This works if you multiply both sides with $bd$ and cancel stuff out... But how does it work?

When I look at it, I would never guess something like that is valid without resorting to the established arithmetic rules.

Maybe this is a nonsense question, should these things be analyzed in such a way or just accepted from the arithmetic rules? Since, that's the reason humanity has developed mathematics. To simplify and abstract things which would otherwise be out of the reach of our mind.

Just a simple example, so, please... Set me straight. Thanks!

  • $\begingroup$ Check out this link that Gerry Myerson gave in the comments to your recent question about multiplication. My suggestion based on this is to give the variables some meaningful physical units, for example let $a$ and $b$ be densities, and let $c$ and $d$ be volumes. Then the equation $ad = bc$ equates two masses. $\endgroup$ – Dan Brumleve Sep 15 '11 at 4:53
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    $\begingroup$ I think this particular implication is not intuitive. Is 50 miles in 1 hour the same speed as 100 miles in 2 hours? Multiply by 2 squared-hours and you get 100 mile-hours on both sides... Wait, what? $\endgroup$ – Rahul Sep 15 '11 at 5:05
  • $\begingroup$ Rahul, I guess you have to choose the right units for it to make sense. Maybe all this method really shows is that it might work because the units are the same on both sides. $\endgroup$ – Dan Brumleve Sep 15 '11 at 5:13
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    $\begingroup$ You do need to assume $b$ and $d$ are nonzero. $\endgroup$ – Robert Israel Sep 15 '11 at 7:14

Would a picture help? enter image description here

Note that $\frac{a}{b} = \frac{c}{d}$ by similar triangles. The blue and green rectangle has area $ad$ while the green and yellow rectangle has area $bc$. These are equal, namely fraction $\frac{a}{b} = \frac{c}{d}$ of the area of the big rectangle.

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    $\begingroup$ Well, I know that the two areas should be equal, but I do not see that at all. In fact, just from the picture I would say that the green and yellow rectangle has a larger area. $\endgroup$ – Srivatsan Sep 15 '11 at 15:00
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    $\begingroup$ @Srivatsan: One way to see the areas are equal: The red line divides the $b\times d$ rectangle in half. The two white triangles have the same area, and the two green triangles have the same area. (Since in both cases they are half of a rectangle) Hence the yellow and blue square have the same area. (Otherwise the biggest two triangles would not have the same area) $\endgroup$ – Eric Naslund Sep 15 '11 at 17:39
  • $\begingroup$ @Eric Nice. That's more intuitive. $\endgroup$ – Srivatsan Sep 15 '11 at 17:40

HINT $\ $ It boils down to putting the two fractions over the common denominator $\rm\:b\:d\:,\:$ or, equivalently, changing the "unit" of measurement on your ruler from $1$ to $\rm\:1/(b\:d)\:.$

On the new ruler $\rm\ \dfrac{1}b\ $ has measure $\rm\ d\ $ since $\rm\ \dfrac{1}b\: =\ d\:\dfrac{1}{b\:d}\ $ hence $\rm\ a\:\dfrac{1}b\ $ has measure $\rm\ a\:d\:.$

Similarly $\rm\ c\dfrac{1}d\ $ has measure $\rm\:c\:b\:.$

Analogously, you can use this ruler to compare any fractions whose denominator divides $\rm\:b\:d\:.$


If $\frac{a}{b}$ represents "the solution to the equation $bx=a$", then saying that $\frac{c}{d}=\frac{a}{b}$ means that any solution to $bx=a$ is a solution to $dy=c$, and vice-versa. So if $x$ is a solution to $bx=a$, then multiplying by $d$ we have $ad = dbx = b(dx)$. But since $x$ is also a solution to $dy=c$, that means that $dx=c$, so $ad=b(dx) = bc$.

So if $\frac{a}{b}=\frac{c}{d}$, then $ad=bc$.

Conversely, if $ad=bc$, and $x$ is a solution to $bx=a$, then it is also a solution to $dbx = da=bc$. Since $b\neq 0$, $dbx = bc$ if and only if $dx=c$, so $x$ is a solution to $bx=a$ if and only if it is a solution to $cy=d$.

In short, the equations $bx=a$ and $cy=d$, with $a,b,c,d$ integers, $b$ and $d$ nonzero, have the same solution if and only if $ad=bc$. So if $\frac{r}{s}$ for integers $r,s$, $s\neq 0$, represents "the solution to $sx=r$", then for integers $a,b,c,d$, $b\neq 0$, $d\neq 0$, $$\frac{a}{b}=\frac{c}{d}\text{ if and only if }ad=bc.$$

  • $\begingroup$ Thank you for your answer, I enjoyed it! $\endgroup$ – Curiosity Sep 18 '11 at 5:21

Lately, I've been thinking about it this way:

If $\frac{a}{b} = \frac{c}{d} $ then $ \frac{c}{d} $ must be equal to $\frac{ka}{kb} $. Their ratio is the same if and only if $a$ and $b$ are scaled by a constant $k$.

Therefore $\frac{a}{b} = \frac{c}{d} $ can be rewritten as $\frac{a}{b} = \frac{ka}{kb} $.

If the constant $k$ is the same, which is the condition of equality, we can simply cancel it out, which will leave us with $\frac{a}{b} = \frac{a}{b} $ which is evidently true.

If we didn't want to cancel out the constant $k$, we can try the $ad = bc$ method. Let's move $kb$ and $b$ to the other sides of the equation:

$akb = kab$ -> $akb = akb$

which says that if the constant in the numerator and the denominator is different, the equality fails. Is this also a way of looking at this?

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    $\begingroup$ That'd be a "numerator", not a "nominator". But yes, that's one way to look at it... $\endgroup$ – J. M. is a poor mathematician Sep 18 '11 at 5:18
  • $\begingroup$ Oh, man... I failed so much right now. Thanks for the correction and the confirmation! $\endgroup$ – Curiosity Sep 18 '11 at 5:20
<------a------> <--d-->
<---b---> <-----c----->
         ^     ^
         | a-b |
         | c-d |  

That is a+d=b+c <-> a-b=c-d. Then, assume this was drawn on logarithmic ruler; therefore, substitute "+" for "*" and "-" for "/"

  • $\begingroup$ Very good answer, I wonder why nobody upvoted it! $\endgroup$ – Generic Human Jul 23 '12 at 10:52
  • $\begingroup$ It was down voted initially (due to poor phrasing?) $\endgroup$ – Tegiri Nenashi Jul 26 '12 at 17:23

An intuitive way is to think of a money problem. $a$ is the money that John produced in one hour then $ad$ is the total after $d$ hours. Then let $c$ be the money that Paul produced.

Paul wants to work enough hours $d$ to produce the same rate of money like John. Then Paul needs to work $\frac{a}{c}d$ hours to produce the same amount per hour like John.


You may wish to read through the comments/answers to this related SE question:

Is there any toy for learning algebraic manipulation of fractions?

In particular, I think it helps to think of transforming fractions using the single rule: If a quantity moves across the $=$ sign, it must change positions in the fraction -- numerator becomes denominator, and vice versa.

Hope this helps!

  • $\begingroup$ Though I realize this doesn't answer the OP's question concerning the intuition behind this procedure. I must defer to @Robert Israel's answer for that, which is very nice. $\endgroup$ – Shaun Ault Sep 15 '11 at 12:32

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