Prove using field axioms that $(1/x)+(1/y) = (y+x)/(x*y)$

Question

How can I prove $(1/x)+(1/y) = (y+x)/(x*y)$ with field axioms?

I tried to write it like

$$x{^-}^1+ y{^-}^1 $$

$$x*x{^-}^1+y*y{^-}^1 $$

I am not sure how to process or if this is the right direction..

Answer 1

$$\left(x^{-1}+y^{-1}\right)xy=x^{-1}xy+y^{-1}xy=y+y^{-1}yx=y+x.$$

Thus, $$x^{-1}+y^{-1}=\left(x^{-1}+y^{-1}\right)xy(xy)^{-1}=(y+x)(xy)^{-1}.$$

The first line it's $$\left(x^{-1}+y^{-1}\right)(xy)=x^{-1}(xy)+y^{-1}(xy)=(x^{-1}x)y+y^{-1}(yx)=$$ $$=1y+(y^{-1}y)x=y+1x=y+x.$$

Try to write the second part of the proof by yourself.

Answer 2

Let $x,y \in \mathbb{R}$ \ {$0$}.

RHS: $(y+x)(xy)^{-1}.$

$(xy)^{-1} = y^{-1}x^{-1}.$

$(y+x)[y^{-1}x^{-1}] =$

(distributive law)

$ y[y^{-1}x^{-1}] +x[x^{-1}y^{-1}]=$

(commutative law, associative law)

$(yy^{-1})x^{-1} + (xx^{-1})y^{-1}=$

$1x^{-1} + 1y^{-1} =$

(commutative law)

$y^{-1}+ x^{-1}.$

(LHS)

Left to show: $(xy)^{-1} = y^{-1}x^{-1}.$

Note: $ (xy)$ has a unique inverse element $(xy)^{-1}$.

Can you take it from here?

Answer 3

This is doesn't need to be much different than how you learned to add fractions in elementary school:

$$ \frac{1}{x} + \frac{1}{y} = \frac{y}{xy} + \frac{x}{xy} = \frac{y+x}{xy} $$

Justify the individual arithmetic steps used as needed in a separate argument. (e.g. prove a theorem $\frac{a}{b} = \frac{ac}{bc}$ if $b,c \neq 0$ if that can't be taken as a "known")