What is Irrational Number

Irrational number is a number, that is not rational.

What does that mean?

It means, that we can't represent irrational number as a fraction, decimal with finite number of digits, or repeating decimal.

For, example:

  • Whole number 3 can be represented as 31\frac{{3}}{{1}}.
  • Fraction 510\frac{{5}}{{10}} is already fraction (reduced form is 12\frac{{1}}{{2}}).
  • Decimal 0.16{0.16} is 425\frac{{4}}{{25}}.
  • Repeating decimal 0.33333333=0.3{0.33333333}={0}.{\overline{{{3}}}} is 13\frac{{1}}{{3}}.

Irrational numbers often occur, when we want to find nth root of a number.

For example, 73{\sqrt[{{3}}]{{{7}}}} is irrational, because there is no such fraction mn\frac{{m}}{{n}}, that (mn)3=7{{\left(\frac{{m}}{{n}}\right)}}^{{3}}={7}.

A couple of famous irrational numbers are:square root of 2

  • π3.1415926535898\pi\approx{3.1415926535898} (ratio of circumference of any circle and its diameter)
  • 21.4142135623731\sqrt{{{2}}}\approx{1.4142135623731} (length of diagonal of the square with sides equal 1)

As can be seen, irrational numbers can be represented as decimals, but what then is the difference between repeating decimal and irrational number?

Answer is simple. Irrational number has no pattern in decimal representation.

Compare:

  • 1011=0.909090909090.=0.90\frac{{10}}{{11}}={0.909090909090}\ldots.={0}.{\overline{{{90}}}} (here is pattern, we know what comes next: 90)
  • π=3.1415926535897\pi={3.1415926535897}\ldots (what's next???)

Maybe, you don't believe that 2\sqrt{{{2}}} is irrational, so here is proof.

Fact. 2\sqrt{{{2}}} is irrational.

Proof. Pretend that we can represent 2\sqrt{{{2}}} as irreducible fraction mn\frac{{m}}{{n}}, that is 2=mn\sqrt{{{2}}}=\frac{{m}}{{n}}.

If numbers are equal, then their squares are also equal: (2)2=(mn)2{{\left(\sqrt{{{2}}}\right)}}^{{2}}={{\left(\frac{{m}}{{n}}\right)}}^{{2}}.

This can be rewritten as 2=m2n2{2}=\frac{{{m}}^{{2}}}{{{n}}^{{2}}}.

From last equality we have that m2=2n2{{m}}^{{2}}={2}{{n}}^{{2}}.

This means, that m2{{m}}^{{2}} is even. But if m2{{m}}^{{2}} is even, then m{m} is also even (square of odd number will give odd number).

That's why we can write m{m} as m=2c{m}={2}{c}.

Return to m2=2n2{{m}}^{{2}}={2}{{n}}^{{2}} and plug instead of m{m} 2c{2}{c}: (2c)2=2n2{{\left({2}{c}\right)}}^{{2}}={2}{{n}}^{{2}}.

This means that 4c2=2n2{4}{{c}}^{{2}}={2}{{n}}^{{2}} or 2c2=n2{2}{{c}}^{{2}}={{n}}^{{2}}.

So, n{n} is also even number.

Since n{n} is even and m{m} is even, then mn\frac{{m}}{{n}} is not irreducible (they have at least one common factor 2).

This is contradiction.

So, there is no such fraction mn\frac{{m}}{{n}}, that 2=mn\sqrt{{{2}}}=\frac{{m}}{{n}}.

Thus, 2\sqrt{{{2}}} is irrational.

When does irrational number occur?

  • Some known constants: π\pi, e{e} (Euler number) etc.
  • Nth root of numbers: 2,23\sqrt{{{2}}},{\sqrt[{{3}}]{{{2}}}} etc.

But keep in mind, that not all nth roots are irrational. For example 4=2\sqrt{{{4}}}={2}, 814=3{\sqrt[{{4}}]{{{81}}}}={3}.

Finally, multiplication of irrational numbers can give rational number:

  • 33=(3)2=3\sqrt{{{3}}}\cdot\sqrt{{{3}}}={{\left(\sqrt{{{3}}}\right)}}^{{2}}={3}
  • 28=4\sqrt{{{2}}}\cdot\sqrt{{{8}}}={4}

But that's not always true: π2=ππ=9.869604401{\pi}^{{2}}=\pi\cdot\pi={9.869604401} is still irrational.