1

Is there any rule in algebra or any related rules for this equations?

  • if a is not equal to zero $$ \frac{ab}{c} = \frac{b}{\frac{\displaystyle c}{\displaystyle a}} $$

  • if b is not equal to zero $$ \frac{ab}{c} = \frac{a}{\frac{\displaystyle c}{\displaystyle b}} $$

Gaurav Mishra
  • 284
himynameisjm
  • 41
  • Welcome to MSE. Which equations? – José Carlos Santos Oct 23 '22 at 08:36
  • @JoséCarlosSantos People often confuse the words "equality" and "equation". I am (almost) certain the OP wanted to say "equalities". –  Oct 23 '22 at 08:37
  • These rules follow directly from the field axioms. That is, they are immediate consequences of the definition of the division operation, along with the other arithmetic operations. – Jam Oct 23 '22 at 08:42
  • @himynameisjm Are you interested whether those equalities are true (yes!), provable (yes!) or do you want to see an actual proof? In the last case, as with everything in mathematics, you need to tell us what axioms you use, i.e. what you take as the basis of your deduction. Depending on how elementary your axioms are, the proof may look different - and may be very short or incredibly long. (Recall Russel and Whitehead took 379 pages of prerequisites in their Principia Mathematica to prove $1+1=2$.) –  Oct 23 '22 at 08:45
  • Does this answer your question? Proof of dividing fractions $\frac{a/b}{c/d}=\frac{ad}{bc}$ – Jam Oct 23 '22 at 09:41

2 Answers2

1

I think a common rule encountered at lower levels is $$ \frac{\left(\frac ab\right)}{\left(\frac cd\right)} = \frac ab \cdot \frac dc. $$ (Where I’m from we called it “multiplying by the opposite”). This follows from $$ \frac{1}{\left(\frac cd\right)} = \frac dc, $$ which again follows from $$ \frac cd \cdot \frac dc = \frac{cd}{dc}=1. $$ Does that help at all?

Even more fundamentally, in the end it really boils down to $$ \frac{1}{\left(\frac 1d\right)} =d $$ (together with how fractions and multiplications “play nice” with each other).

Milten
  • 7,031
0

Note that

$$\frac{ab}{c}=\frac{ab}{c}×\frac {\frac 1a}{\frac 1a}=\frac{ab×\frac 1a}{c×\frac 1a}=\frac{b}{\frac ca}$$

Similarly

$$\frac{ab}{c}=\frac{ab×\frac 1b}{c×\frac 1b}=\frac{a}{\frac cb}$$

Here we used the rule

$$\frac AB×\frac CD=\frac {AC}{BD}.$$

and the following definition

$$\frac AB=A×\frac 1B.$$

User
  • 1,671