WJEC Maths for A2 – Pure

1.5 Application of proof by contradiction to unfamiliar proofs 13 1.4 Proof of the infinity of primes There is no maximum prime number. Here is one way this can be proved. Suppose we have the following list of prime numbers: p 1 , p 2 , p 3 , p 4 , p 5 , … p n When all of them are multiplied together and 1 is added, the resulting answer, which we can call q , may or may not be prime. If it is prime it is a prime number that was not in the original list and so is a new prime number. If it is not prime, it must be divisible by a prime number r . Now r cannot be p 1 , p 2 , p 3 , etc., as dividing q by any of these numbers would result in a remainder of 1. This means that r is a new prime number. So there can be a new prime number q or if q is not prime it has a new prime for a prime factor. Hence there are an inϐinite number of primes. 1.5 Application of proof by contradiction to unfamiliar proofs Some of the questions on proofs will refer to proofs you are familiar with. For example, the proof for proving √ 7 is irrational is much the same as the familiar proof that √ 2 is irrational. You will get questions in the exam where proof by contradiction needs to be used in unfamiliar situations. Sometimes these questions are in questions that are not entirely on proof. For example, a question on trigonometry may have a part where you have to prove a particular statement. The examples below give you a good range of these types of questions. Example 1 If a and b are real integers, then a 2 − 4 b ≠ 2. Use proof by contradiction to prove that the above statement is true. Answer 1 Suppose we assume that this statement is false. Assuming there are values for a and b such that a 2 − 4 b = 2 Rearranging gives a 2 = 4 b + 2 As a 2 = 2(2 b + 1) it means that a 2 must be even (as it has 2 as a factor). This also means a must be even as if you square an odd number you always obtain an odd number and squaring even numbers results in even numbers. We can write a = 2 c for a certain integer c . Substituting a = 2 c into the formula, we obtain 4 c 2 = 2(2 b + 1) Rearranging, gives 4 c 2 − 4 b = 2 Dividing both sides by 2 gives 2 c 2 − 2 b = 1 so 1 = 2( c 2 − b ) Now as b and c are both integers this cannot be true as 2( c 2 − b ) is even and 1 is odd. A contradiction has been found, so the original assumption a 2 − 4 b = 2 is false, so a 2 − 4 b ≠ 2 is true. In many ways proofs are difϐicult because the types of questions you get are so variable and it is difϐicult to sometimes see what you have to do. The more questions you do, the better you will get. BOOST G ra d e ⇪⇪⇪⇪ Here we assume that the initial statement in the question is false.