Proof techniques
Apply proof by contradiction to establish irrationality of surds and logarithms; construct existence and uniqueness proofs; prove results involving rational and irrational numbers.
Worked examples
Proof by contradiction: is irrational
Straightforward
Problem
Prove that is irrational.
1
Assume the opposite of what is to be proved.
Suppose is rational. Then where , , and (the fraction is in lowest terms).
2
Square both sides and deduce that is even.
. So , and since squaring preserves parity, . Write .
3
Substitute back and deduce that is also even.
. So , hence .
4
Identify the contradiction.
Both and are even, so . This contradicts .
Answer
is irrational.
Irrationality of a logarithm
Moderate
Problem
Prove that is irrational.
1
Assume the opposite.
Suppose for positive integers and .
2
Rewrite in exponential form.
.
3
Compare the parity of both sides to find the contradiction.
The left side is even (a power of 2); the right side is odd (a power of 5). An even number cannot equal an odd number — contradiction.
Answer
is irrational.
Sum of a rational and an irrational number
Challenging
Problem
Prove that is irrational.
1
Assume the opposite, using the general result that rational + irrational = irrational.
Suppose, for contradiction, that is rational; call it .
2
Rearrange to isolate the irrational quantity.
Then .
3
Identify the contradiction.
The right-hand side is the difference of two rationals ( and ), so it is rational. But is irrational — contradiction.
Answer
is irrational.
Practise
Q1·Straightforward
Prove by contradiction that is irrational.
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Proof by contradiction:**
Assume is rational. Then for integers and with and (the fraction is in lowest terms).
Square both sides:
Therefore is even (it equals ). Since the square of an odd integer is odd, must be even. Write for some integer .
Substitute:
Therefore is even, so is even.
But now both and are even, contradicting .
**Conclusion:** The assumption that is rational leads to a contradiction. Therefore is irrational.
Assume is rational. Then for integers and with and (the fraction is in lowest terms).
Square both sides:
Therefore is even (it equals ). Since the square of an odd integer is odd, must be even. Write for some integer .
Substitute:
Therefore is even, so is even.
But now both and are even, contradicting .
**Conclusion:** The assumption that is rational leads to a contradiction. Therefore is irrational.
Q2·Straightforward
In a proof by contradiction that is irrational, which is the correct first assumption?
Explanation
To prove is irrational by contradiction, you assume it is **rational**.
A rational number can be written as where are integers, , and — importantly — the fraction is in **lowest terms** (i.e., ). This last condition is what delivers the contradiction later, when you show both and must be divisible by 3.
Option (a) assumes exactly what you want to prove — that would be circular.
Option (c) is a true fact but is not the starting assumption of the proof.
Option (d) is a consequence of option (b), not the starting assumption.
A rational number can be written as where are integers, , and — importantly — the fraction is in **lowest terms** (i.e., ). This last condition is what delivers the contradiction later, when you show both and must be divisible by 3.
Option (a) assumes exactly what you want to prove — that would be circular.
Option (c) is a true fact but is not the starting assumption of the proof.
Option (d) is a consequence of option (b), not the starting assumption.
Q3·Straightforward
Which of the following correctly describes a proof of **uniqueness** of the additive inverse of a real number ?
Explanation
**Option (b)** is the correct structure for a uniqueness proof.
The standard technique: assume two elements and both satisfy the defining property, then prove .
Here:
- Suppose and .
- From the first equation: .
- From the second: .
- Therefore .
This shows the additive inverse, if it exists, is unique.
Option (a) is an **existence** proof, not uniqueness. Option (c) is the definition of the additive identity, unrelated. Option (d) is incorrectly stated.
The standard technique: assume two elements and both satisfy the defining property, then prove .
Here:
- Suppose and .
- From the first equation: .
- From the second: .
- Therefore .
This shows the additive inverse, if it exists, is unique.
Option (a) is an **existence** proof, not uniqueness. Option (c) is the definition of the additive identity, unrelated. Option (d) is incorrectly stated.
Q4·Moderate
Prove by contradiction that is irrational.
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Proof by contradiction:**
Assume is rational. Write with , , .
Square both sides:
So . Since 5 is prime, (if a prime divides a square, it divides the base). Write .
Substitute:
So , hence .
But then both and are divisible by 5, contradicting .
**Conclusion:** is irrational.
**Note:** The key step — "" — holds because 5 is prime. For composite numbers this may fail (e.g., but ).
Assume is rational. Write with , , .
Square both sides:
So . Since 5 is prime, (if a prime divides a square, it divides the base). Write .
Substitute:
So , hence .
But then both and are divisible by 5, contradicting .
**Conclusion:** is irrational.
**Note:** The key step — "" — holds because 5 is prime. For composite numbers this may fail (e.g., but ).
Q5·Moderate
Prove that the sum of a rational number and an irrational number is irrational.
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Proof by contradiction:**
Let be rational and be irrational. Assume, for contradiction, that is rational.
Since is rational, is rational (the negative of a rational is rational).
The sum of two rationals is rational, so:
is rational.
But this contradicts the assumption that is irrational.
**Conclusion:** The sum of a rational and an irrational number is irrational.
**Example:** is irrational since is rational and is irrational.
Let be rational and be irrational. Assume, for contradiction, that is rational.
Since is rational, is rational (the negative of a rational is rational).
The sum of two rationals is rational, so:
is rational.
But this contradicts the assumption that is irrational.
**Conclusion:** The sum of a rational and an irrational number is irrational.
**Example:** is irrational since is rational and is irrational.
Q6·Moderate
Prove by contradiction that is irrational.
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Proof by contradiction:**
Assume is rational. Then for some positive integers and .
(Note: since , so we may take .)
Rewrite in exponential form:
The left-hand side is a power of 2, so it is **even**.
The right-hand side is a power of 3, so it is **odd** (since 3 is odd and any power of an odd number is odd).
An even number cannot equal an odd number — contradiction.
**Conclusion:** is irrational.
**Remark:** The same argument applies more generally: is irrational whenever and are positive integers and is not an integer power of .
Assume is rational. Then for some positive integers and .
(Note: since , so we may take .)
Rewrite in exponential form:
The left-hand side is a power of 2, so it is **even**.
The right-hand side is a power of 3, so it is **odd** (since 3 is odd and any power of an odd number is odd).
An even number cannot equal an odd number — contradiction.
**Conclusion:** is irrational.
**Remark:** The same argument applies more generally: is irrational whenever and are positive integers and is not an integer power of .
Q7·Moderate
Which of the following numbers is rational?
Explanation
is a fraction of two integers, so it is **rational** (option c).
Note: is a common approximation of but is not equal to . itself is irrational (and in fact transcendental).
- : irrational (7 is not a perfect square)
- : irrational and transcendental
- : irrational (this requires a proof — see the difficulty 3 question in this sub-topic)
- : rational ✓
Decimal expansions: (repeating) and (non-repeating, non-terminating).
Note: is a common approximation of but is not equal to . itself is irrational (and in fact transcendental).
- : irrational (7 is not a perfect square)
- : irrational and transcendental
- : irrational (this requires a proof — see the difficulty 3 question in this sub-topic)
- : rational ✓
Decimal expansions: (repeating) and (non-repeating, non-terminating).
Q8·Moderate
Prove by contradiction: "If is odd, then is odd."
(This can also be proved by contrapositive — here, use a direct proof by contradiction instead.)
(This can also be proved by contrapositive — here, use a direct proof by contradiction instead.)
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Proof by contradiction:**
Assume is odd but is **even**. Write for some integer .
Then:
This is even (it is times an integer), which contradicts the assumption that is odd.
**Conclusion:** The assumption that is even leads to a contradiction. Therefore, if is odd, then is odd.
**Compare with proof by contrapositive:** The contrapositive approach ("if is even, show is even") leads to essentially the same computation. Both are valid; proof by contradiction is sometimes preferred when the structure is more natural.
Assume is odd but is **even**. Write for some integer .
Then:
This is even (it is times an integer), which contradicts the assumption that is odd.
**Conclusion:** The assumption that is even leads to a contradiction. Therefore, if is odd, then is odd.
**Compare with proof by contrapositive:** The contrapositive approach ("if is even, show is even") leads to essentially the same computation. Both are valid; proof by contradiction is sometimes preferred when the structure is more natural.
Q9·Challenging
Prove that is irrational.
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Proof by contradiction:**
Assume is rational. Call it , so for some .
Isolate :
Square both sides:
Rearrange:
(We need ; since , we have .)
The right-hand side is rational (ratio of rationals). But is irrational — contradiction.
**Conclusion:** is irrational.
Assume is rational. Call it , so for some .
Isolate :
Square both sides:
Rearrange:
(We need ; since , we have .)
The right-hand side is rational (ratio of rationals). But is irrational — contradiction.
**Conclusion:** is irrational.
Q10·Challenging
Prove that the product of a non-zero rational number and an irrational number is irrational.
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Proof by contradiction:**
Let with , and let be irrational. Assume, for contradiction, that is rational.
Since and is rational, is also rational (the reciprocal of a non-zero rational is rational).
The product of two rationals is rational, so:
is rational.
This contradicts the assumption that is irrational.
**Conclusion:** The product of a non-zero rational and an irrational number is irrational.
**Example:** is irrational since is a non-zero rational and is irrational.
**Note:** The condition is essential: , which is rational.
Let with , and let be irrational. Assume, for contradiction, that is rational.
Since and is rational, is also rational (the reciprocal of a non-zero rational is rational).
The product of two rationals is rational, so:
is rational.
This contradicts the assumption that is irrational.
**Conclusion:** The product of a non-zero rational and an irrational number is irrational.
**Example:** is irrational since is a non-zero rational and is irrational.
**Note:** The condition is essential: , which is rational.
Q11·Challenging
Prove that there is no rational number such that .
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
**Proof by contradiction:**
Assume there exists a rational with . Write with , , .
Then:
So , which means (since if is odd, is odd). Write .
So , hence (since 3 is prime). Write .
Then , so . Also:
So , hence .
But then both and are divisible by 3, contradicting .
**Conclusion:** There is no rational with , i.e., is irrational.
Assume there exists a rational with . Write with , , .
Then:
So , which means (since if is odd, is odd). Write .
So , hence (since 3 is prime). Write .
Then , so . Also:
So , hence .
But then both and are divisible by 3, contradicting .
**Conclusion:** There is no rational with , i.e., is irrational.
Q12·Challenging
Proof by **construction** (direct existence proof) means:
Explanation
**Option (c)** is correct.
A proof by construction (also called a direct existence proof or constructive existence proof) proves by explicitly producing an that satisfies .
**Example:** Prove there exists an irrational number such that .
*Construction:* Let . Then , and is irrational (proved separately by contradiction).
**Compare with:**
- Option (a): proof by contradiction (non-constructive — it shows the object must exist without naming it).
- Option (d): mathematical induction (proves statements about all positive integers, not existence in general).
Construction is often the cleaner option when an explicit example is available. Contradiction is used when no explicit example is apparent.
A proof by construction (also called a direct existence proof or constructive existence proof) proves by explicitly producing an that satisfies .
**Example:** Prove there exists an irrational number such that .
*Construction:* Let . Then , and is irrational (proved separately by contradiction).
**Compare with:**
- Option (a): proof by contradiction (non-constructive — it shows the object must exist without naming it).
- Option (d): mathematical induction (proves statements about all positive integers, not existence in general).
Construction is often the cleaner option when an explicit example is available. Contradiction is used when no explicit example is apparent.
Open Math
Proof techniques
Proof · MEX-12-01
Name:
Date:
Q1Straightforward
Prove by contradiction that is irrational.
Q2Straightforward
In a proof by contradiction that is irrational, which is the correct first assumption?
- A.Assume is irrational.
- B.Assume for integers with and .
- C.Assume is not a perfect square.
- D.Assume for some integers and .
Q3Straightforward
Which of the following correctly describes a proof of **uniqueness** of the additive inverse of a real number ?
- A.Show that exists and that .
- B.Show that if and , then .
- C.Show that for all real .
- D.Show that there is no real number such that .
Q4Moderate
Prove by contradiction that is irrational.
Q5Moderate
Prove that the sum of a rational number and an irrational number is irrational.
Q6Moderate
Prove by contradiction that is irrational.
Q7Moderate
Which of the following numbers is rational?
- A.
- B.
- C.
- D.
Q8Moderate
Prove by contradiction: "If is odd, then is odd."
(This can also be proved by contrapositive — here, use a direct proof by contradiction instead.)
(This can also be proved by contrapositive — here, use a direct proof by contradiction instead.)
Q9Challenging
Prove that is irrational.
Q10Challenging
Prove that the product of a non-zero rational number and an irrational number is irrational.
Q11Challenging
Prove that there is no rational number such that .
Q12Challenging
Proof by **construction** (direct existence proof) means:
- A.Assuming the desired object does not exist, then deriving a contradiction.
- B.Proving that the object's properties follow logically from axioms.
- C.Directly exhibiting or building an example that satisfies the required condition.
- D.Using mathematical induction to show existence for all natural numbers.
Worked solutions and answers at openmath.au/year-12/extension-2/the-nature-of-proof/proof-techniques