Proof by contradiction ( Algebratic Methods ) Flashcards
1
Q
Real numbers
A
- A number that can be expressed as an infinite decimal expansion
- e.g positive and negative integers, rational ( fractions ) and irrational numbers
2
Q
Rational numbers
A
- can be written as a / b, where b is not equal to 0
- a and b don’t have any common factors
3
Q
Irrational numbers
A
- all numbers that aren’t rational
- so cannot be expressed as a ratio of two integers
4
Q
Non-real numbers
A
- Imaginary numbers
- Cannot be represented on the number line
5
Q
Use proof by contradiction to prove the statement: “The product of two odd numbers is odd.”
A
Assumption:
- Products of two odd numbers is even
- ( 2n+1 ) ( 2m+1 )= 4nm + 2m + 2n + 1
- ( expand odd no. )
- 2( 2nm+m+n ) + 1 is an odd number
- ( factorise )
- hence the product of two odd numbers is odd
- Contradicts the above assumption, so statement is correct
6
Q
Prove that root 2 is irrational.
A
Assumption:
- root 2 is rational
- root 2 = a / b
- ( equating and solving )
- 2 = a^2 / b^2
- 2b^2 = a^2
- ( a^2 is an even no. so a is an even no. )
- let a = 2c
- 2b^2 = ( 2c )^2
- ( substitution )
- 2b^2 = 4c^2
- b^2 = 2c^2
- ( b^2 is an even no. so b is an even no. )
- rational numbers don’t have common factors, therefore this assumption has been contradicted
- hence root 2 is irrational
7
Q
Prove that if x^3 - 3 = 0 then x is irrational
A
- Assumption: if x^3 - 3 = 0 then x is rational
- x = a / b
- ( a/b )^3 - 3 = 0
- ( substitution )
- a^3 / b^3 - 3 = 0
- a^3 / b^3 = 3
- a^3 = 3b^3
- So a is a multiple of 3
- ( a = 3p ( or any multiple of 3 as b^3 is an integer ) )
- ( 3p / b )^3 - 3 = 0
- ( substitution )
- 27p^3 = 3b^3
- 9p^3 = b^3
- b is also a multiple of 3
- therefore a and b are both multiples of 3, hence x^3 - 3 = 0 cannot be rational
- so by proof of contradiction, x is irrational when x^3 - 3 = 0
8
Q
Root 2 + root 3 is irrational
A
Assumption:
- root 2 + root 3 is rational
- root 2 + root 3 = r
- ( r = rational number )
- ( root 2 + root 3 )^2 = r^2
- 2 + 2 root 6 + 3 = r^2
- root 6 = ( r^2 - 5 ) / 2
- root 6 is irrational
- ( r^2 - 5 ) / 2 is rational, since r is rational
- therefore root 2 + root 3 is irrational
9
Q
The sum of a rational number and an irrational number is an irrational number.
A
Assumption:
- The sum of a rational and an irrational number is a rational number.
- a / b = rational
- x = irrational
- m / n = rational
- a / b + x = m / n
- x = m / n - a / b = ( mb - na ) / nb
- m,b,n and a are all rational numbers so, x = ( mb - na ) / nb is rational, hence x is a rational number
- therefore assumption is incorrect
- thus rational number + irrational number = irrational
10
Q
The sum of two positive numbers is always positive.
A
Assumption:
- The sum of two positive numbers is always negative
- a + b < 0
- a < - b
- When both a and b are positive, a < - b, however this means that a is also negative
- Therefore assumption is wrong, sum of two positives cannot be a negative
- Thus the sum of two positives is always positive
11
Q
For all integers a,b,c, if a^2 + b^2 = c^2 then a or b is even.
A
Assumption:
- for all integers a,b,c if a^2 + b^2 = c^2 then a and b are odd
- a = 2m+1
- b = 2n+1
- ( 2m+1 )^2 + ( 2n+1 )^2 = c^2
- 4m^2 + 4m + 1 + 4n^2 + 4n + 1 = c^2
- 4( m^2 + m + n^2 + n ) + 2
- 4k + 2 isn’t a square
- if c = 2a ( even ), c^2 = 4a^2 which is not equal to 4k + 2
- if c = 2a + 1 ( odd ), c^2 = 4a^2 + 4a + 1 which is not equal to 4k + 2
- so 4k + 2 = a^2 + b^2 is not a square
- so assumption that a and b are both odd is false
- hence either a or b are even
12
Q
Prove by contradiction that there are no integers a and b such that 12a + 16b = 1
A
Assumption:
- There are integers such that 12a + 16b = 1
- Highest common factor between 12 and 16 is 4
- 3a + 4b = 1 / 4
- ( Divided by 4 )
- 3a and 4b are still integers as a and b are integers
- The sum of the two integers are always an integer
- however 1 / 4 isn’t an integer
- Therefore contradicts the statement that 3a + 4b = 1 / 4
- So there’s no integers a and b such that 12a + 16b = 1
