Determining if the statement is true

[ver_mathstats]

Homework Statement

Let n be fixed positive integer, is it true regardless of n?

(a) If n is divisible by 6, then n² is divisible by 6.

Homework Equations

The Attempt at a Solution

For (a) so far I have n=6k and n²=6k, however I am unsure of where exactly I have to go from there?
I think this statement is true because I tested numbers in place of n so I think it is true, but I know that is not enough to determine if it really is true and I am having trouble figuring out the next step.

Thank you.

 

[fresh_42]

Homework Statement

Let n be fixed positive integer, is it true regardless of n?

(a) If n is divisible by 6, then n² is divisible by 6.

Homework Equations

The Attempt at a Solution

For (a) so far I have n=6k and n²=6k, however I am unsure of where exactly I have to go from there?
I think this statement is true because I tested numbers in place of n so I think it is true, but I know that is not enough to determine if it really is true and I am having trouble figuring out the next step.

Thank you.

If $n=6k$ then what is $n^2?$

 

[ver_mathstats]

If $n=6k$ then what is $n^2?$

Would I have to take the square root of n² and 6k? Which then would become n = √6k. Or would I have to do n²=6²k²?
Or is n² = √36k²?

 

[fresh_42]

It is always a good strategy to list what is given, and maybe to write down what has to be achieved. Maybe, because that can also be a distraction. Some students tend to use what they have to show, which of course is forbidden. At least as long you don't perform an indirect proof, in which case the opposite of what has to be shown is assumed to be true.

1. Here we have $6\,|\,n$ or as you correctly said: $n=6\cdot k$ for some $k$.
2. We want the same for $n^2$, i.e. $6\,|\,n^2$ resp. $n^2=6 \cdot m$ for some $m$.

Now use what is given: $n=6\cdot k$. Then $n^2=n\cdot n= 6^2\cdot k^2$ and the only thing left is to write it as $n^2=6 \cdot m$.

 

[ver_mathstats]

It is always a good strategy to list what is given, and maybe to write down what has to be achieved. Maybe, because that can also be a distraction. Some students tend to use what they have to show, which of course is forbidden. At least as long you don't perform an indirect proof, in which case the opposite of what has to be shown is assumed to be true.

1. Here we have $6\,|\,n$ or as you correctly said: $n=6\cdot k$ for some $k$.
2. We want the same for $n^2$, i.e. $6\,|\,n^2$ resp. $n^2=6 \cdot m$ for some $m$.

Now use what is given: $n=6\cdot k$. Then $n^2=n\cdot n= 6^2\cdot k^2$ and the only thing left is to write it as $n^2=6 \cdot m$.

I am unsure of how to get to n²=6m but I thought that if nn=6²m² then it would also become (6m)(6m).

 

[fresh_42]

I am unsure of how to get to n²=6m but I thought that if nn=6²m² then it would also become (6m)(6m).

Sure. There is nothing more to do. It was an easy exercise. In such cases it is important to be precise, e.g. you can write $n^2=n\cdot n=(6k)\cdot(6k)=6\cdot(6k^2)$ and so $6\,|\,n^2$.

 

[ver_mathstats]

Sure. There is nothing more to do. It was an easy exercise. In such cases it is important to be precise, e.g. you can write $n^2=n\cdot n=(6k)\cdot(6k)=6\cdot(6k^2)$ and so $6\,|\,n^2$.

Okay I understand now, thank you, also if n² is divisible by 6, then n is divisible by 6 it would be the same as the other statement but just now we are working backwards?

 

[fresh_42]

Okay I understand now, thank you, also if n² is divisible by 6, then n is divisible by 6 it would be the same as the other statement but just now we are working backwards?

It doesn't work backwards. Given $n^2=6k$, how can we know that there is another factor $6$? It is true, but the proof is a bit more complicated.

Say we have $6\,|\,n^2$ and $n^2=6k=2\cdot 3\cdot k$. Now $2\,|\,n^2=n\cdot n$. As $2$ is a prime number, it has to divide one of the factors, by the definition of a prime number. So $2\,|\,n$. The same is true for $3$, and if $2\,|\,n$ and $3\,|\,n$ then $2\cdot 3 = 6 \,|\,n$.

But from $6\,|\,n^2$ alone it is not clear how $6$ is in $n$; it could have been split. O.k. not in this case, but consider the following example: $25\,|\,100=10^2$ but $25\,\nmid\,10$.

Do you see what makes the proof for $6\,|\,n^2$ different from the example $25\,|\,100\,?$ Why doesn't the proof work for $25\,|\,100$ although $5$ is a prime?

 

[WWGD]

If n=6k , then ## n^2 =(6k

Okay I understand now, thank you, also if n² is divisible by 6, then n is divisible by 6 it would be the same as the other statement but just now we are working backwards?

This has to see with the general concept of prime ( and not just for natural numbers) . P is a prime if p|ab implies p|a or p|b (p|x:= " p divides x " )