Question

prove that root 3 is irrational number​

Answer 1

Answer:

Let us assume on the contrary that 3

​ is a rational number. Then, there exist positive integers a and b such that

3

​=ba​ where, a and b, are co-prime i.e. their HCF is 1

Now,

3

​=ba​

⇒3=b2a2​

⇒3b2=a2

⇒3∣a2[∵3∣3b2]

⇒3∣a...(i)

⇒a=3c for some integer c

⇒a2=9c2

⇒3b2=9c2[∵a2=3b2]

⇒b2=3c2

⇒3∣b2[∵3∣3c2]

⇒3∣b...(ii)

From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

Hence, 3

​ is an irrational number.

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Answer 2

Answer:

Let us assume that √3 is a rational number.

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

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

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

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

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number