Read Professor Stewart's Hoard of Mathematical Treasures Online

Authors: Ian Stewart

Tags: #Mathematics, #General

Professor Stewart's Hoard of Mathematical Treasures (58 page)

BOOK: Professor Stewart's Hoard of Mathematical Treasures
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All very well, but why does the multiplication rule work, and what should we use for addition instead?
The easy way to see why the rules differ - and what they should be - is to use pictures. Here’s a picture for
×
.
Multiplying fractions.
The vertical bar shows a line of five equal pieces, with two of them shaded grey. That represents
: two parts out of five. Similarly the horizontal bar represents
. The rectangles represent multiplication, because the area of a rectangle is what you get when you multiply the two sides. The big rectangle contains 5 × 7 = 35 squares. The shaded one contains 2 × 3 = 6 squares. So the shaded rectangle is
of the big one.
When it comes to addition, the corresponding picture looks like this:
Adding fractions.
We get
of the big rectangle by taking the top two rows out of five, and
by taking the left-hand three columns out of seven. These regions are shown in the left-hand picture, with different shading, and they overlap. To count how many squares there are altogether, we either count the overlapping squares twice, or make an extra copy as in the right-hand picture. Either way, we get 29 squares out of 35, so the sum must be
To see how the 29 relates to the original numbers, just count the squares in the top two rows, 2 × 7, and add those in the left-hand three columns, 3 × 5. Then 2 × 7 + 3 × 5 = 29. So the addition rule is
This is where the usual recipe ‘put both fractions over the same denominator’ comes from.
Pooling Resources
It was a bad idea. Separately, Christine would make £150 and Daphne £100, a total of £250. Together, their total income would be £240 - which is smaller.
Both pairs of ladies are making an unwarranted assumption; it happens to work in favour of the first pair and against the second pair. The assumption is that the way to combine prices of a for £b and c for £d is to add the numbers, getting
a
+
c
for £(
b
+
d
). This boils down to trying to add the corresponding fractions using the rule
and we have seen with the previous puzzles that this doesn’t work. Sometimes it is an underestimate, sometimes an overestimate. It is correct when the two fractions concerned are the same.
BOOK: Professor Stewart's Hoard of Mathematical Treasures
11.6Mb size Format: txt, pdf, ePub
ads

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