rational then it is irrational. Therefore we can say that √2 is irrational and if we found √2 isn't rational then √2 will irrational.

Prove:

Let us assuming that $\small{\tt{\sqrt{2}}}$ is rational number.

Now,

If $\small{\tt{\sqrt{2}}}$ is rational number then we can write it in form of $\large{\tt{\frac{p}{q}}}$ where p and q are co - prime and q ≠ 0

Therefore,

√2 = p/q where p and q are co - prime.

Squaring both sides:

(√2)² = (p/q)²

=> 2 = p²/q²

=> 2q² = p²

=> q² = p²/2 •••••(1)

Here, we found that 2 is dividing p² so 2 will also divide p.

∴ 2 is factor of p •••••••(2)

Now,

If 2 is dividing p then supposing that p = 2a •••••••••(3)

Therefore,

Putting value of p from equation 3 in equation 1.

=> $\large{\tt{ {q}^{2} = \frac{ {p}^{2} }{2}}}$

=> $\large{\tt{{q}^{2}=\frac{({2a})^{2}}{2}}}$

=> $\large{\tt{{q}^{2}=\cancel\frac{4{a}^{2}}{2}}}$

=> $\small{\tt{{q}^{2} = 2a²}}$

=> $\large{\tt{ a² =\: \frac{q²}{2}}}$

Here, we found that q² is divisible by 2 so, q is also divisible by 2.

∴ 2 is factor of q. ••••••(4)

From equation 2 and 4 , we found that 2 is common factor of both p and q. But, it is contradicting that p and q have no common factor.

∴ √2 is irrational ••••••• proved