Question

prove that \sqrt{ 2} is irrational?

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Answer 1

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Let us assume that √2 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

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

√2q = p

Now, by squaring both the side

we get,

(√2q)² = p²

2q² = p² ........ ( i )

So,

if 2 is the factor of p²

then, 2 is also a factor of p ..... ( ii )

=> Let p = 2m { where m is any integer }

squaring both sides

p² = (2m)²

p² = 4m²

putting the value of p² in equation ( i )

2q² = p²

2q² = 4m²

q² = 2m²

So,

if 2 is factor of q²

then, 2 is also factor of q

Since

2 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

Hence √2 is an irrational number

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Answer 2

Step-by-step explanation:

Let's suppose √2 is a rational number. Then we can write it √2  = a/b where a, b are whole numbers, b not zero.

We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.

From the equality √2  = a/b 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 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 it if you don't believe me!

Okay, if a itself is an even number, then a is 2 times 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.