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Lecture 1: Inventing numbers
Julia Collins
25th September 2013
Proof of the irrationality of
√
2
√
We are going to prove that 2 is irrational, which means it can’t be written as a fraction pq
where both p and q are whole numbers. The proof given below was first written in Euclid’s
Elements in about 300BC, but was probably known to people even earlier than that.
The method of proof is known as a proof by contradiction. To prove that something is
true, we assume that it is false and show that this leads to an absurdity. If the statement is
not false, we must conclude that it is true.
√
So suppose that 2 is a rational number; that is, we can write
√
p
2=
(1)
q
where p and q are positive whole numbers. We can further assume that pq is a fraction in its
lowest terms, so that p and q have no common factors. For example, if we had the fraction
4
, we could cancel a factor of 2 from top and bottom to give the equivalent fraction 23 . In
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particular, this assumption means that p and q cannot both be even.
Take equation 1 and square both sides:
2=
p2
q2
Multiply both sides by q 2 :
2q 2 = p2
(2)
Since p2 is two times another number, this means that p2 is even. An odd number squared
is always odd (you should check this for yourself!), so we can deduce that p is even. Since p
is even, we can write it as two times some other number, say p = 2m for some number m.
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Now substitute this formula for p into equation 2:
2q 2 = p2 = (2m)2 = 4m2
Cancel a factor of 2 from each side:
q 2 = 2m2
As before, we can deduce that q 2 is even (because it’s a multiple of 2), and therefore that q
is even.
Now we have shown that both p and q must be even,√but this contradicts the assumption
we made at the beginning! Thus we have shown that 2 cannot be rational, proving that it
must be irrational.
√
For more proofs of the irrationality of 2, see the Wikipedia article
http://en.wikipedia.org/wiki/Square_root_of_2
which gives 5 different proofs! It also gives some nice properties of this interesting number.
Why an irrational number has a decimal which never
repeats
Suppose we had a number with a repeating decimal expansion. For example, let’s take
x = 0.127127127 . . . . This number repeats after every 3 decimal places, but in general we
might pick a number that repeats after n decimal places.
Multiply your number by 10n . In our case we multiply by 103 = 1000 to get
1000x = 127.127127127...
and in general it will have the effect of shifting the digits so that the repeated part is now
in front of the decimal point.
Now subtract the two numbers:
1000x − x = 999x = 127
since everything after the decimal place disappears. But now the equation tells us that
x = 127
, so that x is a rational number.
999
So any decimal with recurring digits has to be rational. This also means that no irrational
number can have a recurring decimal.
2
Why is negative times negative positive?
The definition of a negative number −x is that it is the number which needs to be added to
x to get zero. So x + (−x) = 0.
There are a few different proofs of the fact that negative times negative is positive. For some
people it will be intuitive: the negation of a debt is a credit. Or if you think of a negative
number as a reflection of the number line, then two reflections gets you back to the start.
Using the definition of a negative number, we could say that −(−1) is the number which,
when added to −1, makes zero. We know that −1 + 1 = 0, so −(−1) = 1.
Here is a more involved proof which uses the fact that positive times negative is negative.
This fact is quite intuitive; for example
3 × (−2) = (−2) + (−2) + (−2) = −6 = −(2 × 3)
The proof is going to work by multiplying together zeros in a clever way.
0 =
=
=
=
=
0×0
(1 − 1) × (1 − 1)
by definition of a negative number
(1 × 1) + (−1 × 1) + (1 × −1) + (−1 × −1)
by the distributive law of multiplication
1 + (−1) + (−1) + (−1 × −1)
−1 + (−1 × −1)
because 1 + (−1) = 0, and 0 + x = x
Therefore (−1 × −1) must be equal to 1. This type of proof will work with any two negative
numbers to show that their product is positive.
Using complex numbers to find real solutions of a cubic
A cubic equation is anything of the form
ax3 + bx2 + cx + d = 0
where a, b, c, d are any (real) numbers. Some cubic equations were studied by the ancient
Greeks and Babylonians, and many civilisations found numerical methods to get the solution,
but it was 15th century Italians who finally found a formula for the answer. A nice description
of the main characters involved can be found at
http://www.storyofmathematics.com/16th_tartaglia.html
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For special kinds of cubic equations of the form
x3 + cx + d = 0
Cardano/Fiore/Tartaglia had developed a formula that would give the answer. The formula
was quite horrible! It turned out that x = u + v where
r
d2
c3
d
3
+
u =− +
2
4
27
and
d
v3 = − −
2
r
d2
c3
+
.
4
27
Consider the cubic
x3 − 15x − 4 = 0
This cubic has got an obvious whole number answer, namely x = 4. What happens when
we try to work this out using the formula above? We get
r
(−4)2 (−15)3
−4
+
+
u3 = −
2√
4
27
= 2 + 4 − 125
√
= 2 + −121
At this point, Cardano stopped and said that the equation was nonsense, because it was
impossible to take the square root of a negative number. It was a hydraulic engineer called
Rafael Bombelli who stopped to ask the question, “What happens if we assume that we can
take negative roots
and just continue with the algebra?”
√ √
√
√
Let us denote −1 by the letter i. (‘i’ for imaginary!) Then −121 = 121 −1 = 11i. So
we have
u3 = 2 + 11i
and
v 3 = 2 − 11i .
It is not too difficult to show that u = 2 + i and v = 2 − i. Try cubing these complex
numbers and check that they give you the answers above! (Remember that i2 = −1 and
i3 = i × i2 = −i.)
Now Cardano’s formula says that x = u + v = (2 + i) − (2 − i) = 4, which is exactly the
answer we were expecting! As if by magic, all the imaginary numbers cancel out to give
the real answer, and nobody need ever know that complex numbers were used to find the
solution...
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Homework problem
The challenge for Week 1 is to find a pair of numbers, x and y, such that the sum x + y is
equal to 10, and the product xy is equal to 40. The answer is on the next page, so don’t
peek until you’ve had a go!
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Homework solution
The solution is that x = 5 +
√
−15 and y = 5 −
√
−15.
Here’s how you work it out. You have
x + y = 10
and
xy = 40.
Rearrange the first equation to get y = 10 − x, and the substitute this for y in the second
equation to get
x(10 − x) = 40.
Multiply this out to get 10x − x2 = 40 and rearrange to put all the terms on the same side
of the equation: x2 − 10x + 40 = 0.
At the point you have to remember some high-school maths: the infamous quadratic formula!
If you have a quadratic equation ax2 + bx + c = 0 then the two solutions are given by
√
−b ± b2 − 4ac
.
x=
2a
In our situation, the coefficients are a = 1, b = −10 and c = 40. Stick the numbers in:
√
√
√
10 ± 100 − 160
10 ± −60
x=
=
= 5 ± −15
2
2
√
√
√
√
(remembering that −60 = −15 × 4 = −15 × 4). Ta daa!
The final thing to do is check that the two solutions actually answer the question correctly.
x + y = (5 +
√
−15) + (5 −
√
−15) = 10
because the square roots cancel out. And
√
√
√
√
√
xy = (5 + −15)(5 − −15) = 25 + 5 −15 − 5 −15 − ( −15)2 = 25 − (−15) = 40
because if you square a square root you get the number you first thought of.
(Please email me or let me know if you find any mistakes!)
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