Plot any irrational number on number line.

Question

I have a basic question:
Can we plot any irrational number on a number line? I can plot all integers and rational numbers on it but how can I plot any irrational number, such as $\sqrt2$,$\sqrt3$ etc.?

Answer 1

Hints:

Draw a line of length $\,1\,$ , from one of its end points draw another line perpendicular to the first one (how? Can you use a compass...?), and now draw the diagonal between the other end of the original line and the farthest end of the perpendicular one. This perpendicular has length $\,\sqrt 2\,$ (why?) , so with a compass measure it and then place the compass on the origin *the "zero: of your line) and "plot" $\,\sqrt 2\,$ ...you can do something similar with "any" non-rationa square root.

Answer 2

... as in how do you physically carry out the plotting? Same as plotting any other number.

Pick up a pencil and use a ruler to locate the point at about the right distance from the origin. Use the pencil to apply a short straight mark on the paper, perpendicular to the number line and intersecting it at the point you found with the ruler.

It your pencil is of a somewhat orthodox design, the pencil mark will be at least 0.1 milimeters wide, corresponding to an uncountable infinity of different numbers on the line. If one of those points is the one you set out to plot, you're good.

Answer 3

Choosing a unit length, you can use the spiral of Theodorus to find any length of the form $\sqrt{n}$ in which n is an integer:

As a matter of fact, as shown in the picture below, using a similarity argument we can deduce that $CH^2=AH\times BH$, which means by choosing AH to be the unit length, we can construct $\sqrt{BH}$. You just have to find a way to construct such a triangel (Hint: Use a semicircle!).

For other irrational numbers, it depends on the number AND the tools you are allowed to use. For example, it has been proved that these numbers are not constructible using just a straight edge and compass: $2^{\frac{1}{3}}$, $\pi$, $e$. But, if you're allowed to use a tool to construct a hyperbola and another for a parabola, then you can construct $2^{\frac{1}{3}}$.