__EDEXEL A- LEVEL MATHS__

__Paper 1 and paper 2 (Pure maths)__

**1. Proof**

**– Proof by deduction**

**P****roof by deduction** is the proof that something is true by showing that it must be true for all instances that could possibly be considered.

Example : Prove that f(x) = x^{2} -4x +10 is always positive

Answer : This can be rewritten as f(x) = (x – 2)^{2} + 6 and

f(x)>0 ∀ x ∈ R

**– Proof by Exhaustion**

**Proof by Exhaustion** is the demonstration that something is true by showing that it is true for each and every case that could possibly be considered.

Example : Prove that (n +1)^{3 } ≥ 3^{n} for n ∈ N and n ≤ 4

Answer : Prove that the statement is true for n=0, n=1, n=2, n=3 and n=4

**– Disproof by Counter example**

**Disproof by counterexample** is the technique used in mathematics where a statement is shown to be wrong by finding a single example whereby it is not satisfied.

Example : Prove that all prime numbers are odd

Answer : 2 is considered a prime number, and this disproves the statement by a counter example

Example 2: If sin x = \(\sqrt{1 - cos^2 x}\)

Answer 2 : For x = \(\frac{3π}{2}, sin x = -1\) and

\(\sqrt{1 - cos^2 (\frac{3π}{2})}\) = \(\sqrt{1 - 0}\) = 1

Example 3: If x≥\(\sqrt{x}\) for x>0

Answer 3 : x = \(\frac{1}{4}\) => \(\sqrt{\frac{1}{4}}\) = \(\frac{1}{2}\)

Example 4 : If x^{2} = y^{2} then x = y

Answer 4 : x = -4 => y = 4 => x≠y

** – Proof by Contradiction**

**P****roof by ****contradiction** is the technique used in mathematics to prove a statement is correct *by assuming that it is wrong, then proving that the assumptions are wrong by using logical steps to show that the assumption leads to something impossible. Then conclude that the assumption is wrong and the original statement is correct.*

Example 1 : Prove that the primes are finite in number

Answer 1 : Let the set of prime numbers de {p_{0} , p _{1}, p_{3}…………p_{n}};

The product of the prime numbers is p_{0} , p _{1}, p_{3}…..p_{n }; Add 1 and you obtain +1, this number is not divisible by any of the prime numbers , since there is a remainder of 1. There is a contradiction since there is a number greater than p_{n}

Example 2 : Prove that \(\sqrt{2}\) is an irrational number.

Answer 2 : Let \(\sqrt{2}\) = \(\frac{p}{q}\) (irreducible) => 2 = \(\frac{p^2}{q^2}\) => \(2q^2\) = \(p^2\) => \(p^2\) is even and p is even, then there is k such that p = \(2c^2\) . There is a contradiction, p and q cannot be both even.

Example 3 : Prove that there are no integer solutions to

24p + 12q = 1

Answer 3 : 2p + q = \(\frac{1}{12}\) ,Two integers cannot add together to give a fraction

Example 4 : If n^{2 } is even, then n must be even.

Answer 4 : Let n be odd then, n = 2k + 1 => n^{2 } = (2k+ 1)^{2} => n^{2 } = 4k^{2 } + 4k +1 = 4(k^{2 } + k) + 1 ; so n^{2} is odd which is a contradiction.