- #1

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[tex]\frac{a}{b}*\frac{c}{d}=\frac{ac}{bd}[/tex]

[tex]\frac{a}{b}:\frac{c}{d} =\frac{ad}{bc}[/tex]

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- Thread starter poutsos.A
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- #1

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[tex]\frac{a}{b}*\frac{c}{d}=\frac{ac}{bd}[/tex]

[tex]\frac{a}{b}:\frac{c}{d} =\frac{ad}{bc}[/tex]

- #2

quasar987

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- #3

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

- #4

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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.

- #5

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commutativity and associativity

- #6

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commutativity and associativity

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

- #7

quasar987

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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

[tex]

\frac{a}{b}*\frac{c}{d}= a\frac{1}{b}*c\frac{1}{d}=ac*\frac{1}{bd}=\frac{ac }{bd}[/tex]

better is

[tex]\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}[/tex]

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.

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

[tex]

\frac{a}{b}*\frac{c}{d}= a\frac{1}{b}*c\frac{1}{d}=ac*\frac{1}{bd}=\frac{ac }{bd}[/tex]

better is

[tex]\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}[/tex]

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.

Last edited:

- #8

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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:

[tex]\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[/tex].

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

- #9

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

[tex]\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[/tex].

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.

- #10

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[tex]\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b}.\frac{d}{c}[/tex], be consider as a rigorous proof??

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

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

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

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

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

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

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

= [tex]\frac{a}{b}\frac{1}{c}.d [/tex] = .........................................................................again by the axiom 1.x =x

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

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