So far we have seen different types of numbers like fractions (proper, improper, mixed), decimals (recurring and non-recurring), integers. All these numbers have one property in common, which is that these numbers can be converted to a fraction with an integer as the numerator and another integer as a denominator, in other words a
ratio of two integers
. Even a recurring decimal, which has an infinite sequence of places after the decimal point can be converted to a fraction.
But there are numbers that cannot be converted exactly into fractions, using an integer numerator and an integer denominator.
These are called $Irrational\ Numbers$.
Let us take some examples. We know that the numbers $1$, $4$ or $9$ are perfect squares, and the square root of $1$ is $1$, square root of $4$ is $2$ and the square root of $9$ is $3$. But all the number in between $1$, $4$ and $9$, like $2$, $3$, $5$, $6$, $7$ and $8$ are not perfect squares, but it is still possible to calculate their square roots fairly accurately.
Let us take the case of $\sqrt{2}$.
If we calculate $1.4 \times 1.4$, we get $1.96$. This is close to $2$, but still not equal to $2$. Likewise,
$1.41 \times 1.41 = 1.9881$. This is closer to $2$ than before, but still not same.
$1.414 \times 1.414 = 1.999396$. Again this is much closer to $2$, but still not equal.
$1.4142 \times 1.4142 = 1.99996164$.
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If you go on like this with a good calculator, that can calculate more upto more than 30 digits, like most modern day desktop calculator, you will see that
This is very close to $2$, but still not equal to $2$.
It is possible to go on like this for ever, and come as close as possible to $2$, but still the value will not be
exactly equal
to $2$.
Also, we see that there is no repeating pattern in the numbers after the decimal place, so this is not a recurring fraction.
Had there been a repeating pattern, then we could have converted this into a proper fraction, that is a ratio of two integers.
This is true for square roots of all numbers that are not perfect squares, so $\sqrt{144} = 12$, which is an integer, whereas $\sqrt{145}$ is an irrational number.
Similarly, any $n^{th}$ root of a number, which is not a perfect $n^{th}$ power of an integer or a proper fraction, will be an irrational number.
There are many irrational numbers that are used extensively in mathematics and science. One of the most famous being the number called $\pi$ (pronounced like $pie$ named using the greek alphabet $pi$), which is the ratio of the circumference of a circle and its diameter. The approximate value of this number is $3.14159265$. There are many different approximations of $pi$ as fractions, some of them are $\dfrac{22}{7}$ which gives the value of $\pi$ correct to two places of decimal, $\dfrac{333}{106}$ (correct to $4$ places of decimal), $\dfrac{355}{113}$ (correct to $6$ places of decimal).
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Some people have used powerful computers to calculate the value of $\pi$ upto many trillions of digits after the decimal. But that is still not the exact value of $\pi$.
As we have learned today, $\pi$ will have a never ending, non-recurring series of digits after the decimal point.
Let us try the following question:
--------- Reference to question: c93ad03b-b890-4b3d-9710-c54e010e5b72 ---------
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In the subsequent sections we will see the proof of irrationality of $\sqrt{2}$, $\sqrt{3}$ and square roots of other prime numbers.
Before, we proceed with the proofs, we need to understand a few things about perfect squares.
If a perfect square $s$ has the prime factor $p$, then $s$ is divisible by $p^2$, that is, all prime factors of $s$ appear in pairs.
For example the perfect square $36$ can be written in terms of its prime factors as:
$36 = 2 \times 2 \times 3 \times 3$ has one pair of $2$s and one pair of $3$s
or the perfect square $81$ as:
$81 = 3 \times 3 \times 3 \times 3$, has two pairs of $3$s
II. Proof - $\sqrt{2}$ Is Irrational
We will use a technique of proof which is well known as proof by contradiction. That is we is assume the contrary of what we are trying to prove and show that it is not possible.
Let us assume that $\sqrt{2}$ can be expressed as a ratio of two integers $\dfrac{a}{b}$ where $a$ and $b$ are co-primes, that is the fraction $\dfrac{a}{b}$ is in its completely reduced form.
Therefore,
$(\sqrt{2})^2 = \left(\dfrac{a}{b}\right)^2$
$\Rightarrow 2 = \dfrac{a^2}{b^2}$
$\Rightarrow a^2 = 2 b^2$
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Since $2b^2$ is equal to $a^2$, $2b^2$ is a perfect square.
The number $2b^2$ has $2$ as a factor. Since $2$ is a prime factor of a perfect square, it must occur in pairs in the number.
Therefore, the square must be divisible by $4$. Therefore, $b^2$ must have at least one $2$ as a factor.
If $b^2$ is even, then its square root $b$ is also even.
Similarly, $a^2$ is even because it is equal to $2b^2$, which means $a$ is even.
Since both $a$ and $b$ are even, they share the common factor $2$. This contradicts our initial statement that the fraction $\dfrac{a}{b}$ is in its completely reduced form and does not share a common factor.
Every fraction can be reduced completely to a form where the numerator and denominator are co-prime, but $\sqrt{2}$ cannot be expressed as a ratio of two co-prime integers, it is an irrational number.
III. Proof - $\sqrt{3}$ Is Irrational
Like before, we will start by assuming the contradiction, that is, $\sqrt{3}$ can be expressed as a ratio of two co-prime integers, $\dfrac{a}{b}$:
$\sqrt{3} = \dfrac{a}{b}$
$\Rightarrow 3 = \dfrac{a^2}{b^2}$
$\Rightarrow a^2 = 3b^2$
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Since $3b^2$ is equal to $a^2$, it is a perfect square and has $3$ as a prime factor.
Since all prime factors of a perfect square occur in pairs, it must have one more $3$ as a factor, which implies $b^2$ is divisible by $3$.
Since $a^2$ is equal to $3b^2$, it is also divisible by $3$, hence its square root $a$ is divisible by $3$.
Since $a$ and $b$ are both divisible by $3$, this contradicts our initial assertion that $a$ and $b$ are co-primes, and we can see that it is not possible to express $\sqrt{3}$ as a fraction of two integers in its completely reduced form.
We can use this approach to prove the irrationality of square roots of all prime numbers.
IV. Things To Remember
Any number of the form $\sqrt[n]{a}$ where $a$ is not the $n^{th}$ power of an integer, is an irrational number, for example: $\sqrt{5}$, $\sqrt[3]{10}$, $\sqrt[8]{100}$ are irrational numbers.
Irrational number cannot be expressed as a ratio of two integers. Rational number including decimals, recurring decimals can be expressed as a ratio of two integers, of the form $\dfrac{a}{b}$ where $b \ne 0$.
Any of the operations like $addition,\ subtraction,\ multiplication\ or\ division$, when performed between a rational and an irrational number will always give an irrational number as a result.
There are other famous irrational numbers, $\pi$ (pi) being the most common of them.
It is impossible to calculate the exact value of an irrational number, they can only be approximated upto some finite decimal places.