EDEXEL A- LEVEL MATHS
Paper 1 and paper 2 (Pure maths)
1. Proof
– Proof by deduction
Proof 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) = x2 -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 ≥ 3n 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 x2 = y2 then x = y
Answer 4 : x = -4 => y = 4 => x≠y
– Proof by Contradiction
Proof 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 {p0 , p 1, p3…………pn};
The product of the prime numbers is p0 , p 1, p3…..pn ; 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 pn
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 n2 is even, then n must be even.
Answer 4 : Let n be odd then, n = 2k + 1 => n2 = (2k+ 1)2 => n2 = 4k2 + 4k +1 = 4(k2 + k) + 1 ; so n2 is odd which is a contradiction.