Answer :

Let us consider that 1/√2 is a rational number.

Let 3√2 = a/b for b ≠ 0 ……………… (i)

Where a and b are co-prime integer numbers.

From (i) we can write as follows:

a = 3b√2

a^{2} = 18b^{2} …………………………….. (i)

Now since 18b^{2} is divisible by 18, so a^{2} has to be divisible by 18. From theorem 2.3 we can clearly states that if a^{2} is divided by 18 so a is also divisible by 18. So we conclude that a divides 18.

Now we can write the integer b in following format,

a = 18c

a^{2} = 324c^{2} …………………………… (ii)

By comparing (i) and (ii) we can state as follows:

324c^{2} = 18b^{2}

b^{2} = 18c^{2}

From the above equation we can conclude that b^{2} is divisible by 18 and also by 18.

From the values of a and b it is seen that a and b has a common factor and it is clearly indicates that 18 is a common factor. But it is assumed in the beginning that a and b has no common factors.

So our assumption made at the beginning of the problem is wrong.

Hence it is proved that 3√2 is an irrational number.

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