Induction, contradiction, counterexample (HL)
Contents
Induction
The base case is
- for ,
- for ,
- for
The steps:
- Show true for $n = $ base case.
- Assume true for .
- Show true for
- is true; is true is true; is true for
Steps 1 and 3 should be done using LHS to RHS proofs.
In step 3, try to involve the assumption from step 2 as early as possible.
If need to work backwards, be sure to cross out rough work. The final solution presented to examiners must be LHS to RHS proofs for steps 1 and 3.
Example: Using induction, prove that for all and .
The base case is when .
Thus true for .
Note that implies .
Assume true for , ie
Note: “Assume true for ” means “if we know the question statement is true for some ”. IB explicitly does not give marks for “let ” or “assume ”, which means something else.
Then for
Since , this shows that
is true is true; is true means is true for
A stronger version of the above example is by breaking into even and odd , then using two base cases of and , and showing means that for and (other than the case which is or indeterminate.)
Example: (From The Art and Craft of Problem Solving) Prove using induction that is divisible by for all .
For , is , which is a multiple of .
Assume true for , or , for some .
Then for
is true is true; is true means is true for
As an extension (beyond HL), there is also strong induction, in which we assume true for .
Contradiction
Contradiction works by assuming the complement (opposite) of what you are trying to show, reaching a false conclusion, which invalidates the assumption.
There are two cases.
single proposition
The first case is “show [some statement] is true”, with the first step to “assume [some statement] is false”.
- “There exists … ” becomes “Assume there is no …”, vice versa.
- “This is an infinite number of …” becomes “Assume there is a list of all …”
- ” is irrational” becomes “Assume and and have no common factors.”
- ” becomes “Assume .”
Example: (Adapted from video by blackpenredpen ) Prove by contradiction that has no integer solutions.
Assume there is a solution. Since right side is odd (EVEN + ODD = ODD), it follows that is odd, and that one of and is even, and other one is odd.
In the case is even and is odd, and is a solution for . It follows that
This is a contradiction because left side is not a multiple of but right side is. (More explicitly, when dividing both sides by 2, we have ODD on the left side and EVEN on the right). Hence is not a solution.
Since has the same solutions as , is also not a solution. Both cases have no solution
Example: Prove by contradiction that is irrational.
Assume and are coprime, ie have no common factors.
Since is even, then and are even, so .
Since is even, then and are even.
This contradicts the assumption that and are coprime. In other words, the contradiction is there is a more simplified fraction than the simplest fraction. Hence is irrational
if A then B
The second case is “if then ”, with the first step to “assume AND not ”. It is at times interchangeable with the first, but this helps with longer propositions or statements.
Example: Prove by contradiction that if and , then at least one of and is even.
Assume and and are both odd. This means for , let , .
is even. Let
This is a contradiction as left side is not a multiple of (remainder ), but right side is.
Hence by contradiction the statement requires and/or to be even
Example: Prove using contradiction that for all .
In quadrant II, and , and they cannot be zero at the same time. So we can immediately show that , and .
Assume that for some .
Then
Since we squared , we see that
This contradicts .
Therefore, the assumption is wrong, and for all
Note: It is insufficient to show that results in no inconsistencies, rather we needed to show led to an inconsistency.
tips
In your working, you can use the following without proof:
- parities (odd vs even) when adding or multiplying two integers
- odd odd is even, even even is even, odd even is odd
- odd odd is odd, even (odd or even) is even
- even numbers are ; odd numbers are
- algebra rules, for both equations and inequalities
Common contradictions look like
- An equation is odd on one side but even on the other.
- (Extension:) A equation has a multiple of on one side but not on the other.
- A statement that is algebraically false.
- Existence of a smaller number than the smallest. (Existence of a large number than the largest.)
Disprove by counterexample
Provide a specific case and explain that it shows the statement is not always true.
Example: Kim claims that if an integer in base-
- is greater than , AND
- each digit is or , AND
- the digits are all different, AND
- the digits do not sum to a multiple of
then the integer is prime.
Provide a counterexample and explain why Kim is incorrect.
is greater than one; its digits are and ; which is not a multiple of , but it is not prime as
is not a counterexample, because it is not greater than .
is not a counterexample, because the statement does not apply to integers with a digit . The statement does not say what happens if an integer is prime.
is not a counterexample, but rather an example for which the statement holds true.