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Prove that √2 is a rational or an irrational

3 Answers

Let âˆš2= ab, where gcd(a,b)=1

so 2 = a²b²

or 2b²=a²

so 2|a² â†’2|a

and 2|b²â†’2|b

which contradicts gcd(a,b) = 1

so its irrational;)

If possible let âˆš2 be rational

so, âˆš2 = a/b where a and b are integers in the simplest form having no common factors other than 1

=> a²/b² = 2

=> a² = 2b² ----- (i)

Now, 2 divides 2b² => 2 divides a²

and, since 2 is prime and divides a² => 2 divides a [result 1]

since, 2 divides a so, let a = 2c, for any integer c

putting, a = 2c in (i) we get,

(2c)² = 2b²

=> 4c² = 2b²

=> b² = 2c²

Now, since 2 divides 2c² => 2 divides b²

and since 2 is prime and divides b² => 2 divides b [result 2]

But we get a contradiction from results 1 and 2 that, a and b are integers in the
simplest form having no common factors other than 1.

Hence, âˆš2 is irrational.

Dhanyavad

Thanks a lot .

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