Y2, C1 - Algebraic Methods

Question 1

Steps for proof by contradiction

Answer 1

1) Assume statement is false
2) Prove this would lead to a contradiction
3) Therefore we would be wrong in assuming the statement was false and therefore it must be true

Question 2

Use proof by contradiction to show that there are no integers a and b for which 25a + 15b = 1

Answer 2

Assume there are integers a and b which satisfy equation
25a + 15b = 1
5(5a + 3b) = 1
5a + 3b = 1/5
Contradiction as no integers can multiply to create a fraction
Hence NO integers for a and b satisfy equation

Question 3

What are rational numbers

Answer 3

Numbers which can be expressed as fractions (in their simplest form)
a / b

Question 4

What are irrational numbers

Answer 4

Numbers that cannot be expressed as fractions

Question 5

Prove that sqr(2) is irrational

Answer 5

Assume sqr(2) is rational
sqr(2) = a / b
2 = a^2 / b^2
2b^2 = a^2, therefore a^2 is even and thus a is even
Set a = 2k
2b^2 = (2k)^2
b^2 = 2k^2, therefore b^2 is even and thus b is even
a and b are both even and so a / b cannot be simplest form and therefore sqr(2) is irrational

Question 6

Use proof by contradiction to show that, given a rational number a and an irrational number b, a - b is irrational

Answer 6

Assume rational a, irrational b, a - b is rational
a = x / y (rational)
a - b = w / z (rational)
(x / y) - b = w / z
b = (x / y) - (w / z)
All components of b are rational and thus b is rational and not irrational therefore assumption is contradicted

Question 7

Prove by contradiction that there are infinitely many primes

Answer 7

Assume there are a finite number of primes
List finite primes as p1, p2, p3, … , pn
Consider number N = (p1 x p2 x p3 x … x pn) + 1
When you divide N by any primes (p1, p2, …), the remainder will always be 1
Therefore N is not divisible by and primes which is a contradiction of assumption