- #1

- 102

- 1

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

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 102

- 1

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

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

- #2

Science Advisor

Homework Helper

Gold Member

- 4,809

- 31

- #3

- 102

- 1

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

- #4

- 1,015

- 70

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

- 102

- 1

commutativity and associativity

- #6

- 1,015

- 70

commutativity and associativity

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

- #7

Science Advisor

Homework Helper

Gold Member

- 4,809

- 31

The whole point of these exercices is to justify each of your steps with the proper axioms. As Slider142 said, now all that's 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

- 102

- 1

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

- 1,015

- 70

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

- 315

- 0

[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

Share:

- Replies
- 12

- Views
- 655

- Replies
- 7

- Views
- 579

- Replies
- 11

- Views
- 771

- Replies
- 3

- Views
- 869

- Replies
- 8

- Views
- 1K

- Replies
- 13

- Views
- 1K

- Replies
- 11

- Views
- 898

- Replies
- 9

- Views
- 1K

- Replies
- 11

- Views
- 865

- Replies
- 19

- Views
- 1K