Equivalent Fractions

[jpanknin]

I'm going through some proofs discussing equivalent fractions. The book gives the following scenario:

[math]\frac{k}{l} = \frac{m}{n} \implies \frac{kn}{ln} = \frac{ml}{ln} \implies kn = ml[/math]
And likewise:

[math]\frac{k}{l} < \frac{m}{n} \implies \frac{kn}{ln} < \frac{ml}{ln} \implies kn < ml[/math]
I can follow that the equality and inequality holds because they both have same denominator, so the equivalence or inequality would be determined by the numerator. That part makes sense. What still seems like magic is how [imath]kn = ml[/imath] or [imath] kn < ml[/imath]. For example, why does [imath] 3 * 8 = 4 * 6[/imath] in the following fraction:

[math]\frac{3}{4} = \frac{6}{8}[/math]
In other words, why does numerator_1 * denominator_2 = numerator_2 * denominator_1?

 

[lookagain]

I can follow that the equality and inequality holds because they both have same denominator, so the equivalence or inequality would be determined by the numerator. That part makes sense. What still seems like magic is how [imath]kn = ml[/imath] or [imath] kn < ml[/imath]. For example, why does [imath] 3 * 8 = 4 * 6[/imath] in the following fraction:

[math]\frac{3}{4} = \frac{6}{8}[/math]
In other words, why does numerator_1 * denominator_2 = numerator_2 * denominator_1?

The first two sentences of your longer paragraph indicate to people that you
understand why, so then you should not be asking why in a specific case.

\(\displaystyle \dfrac{3}{4} \ = \ \dfrac{6}{8} \)

\(\displaystyle \dfrac{3*8}{4*8} \ = \ \dfrac{4*6}{4*8} \)

The denominators are the same, so the numerators are the same.

\(\displaystyle 3*8 \ = \ 4*6 \)

 

[Agent Smith]

From [imath]\frac{a}{b} = \frac{c}{d}[/imath], you may infer [imath]\frac{a}{b} - \frac{c}{d} = 0[/imath]

 

[Steven G]

If a/b = c/d, then what do YOU get when you multiply both sides by bd? You do not need to get common denominators to arrive at ad=bc. Rather, you want to get rid of denominators.

Having said that, I do believe that if you have two fractions with the same denominator (or numerator), then you should be able to tell which fraction is larger. For example, if you are given 7/-3 and 8/-3 then which fraction is larger?

\(\displaystyle \dfrac{kn}{ln}<\dfrac{ml}{ln}\) implies kn < ml is total nonsense. Just consider what what happens if ln<0 and you multiply both sides by ln.

I suspect that you meant to say than k, l, m a nd n are all positive.

 

[Steven G]

[math]\frac{3}{4} = \frac{6}{8}[/math]
In other words, why does numerator_1 * denominator_2 = numerator_2 * denominator_1?

Multiply both sides by 4*8 and see what you get!! You gotta try!

 

[Agent Smith]

If a/b = c/d, then what do YOU get when you multiply both sides by bd? You do not need to get common denominators to arrive at ad=bc. Rather, you want to get rid of denominators.

Having said that, I do believe that if you have two fractions with the same denominator (or numerator), then you should be able to tell which fraction is larger. For example, if you are given 7/-3 and 8/-3 then which fraction is larger?

\(\displaystyle \dfrac{kn}{ln}<\dfrac{ml}{ln}\) implies kn < ml is total nonsense. Just consider what what happens if ln<0 and you multiply both sides by ln.

I suspect that you meant to say than k, l, m a nd n are all positive.

Is this a reply to me?

I was trying to suggest (true/false?) [imath]\frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd} = 0 \implies ad = bc \text{ or } kn = ml[/imath]

 

[Agent Smith]

I think more difficult to wrap your head around is [imath]\frac{a}{b} = \frac{ma}{mb}[/imath]

 

[Agent Smith]

For example, if you are given 7/-3 and 8/-3 then which fraction is larger?

I don't think [imath]-a < a[/imath]

 

[Steven G]

I don't think [imath]-a < a[/imath]

You are correct as long as a<0, but what if a>0.

 

[Agent Smith]

You are correct as long as a<0, but what if a>0.

I don't think [imath]-2 < 2[/imath]

 

[fresh_42]

In other words, why does numerator_1 * denominator_2 = numerator_2 * denominator_1?

This is basically due to the definition of fractions. What is a fraction and why is e.g. [imath] 1/2=2/4\;[/imath]? The answer lies in the construction of rational numbers from the integers. The integers have some important properties. Firstly, they have an invertible addition where [imath] a+x=b [/imath] has always a solution. Secondly, they have a commutative and distributive multiplication, i.e. [imath] a\cdot b=b\cdot a [/imath] and [imath] a\cdot (b+c) = a\cdot b + a\cdot c. [/imath] Most important in this case is, however, that [imath] a\cdot b=0 [/imath] only if at least one of the two factors equals zero. The last property allows us to define fractions. We first establish an equivalence relation on [imath] \mathbb{Z} \setminus \{0\} [/imath] by
[math] (a,b) \sim (c,d) \Longleftrightarrow a\cdot d = b\cdot c. [/math]which we abbreviate as [math] \dfrac{a}{b}=(a,b). [/math]
That this is an equivalence relation (reflexive, symmetric, transitive) has to be proven, as well as [imath] 0\cdot \dfrac{a}{b}=\dfrac{0}{1}\cdot \dfrac{a}{b}=\dfrac{0}{b}=\dfrac{0}{1}=0. [/imath] We call the equivalence classes a rational number. [imath] 1/2 [/imath] and [imath] 2/4 [/imath] are thus only two different representatives of the same equivalence class. I like to say that half a pie is different from two-quarters of a pie, but it's the same amount of a pie.

That's how rational numbers are formally defined, and which is why the answer to your question is: because of definition/construction.