Square Root of 2

πŸ“… 2024-05-28

Many everyday things seem trivial to me, but in reality, I know very little about them, such as 2\sqrt{2}.

Proof of the Irrationality of 2\sqrt{2}

There are many methods to prove that 2\sqrt{2} is irrational. You can directly check wiki for these proofs.

Calculation of 2\sqrt{2}

During our school days, we used the approximate value of 2\sqrt{2}, which is 1.414, for calculations. However, what interests me now is how to calculate 2\sqrt{2} to a higher precision.

1 1 ?

Geometric Method

Adding auxiliary lines, we get a smaller triangle that is proportional to the larger right-angled triangle. Based on the proportional relationships of the line segments in the diagram, we construct an equation:

12=2βˆ’11βˆ’(2βˆ’1)=2βˆ’12βˆ’2\frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{1 - (\sqrt{2} - 1)} = \frac{\sqrt{2} - 1}{2 - \sqrt{2}}

Next, we transform this equation:

2βˆ’1=122βˆ’2=12(2+2)2=122+22=11+2\begin{align} \sqrt{2} - 1 &= \frac{1}{\frac{\sqrt{2}}{2-\sqrt{2}}} \notag \\ &= \frac{1}{\frac{\sqrt{2}(2+\sqrt{2})}{2}} = \frac{1}{\frac{2\sqrt{2} + 2}{2}} \notag \\ &= \frac{1}{1+\sqrt{2}} \notag \end{align}

Move 1 to the right side of the equation:

2=1+11+2\sqrt{2} = 1 + \frac{1}{1+\sqrt{2}}

Here, we have a very important form, which is a continued fraction. Because in this expression, we can infinitely iterate by replacing 2\sqrt{2} on the right side of the equation.

2=1+11+1+11+1+11+1+…=1+12+12+12+…\begin{align} \sqrt{2} &= 1 + \frac{1}{1+1 + \frac{1}{1+1 + \frac{1}{1+1 + \dots}}} \notag \\ &= 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \dots}}} \notag \end{align}

We can see that the omitted part becomes less important as the depth of iteration increases, but the 2 that repeatedly appears in the denominator follows a fixed pattern. Whether we assume the value at the deepest iteration is infinitely large or infinitely small, we will ultimately get a value very close to the initial value of 12\frac{1}{2}. When we use 12\frac{1}{2} in this continued fraction for continuous calculation,

2βˆ’1β‰ˆ12=0.5β‰ˆ12+12=25=0.4β‰ˆ12+12+12=512=0.416β€Ύβ‰ˆ12+12+12+12=1229=0.4137931034482758620689655172β€Ύβ‰ˆ12+12+12+12+12=2970=0.4142857β€Ύβ‰ˆ12+12+12+12+12+12=70169β‰ˆ0.41420118343195266272\begin{align} \sqrt{2} -1 &\approx \frac{1}{2} = 0.5 \notag \\ &\approx \frac{1}{2 + \frac{1}{2}} = \frac{2}{5} = 0.4 \notag \\ &\approx \frac{1}{2 + \frac{1}{2 + \frac{1}{2}}} = \frac{5}{12} = 0.41\overline{6} \notag \\ &\approx \frac{1}{2 + \frac{1}{2 + \frac{1}{2+\frac{1}{2}}}} = \frac{12}{29} = 0.\overline{4137931034482758620689655172} \notag \\ &\approx \frac{1}{2 + \frac{1}{2 + \frac{1}{2+\frac{1}{2+\frac{1}{2}}}}} = \frac{29}{70} = 0.4\overline{142857} \notag \\ &\approx \frac{1}{2 + \frac{1}{2 + \frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}}} = \frac{70}{169} \approx 0.41420118343195266272 \notag \end{align}

After a few iterations, we see that the decimal part 0.4142 is already quite accurate. For a more precise result, we can continuously iterate the numerator and denominator of the fraction until we reach the desired precision.

Here, we have basically solved how to calculate 2\sqrt{2} to a high precision. However, some more interesting questions have also emerged:

  1. Why can an equation be substituted into itself? Is this method correct?
  2. For other square root numbers, like 3\sqrt{3}, how can we construct an equation that is similar to the equation for calculating 2\sqrt{2}?
  3. The form of the continued fraction seems similar to the iterative method for finding the greatest common divisor. What is their relationship?

Let's solve these questions next time. ❀️