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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.?

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  • $\begingroup$ "$\sqrt2,\sqrt3$ etc" is far from representing "any irrational number". You can plot these two, but no $\pi$ nor $e$, nor the vast majority of the reals. $\endgroup$ – Yves Daoust May 11 at 20:23
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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.

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  • $\begingroup$ looks like using right angle triangles propertiess $\endgroup$ – iostream007 May 12 '13 at 14:36
  • $\begingroup$ Well, yes: in fact, using the fact that a square's diagonal's length equals the square's side's length times $\,\sqrt 2\,$ ... $\endgroup$ – DonAntonio May 12 '13 at 14:39
  • $\begingroup$ This works for constructable numbers, like square roots. For things like $\pi$, you just have to plot a rational that is close enough. $\endgroup$ – Ross Millikan May 12 '13 at 15:11
  • $\begingroup$ @DonAntonio But OP asked how can we plot it on Number Line. The answer should be no, we can't accurately plot irrational number on a number line. $\endgroup$ – Quazi Irfan Jan 24 '15 at 12:46
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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:

enter image description here

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 lenght, we can construct $\sqrt{BH}$. You just have to find a way to construct such a triangel (Hint: Use a semicircle!).

enter image description here

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 allowd to use a tool to construct a hyperbola and another for a parabola, then you can construct $2^{\frac{1}{3}}$.

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By definition, you cannot plot most of the numbers on the real line. Only a countably infinite subset of irrational numbers can be plotted (in a sense that there exists a procedure (finite or infinite) that leads you to the point you like).

Using a ruler and a compass, you can construct any constructible number (no kidding), which includes all solutions of quadratic polynomials with rational coefficients (and iterated application of solutions of quadratic polynomials).

Then there are larger sets - solutions of higher-order polynomials (angle trisector helps you with 3rd order, then you need more sophisticated plotting devices), then you have computable transcendental numbers, where you usually need an infinite summation or iteration or something like that. However, that can still be argued to produce digits that tell you which point you mean, even if you cannot get the exact result in a finite time.

Then there are incomputable numbers. These are the ones that are incountably numerous (which means most of the reals). Incomputable? That means, there is no way of even saying which number you mean, but it is there. By "saying which number you mean", I'm referring to giving an algorithm - a way of defining it, computing it,... most of the real numbers are such, that you cannot "name" them - because there are only countably infinite "names" (formulas/computer programs) to tell which number you are referring to. That's quite obvious if you think about it right. If computer programs (which are a generalization of mathematical formulas - more powerful in the sense of being Turing machines) are just lists of 0s and 1s, that's just an integer number in binary form (long, but still...). So... countably infinite number of programs/formulas/algorithms/recipes, and a continuum of incountably many real numbers. That's what doesn't add up. So in short... most of the real numbers, you cannot even point out or refer to in any computable way. They are just there, nameless and hidden forever.

But... you can plot any square root with a ruler and a compass, and you can plot any computable real number to a reasonable finite precision (truncation just means you get a rational approximation). So it's not that bad really.

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... 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.

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(This very first one was made without a compass) enter image description here By using the Pythagorean theorem and intercept theorem

I have In last time uploaded geometrical solutions and approximations for :

$3+\frac{\sqrt{2}}{10}<\pi<\sqrt{2}+\sqrt{3} $ and $\sqrt{\pi}$

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First of all draw the number line. Mark point 'A' as 0 and ponit 'B' as 1. This means A,B = 1 unit.Now, at B, draw BX perpendicular to AB. Cut off BC = 1 unit. Join AC. By Pythagoras Theorem in right triangle ABC, we get AC= root 2 units. Now with radius AC and center A, mark a point on the number line. Let the marked point be M.M represents root 2 on the number line.

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    $\begingroup$ Maybe including a picture would make it easier for people to understand the answer. $\endgroup$ – Indominus Jun 20 '15 at 8:19
  • $\begingroup$ @Indominus and this literally ( by terms like point "A") but multiple times on differenent posts- like here $\endgroup$ – Krzysztof Myśliwiec Sep 24 '18 at 18:56

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