Question

प्रूफ दैट रूट ऑफ थ्री इज इरेशनल
$prove \: that \sqrt{3 \: is \: irrational \: }$
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Answer 1

Let us assume that √3 is a rational number.

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

where a and b are co- prime number.

So,

√3 = a/b { where a and b are co- prime}

√3b = a

Now, by squaring both the side

we get,

(√3b)² = a²

3b² = a² ........ ( i )

So,

if 3 is the factor of a²

then, 3 is also a factor of a ..... ( ii )

=> Let a = 3m { where m is any integer }

squaring both sides

a² = (3m)²

a² = 9m²

putting the value of a² in equation ( i )

3b² = a²

3b² = 9m²

b² = 3m²

So,

if 3 is factor of b²

then, 3 is also factor of b

Since

3 is factor of a & b both

So, our assumption that a & b are co- prime is wrong

hence,. √3 is an irrational number

Answer 2

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