## Rational and Irrational Numbers

The need of men to perform certain mathematical operations led to the birth of different types of numbers. People in the ancient times used only counting numbers to keep track of the number of their belongings such as animals.  The concept of trade led to the invention of 0 and negative numbers. The need to divide led to the invention of rational numbers.

In this article, we are going to take a look at the characteristics of rational and irrational numbers.

Rational numbers are numbers of the form a/b where ab are integers, and b not equal to zero. Rational numbers are called rational not because they are reasonable, but because they are a ratio of two integers.  It is worthy to note the conditions in the definition. That is because not all fractions are rational numbers. For example, $\frac{2 \pi}{3 \pi}$ is a fraction, but it is only a rational number when simplified. From the definition, we can deduce that all integers are rational numbers since {…,-3, -2, -1, 0, 1, 2, 3,…} = {…, -3/1, -2/1, -1/1, 0/1, 2/1, 3/1, …}.

Geometric Interpretation of Rational and Irrational Numbers

In ancient times the Greeks, particularly the Pythagoreans, believed that all quantities are rational; that is, all quantities can be expressed as a ratio of two integers. Geometrically, this can be interpreted as follows. Given any two lengths, a unit length can be found that can measure the two lengths exactly without gaps or overlaps. In the example in Figure 1, we have two segments a and b, and we found a unit length that would fit exactly a whole number of times in both segments. The ratio of a:b is 7:6, or we can express it in a fraction that a is 7/6th of b.

Figure 1 – The division of segments a and b into unit lengths.

This belief, as most of us now know, was proven to be false. The Pythagoreans later discovered that given a square with length 1 unit, no unit length, however short, can be found to measure both the side of the square and its diagonal like what we have done above. They have concluded that the length of the diagonal cannot be expressed as the ratio of two integers and hence not rational.  Today, numbers that are not rational are called irrational numbers. Hence, we define irrational numbers as numbers that cannot be expressed as a ratio of two integers.

Figure 2 – The square with length 1 unit has irrational diagonal.

Using the Pythagorean Theorem, we now know that the length of the diagonal of a square with side length 1 unit is equal to $\sqrt{2}$. We have already discussed and proved that $\sqrt{2}$ is irrational.

The collection of all rational and irrational numbers is called real numbers. Geometrically, real numbers are represented by the real line as shown in Figure 3.

Figure 3 – The real number line.

Each real number can be represented by a point on the real number line and every point on the number line has a corresponding real number.

Another Representation of Rational and Irrational Numbers

Aside from fractions, we can also represent rational numbers with decimals.  For example, 1/5 = 0.2  and 1/3 = 0.333….  Observe that 0.2 has a finite number of decimals while 0.333… has infinite.  Irrational numbers can also be represented using decimals.  They are the types of decimals that do not end and do not repeat.

Several irrational numbers are very popular, and we had been using them from elementary school to college. The irrational numbers $\pi$, $e$ and $\phi$ are several of irrational numbers that we are acquainted with.

Figure 4 – The structure of the real number system.

From our discussion above, we can see that real numbers are divided into two main subsets – rational and irrational numbers.

## 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.

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