Question

is √6 a rational or irrational ? any why

Answer 1

irrational because it's not of the form a/b where a,b real numbers

Answer 2

are you trying to prove it?

Answer 3

irrational, b.c. it dosnt have 2 real factors.

Answer 4

let's assume it's rational... then:
\[\sqrt{6}\] is positive => by Archimedian principle there is exist positive integers m & n such:
\[\sqrt{6}=m/n\]
or \[6n ^{2}=m ^{2}\]
=>if assumption/ ration is true, then 6 is divisor of the left hand side (LHS) and also diviser right hand side (RHS)
=> 2 is also diviser of RHS & LHS
this is only possible if 2 is a factor of m

Answer 5

let m=2k (k= positive integer)
=> 4k^2 = 6 n^2
2k^2=3n^2
as 2 is a factor of LHS => 2 is also a factor of RHS.
But this is only possible if 2 is a factor of n...
but we assumed that \[\sqrt{6}\]=m/n presented in lowest term...
if divided by 2 => not in lowest term => contradiction.
=>\[\sqrt{6} \] is not a rational number