WJEC Maths for A2 – Pure

1 Proof 10 1.1 Real, imaginary, rational and irrational numbers You came across this material in AS Pure, but as we will be using the terminology a lot in this topic, here is a reminder: Real numbers are the numbers we are used to using. So 2, 0, 0.98, 1 3 , −3, π , √ 2, etc., are all examples of real numbers. Imaginary numbers are those numbers that when squared give a negative number. So √ −1 is an imaginary number as ( √ −1) 2 = −1. Rational numbers can be expressed as fractions ( i.e. a b , where b ≠ 0 ) . Irrational numbers cannot be expressed as fractions. Numbers where there are numbers after the decimal point that go on forever and do not repeat are irrational, e.g. √ 2 and π . 1.2 Proof by contradiction Proof is concerned with the demonstration of the truth of a conjecture. The essential steps in proof by contradiction are to assume that the conjecture is false and then show that this assumption leads to a contradiction . Sometimes it is difϐicult or almost impossible to prove that a conjecture such as ‘ √ 2 is irrational’ is true. The problem is that there are an inϐinite number of numbers whose squares might equal 2, so we could not test them all. It is easier to consider the contradiction which is that ‘ √ 2 is rational’. We than have to consider logical steps based on what we know is mathematically true and when we meet something we know is mathematically false it means that the assumption made for the contradiction is incorrect and that the original assumption was indeed true. Examples 1 Complete the following proof by contradiction to show that if n is an integer such that n 2 is divisible by 3, then n is divisible by 3. Assume that n does not have a factor 3 so that n = 3 k + r where k and r are integers and 0 < r < 3. Answer 1 Then n 2 = ȍ3 k + r Ȏ 2 = 9 k 2 + 6 kr + r 2 which does not have a factor of 3. But n 2 has a factor of 3 (given). ∴ there is a contradiction and our assumption that n does not have a factor of 3 is false. 2 Prove by contradiction the following proposition. When x is real and positive, 4 x + 9 x ≥ 12 . The ϐirst line of the proof is given below. Assume that there is a positive and real value of x such that 4 x + 9 x < 12 Proofs are difϐicult because they can apply to so many parts of the speciϐication. This is an area where you really need lots of practice before you become proϐicient. Stick with it. BOOST G ra d e ⇪⇪⇪⇪ r = 1 or 2 r 2 = 1 or 4