# Rigorous proof

Give a rigorous proof using the appropriate axioms and the definition ,$$\frac{a}{b}=a\frac{1}{b}$$ of the following:

$$\frac{a}{b}*\frac{c}{d}=\frac{ac}{bd}$$

$$\frac{a}{b}:\frac{c}{d} =\frac{ad}{bc}$$

quasar987
Homework Helper
Gold Member
According to the forum guidelines, you have to show what you've tried and explain what you don't understand before receiving help.

$$\frac{a}{b}*\frac{c}{d}= a\frac{1}{b}*c\frac{1}{d}=ac*\frac{1}{bd}=\frac{ac}{bd}$$

That is how far i can go.But then this is not a rigorous proof, is there??

$$\frac{a}{b}*\frac{c}{d}= a\frac{1}{b}*c\frac{1}{d}=ac*\frac{1}{bd}=\frac{ac}{bd}$$

That is how far i can go.But then this is not a rigorous proof, is there??
You have not stated which axioms/theorems you used in the second equality.

commutativity and associativity

commutativity and associativity
Right. How did you prove the equality:(1/b)(1/d) = 1/(bd) ?

quasar987
Homework Helper
Gold Member
The whole point of these exercices is to justify each of your steps with the proper axioms. As Slider142 said, now all thats remains to be justified is (1/b)(1/d) = 1/(bd).

But you should use more parenthesis, to highlight the use of associativity. That is, instead of

$$\frac{a}{b}*\frac{c}{d}= a\frac{1}{b}*c\frac{1}{d}=ac*\frac{1}{bd}=\frac{ac }{bd}$$

better is

$$\left(\frac{a}{b}\right)\left(\frac{c}{d}\right)=\left(a\frac{1}{b}\right)\left(c\frac{1}{d}\right)=\left(\left(a\frac{1}{b}\right)c\right)\frac{1}{d}=\left(a\left(\frac{1}{b}c\right)\right)\frac{1}{d}=\left(a\left(c\frac{1}{b}\right)\right)\frac{1}{d}=\left(\left(ac\right)\frac{1}{b}\right)\frac{1}{d}=(ac)\left(\frac{1}{b}\frac{1}{d}\right)=(ac)\left\frac{1}{bd}=\frac{ac}{bd}$$

and the justifications are: by definition, by associativity, by associativity, by commutativity, by associativity, by associativity, by <insert justification why (1/b)(1/d) = 1/(bd)>, by definition.

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Right. How did you prove the equality:(1/b)(1/d) = 1/(bd) ?

The only proof i can thing of ,is the following:

$$\frac{1}{b}\frac{1}{d} = \frac{1}{bd}\Longleftrightarrow (bd)\frac{1}{b}\frac{1}{d} = (bd)\frac{1}{bd}\Longleftrightarrow d(b\frac{1}{b})\frac{1}{d} = 1\Longleftrightarrow d\frac{1}{d} = 1$$.

But then again this is not a rigorous proof ,is there??

The only proof i can thing of ,is the following:

$$\frac{1}{b}\frac{1}{d} = \frac{1}{bd}\Longleftrightarrow (bd)\frac{1}{b}\frac{1}{d} = (bd)\frac{1}{bd}\Longleftrightarrow d(b\frac{1}{b})\frac{1}{d} = 1\Longleftrightarrow d\frac{1}{d} = 1$$.

But then again this is not a rigorous proof ,is there??
That's a perfectly rigorous proof, assuming your definition of 1/(bd) is as the multiplicative inverse of bd. This completes the rigor of your proof.

Could the following proof of :

$$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b}.\frac{d}{c}$$, be consider as a rigorous proof??

$$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b}.\frac{1}{\frac{c}{d}}$$ =......................................................by definition of division x/y = x.(1/y) where we put x = a/b and y= c/d

= $$\frac{a}{b}.\frac{1}{c.\frac{1}{d}}$$ =........................................................by the definition of the division again where we put x=c and y= d

= $$\frac{a}{b}\frac{1}{c}.\frac{1}{\frac{1}{d}}$$ =..........................................................by the use of the theorem (1/x).(1/y) = 1/xy where we put x=c and y= 1/d

= $$\frac{a}{b}.\frac{1}{c}.(1.\frac{1}{\frac{1}{d}})$$ =.........................................................by the use of the axiom 1.x = x where we put $$x=\frac{1}{\frac{1}{d}}$$

= $$\frac{a}{b}.\frac{1}{c}.[(d.\frac{1}{d}).\frac{1}{\frac{1}{d}}]$$ =......................................................by the fact x.(1/x) = 1 where we put x=d

= $$\frac{a}{b}.\frac{1}{c}.[d.(\frac{1}{d}.\frac{1}{\frac{1}{d}})]$$ = ....................................................by the associative law

=$$\frac{a}{b}.\frac{1}{c}.(d.1)$$ = .........................................................................again by the fact x.(1/x) = 1 where we put x= 1/d

= $$\frac{a}{b}\frac{1}{c}.d$$ = .........................................................................again by the axiom 1.x =x

= $$\frac{a}{b}\frac{c}{d}$$ = .............................................................................again by the use of the definition x/y = x.(1/y) where we put x=d and y = c