Question

prove that
$\sqrt{6} \: is \: a \: irrational \: no$
​

Answer 1

Answer:

I don't know the correct answer but I have made like this

Answer 2

Answer:

√6 is irrational number because ,,

we observe that a,b of √6 at least have 6as a common factor....but if a,b are both co-prime numbers,,it means that √6is not a rational number.hence it is irrational number....

Assume that square root 6 is irrational then square root 6 is equals to p by q where p and q are coprime number integers. square root of 6 whole square is equals to 6 is equals to p by Q whole square that p square is equals to square therefore p square is an even number even and since even number is multiplied by any other integer is also an even number p square is a even then we must be also and even or odd number X an odd number would be odd so we can replace P with 2K where K is a integer then we get to took a whole square is equal to 6 square 4K square is equals to Vi Q square 2K then cute took a square is equals to 3 q square now we see that PQ square is even for 3 q square to be with q square must be an even number but since 3 is a not and or X anyone is even and same argument about q square is an even then Q is even so both p and q are even which means that is equal by to but the means of they are not grow coprime Contract contract tale of square root 6 is not a rrational number..