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$\dfrac ab= \dfrac cd$ is true precisely when $ad = bc$.

As an example $\frac 26 = \frac 39$ is true because $2*9 = 3*6$.

$\frac 26$ can be thought of as dividing a pie into $6$ equal slices, then taking two of them.

Similarly $\frac 39$ refers to dividing a pie into $9$ equal slices, then taking $3$ of them.

In this particular scenario I'm struggling a bit to understand if the expressions $2*9$ or $6*3$ represent any meaningful quantities, like total number of slices in both pies etc. I would appreciate any help. Thanks!

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    $\begingroup$ In both cases, you're taking a third of the pie. $\endgroup$ – Andrew Li Feb 26 '18 at 5:47
  • $\begingroup$ I would have said the same thing because in all cases they are the same amount of pie as long as the pies are the same size. $\endgroup$ – crowie Feb 26 '18 at 12:29
  • $\begingroup$ You may not be able to say the same thing that quickly when the fractions get messy $$\dfrac{91}{3913}~~~ ?= ~~~\dfrac{39}{677}$$ It seems cross multiplication is the fastest way to figure out if two fractions are equal $\endgroup$ – rsadhvika Feb 26 '18 at 12:36
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    $\begingroup$ @rsadhvika At first glance I'd say these fractions aren't equal, because 91 is less than 3 times 39 where 3913 is way more than 3 times 677, and I didn't use cross multiplication for this $\endgroup$ – Rafalon Feb 26 '18 at 13:30
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    $\begingroup$ Of course, if either b = 0 or d = 0 this whole thing breaks down... $\endgroup$ – twalberg Feb 26 '18 at 16:51
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Suppose you cut the pie into $6 \cdot 9 = 54$ slices, and call each of those a "unit slice". Then both $2 \cdot 9=3 \cdot 6$ would denote the same amount of "unit slices", which is $\dfrac{18}{54}=\dfrac{1}{3}$ of the full pie.

Without the pie analogy, just note that $\,\dfrac{2}{6} \color{blue}{\cdot \dfrac{9}{9}}=\dfrac{18}{54}=\dfrac{3}{9} \color{blue}{\cdot \dfrac{6}{6}}\,$.

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  • $\begingroup$ That makes sense ty but I'm still a bit stuck on this. I guess I'm trying to see the connection between below two $$$$ 2*9 represents : (number of slices taken from first pie)*(total number of slices in second pie) $$$$ 3*6 represents : (number of slices taken from second pie)*(total number of slices in first pie) $$$$ I'm not yet seeing that clearly why they must be equal @dxiv $\endgroup$ – rsadhvika Feb 26 '18 at 5:58
  • $\begingroup$ @rsadhvika 2*9 represents : (number of slices taken from first pie)*(total number of slices in second pie) No, rather think of one slice from the first pie as equal to nine "unit slices". $\endgroup$ – dxiv Feb 26 '18 at 6:32
  • $\begingroup$ Ah okay you want to cut each 1/6th slice from first pie into 9 equal slices. Then taking two 1/6th slices from the first pie is equivalent to taking eighteen 1/9*1/6 th slices. I'll think a bit and get back ty :) $\endgroup$ – rsadhvika Feb 26 '18 at 6:49
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    $\begingroup$ They don't have to be equal. The point is that converting $\dfrac ab$ and $\dfrac cd$ to the common denominator $bd$, gives us numerators of $ad$ and $bc$ respectively. $\endgroup$ – steven gregory Feb 26 '18 at 7:44
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Imagine 9 cakes with 6 pieces each. Both taking 3 of 9 cakes, or 2 pieces of each of the 9 cakes result in 18 pieces of cake.

enter image description here

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    $\begingroup$ Sorry to upvote, you were at 1,337 rep :) $\endgroup$ – NoOneIsHere Feb 26 '18 at 22:58
  • $\begingroup$ duh. The user specifically asks for "pie" and you give him a cake example. +1 though xD $\endgroup$ – I.Am.A.Guy Feb 27 '18 at 15:46
  • $\begingroup$ @I.Am.A.Guy Pies need 6 cuts each, a cake can be done in 3. And mathematicians strive, wherever possible, to be <s>lazy</s> efficient. :P $\endgroup$ – Chronocidal Feb 28 '18 at 12:43
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Imagine 2 cakes with 18 strawberries on top.

  1. Take the first cake, cut it into 6, and take 2 slices. (2/6) Each slice has 3 stawberries
  2. Take the second cake, cut it into 9 and take 3 slices. (3/9) Each slice has 2 strawberries.

Now, you have 9 slices with 2 strawberries each (9∗2), and 6 slices with 3 strawberries each (6∗3), for a total of 18 strawberries per cake

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Cross multiplication using pie algebra is possible (and tasty), but a little clumsy.

We can read $\frac 26 = \frac 39$ as $$ \text{2 out of 6 slices of pie} = \text{3 out of 9 slices of pie}.\tag{1} $$ To get from $\frac 26 = \frac 39$ to $2\cdot9=3\cdot6$ we need to multiply both sides with six and nine. And we want to do this in terms of pies!

To multiply first by six, we can apply the original identity to six pies at the same time. We get $$ \text{2 out of 6 slices of pie for 6 pies} = \text{3 out of 9 slices of pie for 6 pies} $$ or $\frac 26\cdot6 = \frac 39\cdot6$. Now we can simplify this, because $$ \text{2 out of 6 slices of pie for 6 pies} = \text{2 whole pies}. $$ Instead of cutting each pie in six pieces, you can cut your pile of six pies in six smaller piles (each containing one pie). We have found the cancellation law $\frac26\cdot6=2$.

Now we have $$ \text{2 whole pies} = \text{3 out of 9 slices of pie for 6 pies} $$ and we want to multiply both sides by nine, as we would do algebraically. As before, this involves applying the same identity nine times and tallying the desserts: $$ \text{$2\cdot9$ whole pies} = \text{3 out of 9 slices of pie for $6\cdot9$ pies}. $$ We need to simplify the right-hand side. Multiplying the number of pies by six has the same effect as multiplying the amount of slices by six in the end, so $$ \text{3 out of 9 slices of pie for $6\cdot9$ pies} = (\text{3 out of 9 slices of pie for 9 pies})\cdot6. $$ We can apply the previously used law of pie cancellation to the expression in parentheses to get $$ \text{3 out of 9 slices of pie for 9 pies} = \text{3 whole pies}. $$ We can now conclude that $$ \text{$2\cdot9$ whole pies} = \text{$3\cdot6$ whole pies}.\tag{2} $$ Every step of the reasoning is reversible, so (1) and (2) are equivalent. The numbers $2\cdot9$ and $3\cdot6$ don't represent any natural quantity directly; they arise through algebraic manipulations, whether you use pies or not.

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  • $\begingroup$ I think the footnote here is quite a good one, that $2 \cdot 9$ and $3 \cdot 6$ don't really have natural representations. Every answer ends up sounding a bit contrived, or introducing some additional, "magic" information (like numbers arising as least common multiples), and I think this is likely why. $\endgroup$ – pjs36 Feb 26 '18 at 17:29
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    $\begingroup$ @pjs36 The moral seems to be that to cross multiply, you need to, well, multiply. That means somehow taking several instances of the same division, which somehow always messes up the pie interpretation. $\endgroup$ – Joonas Ilmavirta Feb 26 '18 at 18:07
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The first pie has 6 pieces. Divide each piece into 9 smaller pieces. Then 2*9 represents: (number of large pieces taken from first pie) * (number of small pieces in a large piece of the first pie) = (number of small pieces taken from first pie).

Now do the same thing for the second pie, this time splitting each of the 9 pieces into 6 smaller pieces. So 3*6 represents the number of smaller pieces taken from the second pie.

Finally notice that the total number of small pieces is the same in both pies, i.e. each pie is splitted into 6*9=54 small equal pieces. So, assuming the two pies are of the same size, a small piece of the first pie must have the same size with a small piece of the second pie. Thus, if you were to take an equal amount from each pie, you would have to take the same number of small pieces from each pie. I.e. 2*9 = 3*6.

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Suppose you have two pies, each weighing 54 ounces. You divide the first into $6\text{ } 9-$ounce slices and take $2$ of them, getting $2 \cdot 9=18$ ounces of pie. You divide the second pie into $9\text{ } 6-$ounce slices, and take $3$ o them, getting $3\cdot 6=18$ ounces of pie.

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It is about bringing the fractions to the "simplest common denominator" (more often than not it will not be the lowest, but its construction is a single multiplication which one may consider simple) $$\frac ab\stackrel{?}{=}\frac cd\quad\Rightarrow\quad\frac {ad} {bd}\stackrel{?}{=}\frac {bc}{bd}\quad\Rightarrow\quad ad\stackrel{?}{=}bc$$, if $bd\ne0$.

Pie-wise it means that when a pie is divided into $bd$ slices, both fractions can be expressed with taking an integer number of slices (assuming that all $a$, $b$, $c$ and $d$ are integral numbers). So you do not need multiple pies, but you do need a sharp knife, a steady hand (especially since it can be much more than the lowest necessary number of slices) and a good eye.

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  • $\begingroup$ FYI: => looks really clunky, try \Rightarrow to get $\Rightarrow\,$. Even better, \quad\Rightarrow\quad gives a bit of space around it. $\endgroup$ – David Feb 26 '18 at 23:39
  • $\begingroup$ @David Thank you, applied the magic. I also looked up \stackrel in the process $\endgroup$ – tevemadar Feb 27 '18 at 0:18
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You know that 2/6 th of a pie is the same as 3/9 th of a pie. Or, we can say 2/6 * 1 pie = 3/9 * 1 pie. Now, suppose there are 54 such pie specimens. So, by unitary method, what holds true for one specimen, holds true for 54 such specimens. So, 2/6 th of a pie *(times) 54 such pie specimen is same as 3/9 th of a pie *(times) 54 such pie specimen. Now, 2/6 * 54 can be simplified as 18 and 3/9 * 54 can also be simplified as 18, which means you can make 18 complete pie specimens by joining 2/6th of 54 such pie specimens or by joining 3/9 th of 54 such pie specimens. I hope I could extend your logic.

You can also take an example of a Pizza. Suppose, McDonald's and Domino's both make Pizzas of the same size. To make things unique, McDonald's decides to sell it by cutting a Pizza into 9 equal slices and Domino's decides to sell their Pizza by cutting into 6 equal slices. A worker decides to take 3 slices away from the 9 slices of 54 such Pizzas in McDonald's. Another worker at Domino's decides to take 2 slices away from the 6 slices from 54 such Pizzas. The worker at McDonald's decides to join these slices to make complete Pizzas and sell them. He manages to form 18 Pizzas each having 9 equal slices. The worker at Domino's also decides the same and he also comes up with 18 complete Pizzas, each having 6 equal slices.

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One way to prove the cross-multiplication rule is to see that (assuming $x\neq0\neq x'$) $$ \frac{x}{y} = \frac{x'}{y'} \Leftrightarrow \frac{xx'}{yx'} = \frac{xx'}{xy'} \Leftrightarrow yx' = xy' $$ In "cake reasoning", this is equivalent to:

  • as starting point, assume you have one cake split in six parts, and two of them are taken (but not yet eaten), and one cake split in nine parts, three of them taken, as in the traditional cake analogy for fractions.
  • cut each slice -- both taken and not taken -- of the first cake in three parts (you are multiplying both terms of the left-hand fraction by $x'$)
  • cut each slice of the second cake in two parts (multiply the terms of the right-hand fraction by $x$)
  • you will have 6 pieces of cake taken from either cake, hence the amount of "taken" cake in the two cases will be equal if (and only if) the slices are the same size: the first cake will now have $6*3=18$ total slices, the second one will have $2*9$.
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