Prove: if x^2 is divisible by 3 then x is divisible by 3
Thanks so much anybody who can help me on this.
Fundamental Theorem of Arithmetic -- from Wolfram MathWorld
... in normal practice x indicates a real variable, for integer numers usually letters like n, m, i, j or k are used...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
What I say is that for an integer the letter 'n' is usually preferred to tghe letter 'x' and Your question should be written as: if $\displaystyle 3| n^{2}$ then $\displaystyle 3| n$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
If $\displaystyle x^2$ is divisible by 3, then its prime factorization contains at least one 3. Since 3 is a prime number, it follows that x must also have a factor of 3.
(Since the question is about "divisibility", it is clear that x must be an integer so using "x" rather than "n" doesn't bother me.)
You want to prove: If $\displaystyle \displaystyle x^2$ is divisible by 3 then $\displaystyle \displaystyle x$ is divisible by 3.
The contrapositive of this is: If $\displaystyle \displaystyle x$ is not divisible by 3 then $\displaystyle \displaystyle x^2$ is not divisible by 3.
To prove the contrapositive, note that any integer can be written as $\displaystyle \displaystyle 3n, 3n+1$ or $\displaystyle \displaystyle 3n+2$, where $\displaystyle \displaystyle n$ is some other integer. Since we have said that $\displaystyle \displaystyle x$ is not divisible by $\displaystyle \displaystyle 3$, then we can let $\displaystyle \displaystyle x = 3n+1$ or $\displaystyle \displaystyle x = 3n+2$.
Case 1:
$\displaystyle \displaystyle \begin{align*} x &= 3n + 1 \\ x^2 &= (3n + 1)^2 \\ x^2 &= 9n^2 + 6n + 1 \\ x^2 &= 3(3n^2 + 2n) + 1 \end{align*}$
which is not divisible by 3.
Case 2:
$\displaystyle \displaystyle \begin{align*} x &= 3n + 2 \\ x^2 &= (3n + 2)^2 \\ x^2 &= 9n^2 + 12n + 4 \\ x^2 &= 9n^2 + 12n + 3 + 1 \\ x^2 &= 3(3n^2 + 4n + 1) + 1 \end{align*}$
which is not divisible by 3.
Since we have shown that if $\displaystyle \displaystyle x$ is not divisible by 3, then $\displaystyle \displaystyle x^2$ is not divisible by 3, the contrapositive is also true.
You are more than welcome. Just one thing though, you should note that proving a statement by proving the contrapositive is not considered a direct proof, but rather an indirect proof. However, I think it's important to see how often it is easier to prove something indirectly than directly.