Fibonacci and the
method of false position
John Hannah (University of
Canterbury)
Fibonacci is mostly known nowadays
for the famous rabbit problem whose
solution is the sequence
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
But he wrote about many other topics,
and his treatment of false position is a
good example of proportional thinking.
In the method of single false position
you guess a solution, plug it into the
problem to see if it works, and then
scale the guess to achieve the desired
target. Fibonacci calls it the tree
method because his first example asks
for the size of a tree.
Example 1. There is a tree, and ⅓ and
¼ of the tree lie underground. If the
part under the ground measures 21
metres, what is the total height of the
tree?
Fibonacci tells us to guess a number
divisible by all the fractions in the
problem. So we guess the total height
to be 12 metres. But then the part
underground measures
4+3=7 metres, instead of 21. He
encourages us to recite the formula, ‘I
put 12 metres and the result was 7
metres. What should I put to get 21
metres?’ and to arrange the
information in a table
which leads to the answer
12 × 21
= 36 metres.
7
Fibonacci was writing 800 years ago,
long before algebra was invented, so
instead of an algebraic explanation, he
offers a picture using proportional
thinking.
This example would be easy to solve
using algebra, but in many of
Fibonacci’s problems setting up the
algebraic equation can be quite tricky
for beginners. My guess is that many
of your students will find it easier to
use false position for the next problem.
Example 2. A lion can eat a sheep in 4
hours, a leopard can do it in 5 hours,
and a bear in 6 hours. If all three
animals eat together, how long will it
take them to eat one sheep?
Guess a number of hours divisible by
each of the numbers, 60 hours say. In
this time the lion can eat 15 sheep, the
leopard 12 and the bear 10. You say, ‘I
put 60 hours and the result was 37
sheep. What should I put to get just 1
sheep?’ Following the recipe gives
60 × 1
23
= 1 hours
37
37
Fibonacci gives lots of different
examples using the method: pipes
filling barrels at different rates, ships
from different ports meeting at sea,
merchants buying and selling at
various rates, and so on. But he also
explores to see if the method works on
other types of problems not covered by
the proportionality explanation.
Guess values which agree with this
proportion, say 5 and 7, respectively.
Then their product is 35 when it ought
to be 5+7=12.
Example 3. Find a number so that if
you add 1 3 , 1 4 , 1 5 and 1 6 of it, and then
square the result, then you get the
original number back again.
Guess that the number is 60 (divisible
by 3, 4, 5 and 6). But 20+15+12+10
squared is 3249 which should be 60.
The numbers are
According to Fibonacci’s method the
correct number should be
60 × 60 400
=
3249
361
and this is indeed correct. But there is
something fishy here. If you say, ‘I put
60 and the result was 3249. What
should I put to get 60?’ then 400 361
looks wrong because putting 400 361
does not give 60. Yet 400 361 is correct.
Fibonacci is enough of a
mathematician to look for an
explanation. He finds a geometric
2
proof that the number should be (20 19 )
but he never finds out why the false
position method gives the same
answer. (That explanation had to wait
until algebra came along.)
Undaunted, he tries the method on
problems with two unknowns.
Example 4. Find two numbers so that
1 of the first equals 1 of the second,
5
7
and so that the product of the two
numbers equals their sum.
The first condition fixes the proportion
of the two numbers to one another.
5 × 12 12
= and
35
7
7 × 12 12
= .
35
5
Fibonacci helpfully tells us that in such
problems we should always divide by
the product (so the product should
always appear in the top row of the
table). This time he doesn’t seem to
know how to explain the solution, so
he offers lots of similar examples
instead, as if to say, ‘See, it always
works!’
This is perhaps a little naughty,
recommending a method which he
can’t fully justify, but it is a snapshot
of mathematics as it is often done in
practice. He has a method which he
can justify in simple cases –
algebraically, any problem expressible
in the form ax = b . He experiments
and find that it also works for other
2
problems – such as (ax ) = x or even
simultaneous equations like ax = by
and xy = c( x + y ) . Then he looks, not
always successfully, for an
explanation. If all else fails, he falls
back on experimental evidence.
Challenge. Fibonacci tries out the
method on problems like
2 x = 3 y , 4 y = 5 z and xyz = x + y + z
but he finds he has to modify the
scaling procedure. How?