Question

proove that if x^2 is divisible by 6 then x is divisible by 6

Answer 1

we can write x=6y, where y is an integer (as x is divisible by 6)
so x^2=36y^2
can you do the rest?

Answer 2

it's the other way around, prove that if x^2=6y, then x can be written as x= 6w. x,y,w are all natural numbers

Answer 3

prove the contapositive

Answer 4

I knew the idea of proving it is by contapositive, but im having trouble doing it

Answer 5

oh okay, my bad.
so all the prime factors should appear in multiple of 2. for example, if x=x1^a * x2^b * x3^c... where x1, x2, x3.. are prime factors of x
then x^2= x1^2a * x2 ^2b...
notice that all the prime factors appear 2a, 2b, 2c... times in x^2, that means they appear in multiple of 2
now x^2=6y
x^2=2*3*y
2 and 3 should appear at least two times each as the number is a square
so, x^2=2*3*2*3*y1 where y1=y/6
now y1 also has to be a square number
so x=6*sqrt (y1)
or, x=6 * y2 (y2=sqrt y1, y1 is a square number, so y2 is a whole number)
so x is divisible by 6 when x^2 is divisible by 6

Answer 6

so if i wanna prove it by contapositive i say :
if x is not divisible by 6 then x^2 is not divisible by 6
\[\forall k \in \mathbb{N}, x \neq 6 k \]
\[\therefore \forall w \in \mathbb{N} , x ^{2} \neq 6 w \]

Answer 7

is that enough of a proof ?

Answer 8

contrapositive means assume x is divisible by 6 but x^2 is not divisible by 6, prove the opposite

Answer 9

retaining the assumption x is divisible by 6

Answer 10

are you sure? coz my text book indicate
the contrapositive of
\[A =>B\]
is
\[-B => -A\]

Answer 11

if you prove if x is not divisible by 6 implies x^2 is not divisible by 6, does it necessarily prove if x^2 is divisible by 6 then x is divisible by 6 ? i am not sure.
but at least i am sure about the way i solved

Answer 12

the direct proof works, and the contrapositive also works. contrapositive follows the truth value of original statement

Answer 13

Yeah.

Answer 14

\(\large A \implies B\) and \(\large \neg B \implies \neg A\) are equivalent statements

Answer 15

So my question now is, how do you get from
\[x \neq 6 k \]
to
\[x^2 \neq 6 w\]

Answer 16

\[\forall k \in \mathbb{N}, x \neq 6 k \]
\[\therefore \forall k \in \mathbb{N} , x ^{2} \neq (6 k)^2 \]
\[\therefore \forall w \in \mathbb{N} , x ^{2} \neq 6 w \]

Answer 17

i don't see anything to add here, that should end the proof ^

Answer 18

thanks

Answer 19

yw!