# Suppose I've an equation $1/x = 5$, then is it the same as an equation $5x=1$?

This is giving me some headache.

Suppose I have an equation $$\frac{1}{x}=5$$

Then is this equation the same as $$1=5x\quad ?$$

Now the domain of $x$ in the first equation is $\mathbb{R}\setminus \{0\}$, however the domain of $x$ in the second equation is the whole $\mathbb{R}$.

Does rearranging terms change the meaning (I don't know the right word here) of equations?

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The two equations are equivalent for $x\ne 0$. The solution of both equations is $x=\frac{1}{5}\ne 0$. – Américo Tavares Jul 8 '12 at 21:36
To be precise, an equation is meaningless unless you specify a domain; for most applications, though, the domain is so obvious that it is omitted. If you are working over $\mathbb{R}\setminus \{0\}$ then the equations you give are the same, but of course the first equation is not defined for all of $\mathbb{R}$. – Alastair Litterick Jul 8 '12 at 21:38

Yes and no. We can look at some examples: $$e^3=e^x\implies 3=x$$

this is true because $e^x$ is a one to one function. $$2^2=x^2 \implies x=2$$

is not true, because $x^2$ is not one to one.

So applying functions can change certain aspects of the equations. The original solution is still left in there somewhere, but sometimes extra solutions can pop up, or intervals that don't contain the solution won't be valid. In your example:

$$\frac 1 x=5 \implies 5x=1$$

is true. The solution is clearly $\frac 1 5$. However,

$$\frac 1 x=0 \implies 0x=1$$

clearly has no solutions (since the implication is a falsehood), and is the reason that the first equation is over $\mathbb{R}\backslash\{0\}$. What we're saying is that the equations in abstract don't always work on the same numbers, but if the solution exists in the first case, it will in the second as well.

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I disagree with "$\frac 1 x=0 \implies 0x=1$ is nonsense". If $a = b$, then $ax = bx$, which is a conclusion that can be drawn for any well-defined $x$. The fact that the domain on the left is empty (i.e., there are no solutions $x$) does not imply that the implication is false. – TMM Jul 8 '12 at 21:53
fair enough. I more meant that it's nonsense because the object it claims to talk about doesn't exist, but I'll correct it. – Robert Mastragostino Jul 8 '12 at 21:58
You should be more careful with your statements. I still object to the phrase "However, $\frac{1}{x} = 0 \implies 0x = 1$ is clearly never true" as the proposition you are referring to, $$P \equiv \left(\frac{1}{x} = 0 \implies 0x = 1\right),$$ is absolutely true. I assume you want to refer to the fact that $$Q \equiv \left(\frac{1}{x} = 0\right)$$ is never true, but that's not what it reads now. – TMM Jul 8 '12 at 22:03