[014] Propositions, Contrapositives, and Proof by Contradiction

In the world of mathematics, not only proofs by algebraic manipulation but also proof methods using logical reasoning play an important role.

In particular, "propositions and contrapositives" and "proof by contradiction" are powerful weapons in mathematical proofs.

These proof methods have the fascination of logically unraveling seemingly complex problems, much like solving a trick in a detective novel.

What is a proposition?

A proposition is a statement that is clearly either true or false.

In mathematics, we often deal with propositions in the form of "if $${p}$$ then $${q}$$".

This is sometimes written as "$${p \Rightarrow q}$$".

Specific Example 1

"If $${n}$$ is an even number, then $${n^2}$$ is also an even number."

This is a proposition. This is because we can verify whether this statement is true or false.

In fact, when $${n = 2}$$, $${n^2 = 4}$$, which is even, and when $${n = 4}$$, $${n^2 = 16}$$, which is even, so this proposition is true.

Specific Example 2

"It will probably be sunny tomorrow."

This is not a proposition. Because it contains a conjecture like "probably," its truth or falsehood is not clearly determined.

Specific Example 3

"Masaharu Fukuyama is a woman."

This is a proposition. You might think, "No way, he's not a woman, right??", but a proposition is something where "it is clearly determined whether it is true or false," and since it is clearly determined to be "false," it is a proposition.

What is a contrapositive?

For a proposition "$${p \Rightarrow q}$$", the proposition "if not $${q}$$, then not $${p}$$" is called the contrapositive.

Written in symbols, it becomes "$${\overline{q} \Rightarrow \overline{p}}$$".

What is important is that the original proposition and its contrapositive always have the same truth value.

In other words, if the original proposition is true, the contrapositive is also true, and if the original proposition is false, the contrapositive is also false.

Concrete Example 4

Original proposition: "If $${n}$$ is an even number, then $${n^2}$$ is also an even number"

Contrapositive: "If $${n^2}$$ is not an even number, then $${n}$$ is also not an even number"

Rephrasing this: "If $${n^2}$$ is an odd number, then $${n}$$ is also an odd number"

This contrapositive is also true, just like the original proposition.

Proof using the contrapositive

When it is difficult to prove the original proposition directly, you can prove the original proposition by proving its contrapositive.

This is a very powerful proof technique in mathematics.

Concrete Example 5

Let's prove the proposition: "If $${n^2}$$ is an even number, then $${n}$$ is an even number."

Starting from $${n^2}$$ and discussing $${n}$$ is quite a troublesome matter.

It is easy to "find the value of $${n^2}$$ when $${n=7}$$ ", but it is rather tedious to "find the value of $${n}$$ when $${n^2=7}$$ ".

As you can see from this, proving this proposition directly is a bit tedious, but it becomes easy if you use the contrapositive.

Contrapositive: "If $${n}$$ is not an even number, then $${n^2}$$ is also not an even number," which means: "If $${n}$$ is an odd number, then $${n^2}$$ is also an odd number."

[Proof of the contrapositive] If $${n}$$ is an odd number, it can be expressed as $${n = 2k+1}$$ (where $${k}$$ is an integer).

$${n^2 = (2k+1)^2}$$ $${ = 4k^2 + 4k + 1}$$ $${ = 2(2k^2 + 2k) + 1}$$

Since $${2k^2 + 2k}$$ is an integer, it is in the form $${n^2 = 2}$$ × (integer) $${ + 1}$$.

Therefore, $${n^2}$$ is an odd number.

Therefore, since the contrapositive holds, the original proposition also holds.

Example 6

Let us prove the proposition: "If $${ab}$$ is an irrational number, then $${a}$$ or $${b}$$ is an irrational number."

This is also quite difficult to prove directly. This is because "irrational numbers" are defined as "real numbers that are not rational," making them difficult to work with.

Contrapositive: "If both $${a}$$ and $${b}$$ are rational numbers, then $${ab}$$ is a rational number."

[Proof of the contrapositive] Assume that both $${a}$$ and $${b}$$ are rational numbers.

They can be expressed as $${\displaystyle a = \frac{p}{q}}$$ and $${\displaystyle b = \frac{r}{s}}$$ (where $${p, q, r, s}$$ are integers, $${q \neq 0}$$, and $${s \neq 0}$$).

$${\displaystyle ab = \frac{p}{q} \cdot \frac{r}{s} = \frac{pr}{qs}}$$

Since $${pr}$$ and $${qs}$$ are integers and $${qs \neq 0}$$, $${ab}$$ is a rational number.

Therefore, since the contrapositive holds, the original proposition also holds.

What is proof by contradiction?

Proof by contradiction is a method of showing that an original proposition is true by starting from the negation of the proposition to be proved and deriving a contradiction.

The logic is: "I thought this might be wrong, but if it were, wouldn't that lead to an absurdity? Therefore, the original statement must have been correct!"

Think of it like playing in a maze.
You are looking for a path from the start to the goal, but when you come to a fork in the road, which way do you go?
You are not sure about the path to the right, but you find that the path to the left leads to a dead end after a short distance.
In this case, if you asked 100 people, all 100 would likely go to the right.

This is the concept of proof by contradiction.
"Shouldn't I go left? But if I explore that path a little, isn't it a dead end? Then, I should go right!"

The steps for proof by contradiction are as follows:

$${1.}$$ Negate the proposition you want to prove
$${2.}$$ Logically derive a conclusion from that negation
$${3.}$$ Show that a contradiction arises
$${4.}$$ Therefore, since the negation is incorrect, the original proposition is correct

Example 7

Let us prove the proposition: "$${\sqrt{2}}$$ is an irrational number" using proof by contradiction.

[Proof by contradiction]

$${1.}$$ Assume the negation: Assume that $${\sqrt{2}}$$ is a rational number.

$${2.}$$ Derive a logical consequence: If $${\sqrt{2}}$$ is a rational number, it can be expressed as $${\displaystyle \sqrt{2} = \frac{p}{q}}$$ using coprime positive integers $${p}$$ and $${q}$$.}], 2. [{

Squaring both sides: $${\displaystyle 2 = \frac{p^2}{q^2}}$$

Multiplying both sides by $${q^2}$$: $${2q^2 = p^2}$$

From this, $${p^2}$$ is an even number.

From "Example 5", if $${p^2}$$ is even, then $${p}$$ is also even.

Since $${p}$$ is even, it can be expressed as $${p = 2r}$$ (where $${r}$$ is a positive integer).

$${2q^2 = (2r)^2 = 4r^2}$$

Dividing both sides by $${2}$$: $${q^2 = 2r^2}$$

From this, $${q^2}$$ is an even number.

Therefore, $${q}$$ is also an even number.

$${3.}$$ Find a contradiction: The fact that both $${p}$$ and $${q}$$ are even means that $${p}$$ and $${q}$$ have a common divisor of $${2}$$.}], 3. [{

This contradicts the assumption that "$${p}$$ and $${q}$$ are coprime".

$${4.}$$ Conclusion: Since a contradiction has occurred, the assumption that "$${\sqrt{2}}$$ is a rational number" is incorrect.

Therefore, "$${\sqrt{2}}$$ is an irrational number" is correct.

Example 8

Proposition: Let's prove that "there are infinitely many prime numbers" using proof by contradiction.

[Proof by Contradiction]

$${1.}$$ Assume the negation: Assume that "there are only a finite number of prime numbers".

$${2.}$$ Derive a logical consequence: Let all prime numbers be $${p_1, p_2, p_3, \ldots, p_n}$$.}], 2. [{

$${N = p_1 \times p_2 \times p_3 \times \cdots \times p_n + 1}$$

Consider the number $${N}$$.

Since $${N}$$ is greater than $${1}$$, it must be divisible by some prime number.

However, when $${N}$$ is divided by any prime number $${p_i}$$, the remainder is $${1}$$.

$${3.}$$ Finding a contradiction: $${N}$$ is not divisible by any prime number, but since $${N > 1}$$, it must be divisible by some prime number.

This is a contradiction.

$${4.}$$ Conclusion: Since a contradiction has occurred, the assumption that "there are only a finite number of primes" is incorrect.

Therefore, "there are infinitely many primes" is correct.

Choosing the Right Logical Proof

The proof methods we have looked at so far each have their own suitable situations.

Proof by contrapositive is effective when it is easier to deal with the "negation" of the original proposition.

It is particularly powerful in situations where conditions like "is not" or "other than" are easier to prove.

Proof by contradiction is effective when a direct proof is difficult and a clear contradiction can be derived from the negation of the assumption.

It is often used when dealing with concepts such as "infinity" or "non-existence".

Furthermore, proof by contradiction often provides new insights during the proof process, which can sometimes lead to mathematical discoveries.

Summary

Propositions, contrapositives, and proof by contradiction are the foundations of logical thinking in mathematics.

By mastering these proof methods, you will be able to approach complex mathematical problems systematically.

If proof by algebraic manipulation is a "technique of calculation," then logical proof is a "technique of thinking."

By becoming proficient in both, you will be able to experience the true interest and beauty of mathematics.

A proof is not merely a 'confirmation of an answer,' but a creative activity that logically constructs the path to truth.