# Math Help - by direct method

1. ## by direct method

Prove: if x^2 is divisible by 3 then x is divisible by 3

Thanks so much anybody who can help me on this.

2. ## Re: by direct method

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

$\chi$ $\sigma$

3. ## Re: by direct method

how will i use n m i k for the x^2 and x to be divisible by 3... shall i have to multiply each them by 3?
thanks

4. ## Re: by direct method

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 $3| n^{2}$ then $3| n$...

Kind regards

$\chi$ $\sigma$

5. ## Re: by direct method

If $x^2$ 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.)

6. ## Re: by direct method

this still makes me so hard... please can anyone do this proof for me?

thanks.

7. ## Re: by direct method

Originally Posted by rcs
this still makes me so hard... please can anyone do this proof for me?

thanks.
Post #5 gives you the answer. What don't you understand?

8. ## Re: by direct method

how i wish it could be shown how it is being proved..

thanks

9. ## Re: by direct method

Originally Posted by rcs
how i wish it could be shown how it is being proved..

thanks
Consider the fundamental theorem of arithmetic (unique prime factorisation) applied to both x and x^2

CB

10. ## Re: by direct method

This is very sticklery, but I feel it's important to point out that this is a particular case of Euclid's lemma which is instrumental in proving the fundamental theorem of arithmetic.

11. ## Re: by direct method

You want to prove: If $\displaystyle x^2$ is divisible by 3 then $\displaystyle x$ is divisible by 3.

The contrapositive of this is: If $\displaystyle x$ is not divisible by 3 then $\displaystyle x^2$ is not divisible by 3.

To prove the contrapositive, note that any integer can be written as $\displaystyle 3n, 3n+1$ or $\displaystyle 3n+2$, where $\displaystyle n$ is some other integer. Since we have said that $\displaystyle x$ is not divisible by $\displaystyle 3$, then we can let $\displaystyle x = 3n+1$ or $\displaystyle x = 3n+2$.

Case 1:

\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 \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 $\displaystyle x$ is not divisible by 3, then $\displaystyle x^2$ is not divisible by 3, the contrapositive is also true.

12. ## Re: by direct method

There you are my Prove It! you are the best, you make me see what i wanted to see, thanks a lot, thanks for making me understand again.

more power

razel c. summer

13. ## Re: by direct method

Originally Posted by rcs
There you are my Prove It! you are the best, you make me see what i wanted to see, thanks a lot, thanks for making me understand again.

more power

razel c. summer
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.

14. ## Re: by direct method

taken. thanks again