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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 6 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. 1 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 3 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... 4 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! 5 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!) 6