Question

proof that a √ 6 is an irrational number​

Answer 1

$<b>$Here is the answer to your question.





Let us assume that √6 is rational.

$\sqrt{6} = \frac{p}{q}$

=> Where p and q are integers.

=> q ≠ 0

=> HCF (p and q) = 1

Squaring both the numbers,



${ \sqrt{6} }^{2} = { \frac{p}{ {q}^{2} } }^{2} \\ \\ \\ = > 6 = \frac{ {p}^{2} }{ {q}^{2} }$




That implies = p² = 5q² ----------------(1)





Here 6 is a factor of p², so 6 is also a factor of p.




Let p = 6k

=> (p) ² = (6k) ²

=> p² = 36k² -------------------(2)



Comparing (1) and (2)

6q² = 36k²

${q}^{2} = \frac{ {36k}^{2} }{6} = 6 {k}^{2}$

=> q² = 6k²





Here, 6 is a factor of q² so 6 is a factor of q.




From the above discussion, we observe that 6 is a common factor of both p and q.



So HCF of p and q = 6



This is a contradiction to our equation that HCF =1

So √6 cannot be a rational number.

Hence; √6 is irrational number.





Hope it helped!!

Answer 2

let √6 be a rational no.

then √6=a/b(where a and b are integers and b not equal to 0)

Also,let a and b be co prime.

√6=a/b

(√6)^2=(a/b)^2

6=a^2/b^2

6b^2=a^2 (eq.1)

that implies,a^2 is a multiple of 6.

that implies,a is a multiple of 6.

let a=6c(since a is a multiple of 6)

putting a=6c in eq 1

6b^2=(6c)^2

6b^2=36c^2

b^2=6c^2

that implies,b^2 is a multiple of 6.

that implies,b is a multiple of 6.

this contradicts our assumption that a and b are co prime

therefore √6 is an irrational no.