May I divide by number n in order to solve $2n = n^2$ ( even in a case where $n$ is not equal to $0$)?

Question

Suppose I have the equation : $2n = n^2$.

Dividing by $n$ ( provided $n$ is not $0$) , I get (apparently): $n = 2$. However, from another point of view, I have:

$2n = n^2 \rightarrow n^2 = 2n $

$\rightarrow \sqrt{n^2} = \sqrt{2n}$

$\rightarrow |n| = \sqrt{2n}$

$\rightarrow |n| = \sqrt{2} \cdot \sqrt{n}$

$\rightarrow n = + \sqrt{2} \cdot \sqrt{n} \text{ or } n = - \sqrt{2} \cdot \sqrt{n}$.

And I do not think that here $2$ is still a solution (as it appeared to be the case with the first method).

Which method is correct, if any?

Answer 1

A general rule of thumb for equations like these is to only increase or decrease the total order of the equation to make the algebra a little easier - it is important, though to then arrive at the same order in which you began before giving your final solution.

For examples like this it is actually not required to do any multiplying by or dividing by $n$. Generally, alarm bells should ring when you begin to divide by $n$ or multiply by $n$ - this is when you should ask yourself, do I really need to do this.

I'd always recommend picturing the graph. Consider where the graph $y=x$ intersects with the graph $y=x^2$. Alternatively, rearrange the equation as follows.

$$n^2 - 2n = 0 \Rightarrow n(n-2)=0$$

Then we have one of two scenarios which hold the right hand equation true,

$$n=0\text{, or }n=2$$

OVERALL:

Changing the order of the equation can either introduce additional solutions or remove solutions - so tread carefully!

There are some occasions where it might make sense to divide by a variable, or to ignore a solution. For example, if you have a function $u(t)$ which represents the speed of a particle at a given time the function might be 5th order so has at most 5 real roots - but if some of these roots are for negative $t$ then you can discount them, since you have already defined time as starting at $t=0$.

Answer 2

Assuming we work in $\Bbb Z$, I would proceed as follows:

given that

$n^2 = 2n, \tag 1$

we may write

$n(n - 2) = n^2 - 2n = 0; \tag 2$

now since $\Bbb Z$ is an integral domain, we have

$n \ne 0 \Longrightarrow n - 2 = 0 \Longrightarrow n = 2; \tag 3$

this shows that

$n = 0, 2 \tag 4$

are the only solutions to (1).

As far as the query expressed in the title is concerned, I would say that, yes, one may divide by $n$, but only in the event $n \ne 0$; this is one reason I prefer the method presented above. which in fact stresses the cancellation properties of $\Bbb Z$ over division.

As for our OP Ray LittleRock's second proposed solution, it strikes me that it can be made to work but some care must be taken to ensure the steps are all valid; for example, in inferring

$\sqrt{n^2} = \sqrt{2n} \tag 5$

from

$n^2 = 2n, \tag 6$

one should restrict oneself to the case $n \ge 0$, lest $\sqrt{2n}$ be undefined; but then the assertion

$n = -\sqrt 2 \sqrt n \tag 7$

at the end is inadmissible; in any event it asserts $n < 0$ which makes the presence of $\sqrt n$ erroneous.

Answer 3

$2n=n^2$

$n^2-2n=0$

Let $f(x)=x^2-2x$

From the function $f(x)$, it is a quadratic equation. Plot the graph of $y=f(x)$ you can see that it intersects x-axis at $x=0$ and $x=2$. So, theoretically you can't divide the number $n$ straight even thought $n\neq0$.