Question

$prove \: that \: \sqrt{3 \: } \: \: is \: an \: irrational \: no. \: .$


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Answer 1

Step-by-step explanation:

Say 3–√ is rational.

Then 3–√ can be represented as ab, where a and b have no common factors.

$3 = \frac{a {}^{2} }{b {}^{2} }$

$3b {}^{2} = a {}^{2}$

Now a2 must be divisible by 3, but then so must a (fundamental theorem of arithmetic).

$3b {}^{2} =( 3k) {}^{2}$

$b {}^{2} = 3k {}^{2}$

and now we have a contradiction

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Answer 2

Answer:

Co prime:

Two numbers are called coprime or relative primes, if they have only one common factor which is 1 e.g., (7,9) is coprime but 15 and 21 is not coprime.

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