Question

prove that √2 is an irrational number​

Answer 1

Let √2 be a rational number

Therefore, √2= p/q [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0

On squaring both sides, we get

p²= 2q² ...(1)

Clearly, 2 is a factor of 2q²

⇒ 2 is a factor of p² [since, 2q²=p²]

⇒ 2 is a factor of p

Let p =2 m for all m ( where m is a positive integer)

Squaring both sides, we get

p²= 4 m² ...(2)

From (1) and (2), we get

2q² = 4m² ⇒ q²= 2m²

Clearly, 2 is a factor of 2m²

⇒ 2 is a factor of q² [since, q² = 2m²]

⇒ 2 is a factor of q

Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1

Therefore, Our supposition is wrong

Hence √2 is not a rational number i.e., irrational number.

some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

2 = (2k)2/b2

2 = 4k2/b2

2*b2 = 4k2

b2 = 2k2

This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!

WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.

Answer 2

Let's suppose √2 were a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero. We additionally make it so that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction.

It follows that 2 = a2/b2, or a2 = 2 * b2. So the square of a is an even number since it is two times something. From this we can know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a * a would be odd too. Odd number times odd number is always odd. Check if you don't believe that!

Okay, if a itself is an even number, then a is 2 times some other whole number, or a = 2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming the contradiction:

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

2 = (2k)2/b2

2 = 4k2/b2

2*b2 = 4k2

b2 = 2k2.

This means b2 is even, from which follows again that b itself is an even number!!!