Question

prove that root 2 is an irrational number​

Answer 1

Answer:

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.

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A proof that the square root of 2 is irrational.

2 = (2k)2/b2

b2 = 2k2

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Answer 2

Step-by-step explanation:

✪A proof that the square root of 2 is irrational:-

Answer:-

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.

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:

$2 \: = \: ( \: \frac{2k}{b \: } ) \: {}^{2} \\ \\ 2 = \: \frac{4k {}^{2} }{b {}^{2} } \\ \\ 2b {}^{2} = \: 4k {}^{2} \\ \\ b {}^{2} = \: 2k {}^{2}$

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.