## Proof Tutorial 2: Proving Square Root of 2 is Irrational by Contradiction

One of the most difficult proof strategies in mathematics is proof by contradiction. If P, for example, is a statement or a conjecture, one strategy to prove that P is true is to assume that P is not true  and find a contradiction so that the statement not P does not hold. If not P does not hold, it follows that P is true.

One well-known proof that uses proof by contradiction is proof of the irrationality of $\sqrt{2}$.  If we consider P to be the statement “$\sqrt{2}$ is irrational”, then not P is the opposite statement or “$\sqrt{2}$ is rational”.  To use proof by contradiction, we assume that $\sqrt{2}$ is rational, and find a contradiction somewhere. If this happens, then we would have shown that $\sqrt{2}$ is indeed irrational.

Before proceeding, recall that a rational number is a fraction with non-zero denominator.  We know that all fractions can be expressed in lowest term.  A fraction in $\displaystyle\frac{a}{b}$ is said to be in lowest term if $a$ and $b$ have no common divisors except $1$.

On the other hand, irrational numbers cannot be expressed as fractions. They are decimal numbers that do not end and do not repeat. For example, $0.10100100010000...$ is an irrational number (the three dots means and so on which means that the number does not end). The most popular irrational number is $\pi$.

Now, we prove our conjecture.

Conjecture: The $\sqrt{2}$ is irrational.

Proof:

Suppose $\sqrt{2}$ is rational, then it can be expressed in fraction form $\displaystyle\frac{a}{b}$ . Let us assume that our fraction is in lowest term, i.e., their only common divisor is $1$. Then,

$\sqrt{2} = \displaystyle\frac{a}{b}$

Squaring both sides, we have

$2= \displaystyle\frac{a^2}{b^2}$

Multiplying both sides by $b^2$ yields

$2b^2= a^2$*

Since $a^2 = 2b^2$, we can conclude that $a^2$ is even because whatever the value of $b^2$ has to be multiplied by $2$. If $a^2$ is even, then $a$ is also even. Since $a$ is even, no matter what the value of $a$ is, we can always find an integer that if we divide $a$ by $2$, it is equal to that integer. If we let that integer be $k$, then $\displaystyle\frac{a}{2} = k$ which means that $a = 2k$.

Substituting the value of $2k$ to $a$ in *, we have $2b^2= (2k)^2$ which means that $2b^2=4k^2$.  Dividing both sides by $2$, we have $b^2 = 2k^2$. That means that the value $b^2$ is even, since whatever the value of $k$ you have to multiply it by $2$.  Again, if $b^2$ is even, then $b$ is even.

This implies that both $a$ and $b$ are even, which means that both the numerator and the denominator of our fraction are divisible by $2$. This contradicts our assumption that $\displaystyle\frac{a}{b}$ has no common divisor except $1$. Since we found a contradiction, our assumption is, therefore, false. Hence, the theorem is true.

Notice that I have highlighted the word suppose and assume in the proof. This is one unique feature of proof by contradiction. You can always assume, most of the time, the opposite of the conjecture as long as the following statements are logically valid.

## The Algebraic and Geometric Proofs of Pythagorean Theorem

The Pythagorean Theorem states that if a right triangle has side lengths $a, b$ and $c$, where $c$ is the hypotenuse, then the sum of the squares of the two shorter lengths is equal to the square of the length of the hypotenuse.

Figure 1 – A right triangle with side lengths a, b and c.

Putting it in equation form, we have

$a^2 + b^2 = c^2$.

For example, if a right triangle has side lengths $5$ and $12$, then the length of its hypotenuse is $13$, since $c^2 = 5^2 + 12^2 \Rightarrow c = 13$.

Exercise 1: What is the hypotenuse of the triangle with sides $1$ and $\sqrt{3}$?

The converse of the theorem is also true. If the side lengths of the triangle satisfy the equation $a^2 + b^2 = c^2$, then the triangle is right. For instance, a triangle with side lengths $(3, 4, 5)$ satisfies the equation $3^2 + 4^2 = 5^2$, therefore, it is a right triangle.

Geometrically, the Pythagorean theorem states that in a right triangle with sides $a, b$ and $c$ where $c$ is the hypotenuse, if three squares are constructed whose one of the sides are the sides of the triangle as shown in Figure 2, then the area of the two smaller squares when added equals the area of the largest square.

Figure 2 – The geometric interpretation of the Pythagorean theorem states that the area of the green square plus the area of the red square is equal to the area of the blue square.

One specific case is shown in Figure 3: the areas of the two smaller squares are $9$ and $16$ square units, and the area of the largest square is $25$ square units.

Exercise 2: Verify that the area of the largest square in Figure 3 is 25 square units by using the unit squares.

Figure 3 – A right triangle with side lengths 3, 4 and 5.

Similarly, triangles with side lengths $(7, 24, 25)$ and  $(8, 15, 17)$ are right triangles. If the side lengths of a right triangle are all integers, we call them Pythagorean triples. Hence, $(7, 24, 25)$ and  $(8, 15, 17)$ are Pythagorean triples.

Exercise 3: Give other examples of Pythagorean triples.

Exercise 4: Prove that there are infinitely many Pythagorean triples.

Proofs of the Pythagorean Theorem

There are more than 300 proofs of the Pythagorean theorem. More than 70 proofs are shown in tje Cut-The-Knot website. Shown below are two of the proofs.  Note that in proving the Pythagorean theorem, we want to show that for any right triangle with hypotenuse $c$, and sides $a$, and $b$, the following relationship holds: $a^2 + b^2 = c^2$.

Geometric Proof

First, we draw a triangle with side lengths $a, b$ and $c$ as shown in Figure 1. Next, we create 4 triangles identical to it and using the triangles form a square with side lengths $a + b$ as shown in Figure 4-A. Notice that the area of the white square in Figure 4-A is $c^2$.

Figure 4 – The Geometric proof of the Pythagorean theorem.

Rearranging the triangles, we can also form another square with the same side length as shown in Figure 4-B.This means that the area of the white square in the Figure 4-A is equal to the sum of the areas of the white squares in Figure 4-B (Why?). That is, $c^2 = a^2 + b^2$ which is exactly what we want to show. *And since we can always form a (big) square using four right triangles with any dimension (in higher mathematics, we say that we can choose arbitrary $a$ and $b$ as side lengths of a right triangle), this implies that the equation $a^2 + b^2 = c^2$ stated above is always true regardless of the size of the triangle.

Exercise 5: Prove that the quadrilateral with side length C in Figure 4-A is a square.

Algebraic Proof

In the second proof, we will now look at the yellow triangles instead of the squares.  Consider Figure 4-A. We can compute the area of a square with side lengths $a + b$ using two methods: (1) we can square the side lengths and (2) we can add the area of the 4 congruent triangles and then add them to the area of the white square which is $c^2$.  If we let $A$ be the area of the square with side $b + a$, then calculating we have

Method 1: $A = (b + a)^2 = b^2 + 2ab +a^2$

Method 2:  $A = 4(1/2ab) + c^2 = 2ab + c^2$

Methods 1 and 2 calculated the area of the same square, therefore they must be equal. This means that we can equate both expressions.  Equating we have,

$b^2 + 2ab + a^2 = 2ab + c^2 \Rightarrow a^2 + b^2 = c^2$

which is exactly what we want to show.