All of us has actually done geometry construction using ruler and also compass in high school. We understand that many constructions are feasible using just these two straightforward instruments.

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Now, mean that you have been asked to trisect the angle 60 using these two instruments. The is unlikely that you will come up through a solution also after a few days of concentrated efforts. But prior to you start feeling bad around yourself, it is crucial to understand that your situation is not very different indigenous the mighty old Greeks!

The 3 famous building problems of antiquity are the following:

Trisecting an angle: offered any angle, devise a basic algorithm to trisect it using ruler and compass only.Squaring the circle: given a circle, build a square with the same area together the circle, making use of ruler and compass only.Doubling the cube: offered a cube, build the sheet of a cube that has twice the volume the the provided cube. In this case also, we space only enabled to use ruler and compass.

The Greeks tried hard to settle these three problems and after failed attempts hoped that posterity would certainly devise a systems to each of them. However, v the development of abstract algebra, it came to be clear the the above-mentioned problems are in fact unsolvable, i.e. You cannot perform those geometric constructions using just a ruler and a compass.

But accounts of cube ‘doublers’ or angle ‘trisectors’ spam the inbox of eminent mathematicians with their claimed ‘proof’. This is a an outcome of failure to properly recognize the nature that the problem. For example, think about the problem of angle trisection. It says that we should devise an algorithm to build one-third that any given angle, using simply an unmarked ruler and also a compass. Obviously, we can construct one-thirds that some very distinct angles using ruler and also compass. For example, one knows just how to construct one-third that 90, specific 30. However can we do this because that any angle? Can we provide an algorithm come trisect any type of given angle?

Using some ingenious mathematical techniques, one can show that the answer come this question is negative. In fact every one of the aforementioned classical problems of antiquity have to be proven to it is in unsolvable. But exactly how does one prove the a specific geometric building and construction is not possible using an there was no sign ruler and a compass?

This brings united state to the question: Exactly what geometric constructions deserve to we execute using a ruler and also a compass only?

A pursuit of the answer to this concern will lead us to appreciate the tremendous beauty natural in the impressive power of summary algebra, which uniform seemingly different branches of mathematics.

Straightedge and compass constructions

In the following text, the words straightedge and ruler mean the same thing.

In order to ensure the we room on the same page, let us briefly remind the

rules the straightedge and also compass constructions.

At the onset, us are provided two distinctive points in the plane to start with. These points are constructed.One have the right to use a straight edge only to construct a directly line through any two created points or to expand a offered line segment. The line is then claimed to be constructed.

Given two constructed points, one deserve to use a compass only to construct a one with one of those points together the centre and also the other allude lying top top the circle. The circle is then stated to it is in constructed.

New points space constructed from intersections of constructed lines and also circles.

Points, lines and circles are claimed to be constructible if they deserve to be constructed in finitely numerous steps using these rules.

For example, given any kind of two points A and B, one can display that the mid-point the the line segment AB is constructible. Build the line with A and B using a straight edge. Then constructing circles of radius AB, that have actually centre in ~ A and B, one at some point constructs the midpoint the AB. The construction is displayed in the complying with figure.


Figure 1: Bisecting a given line segment.

Arithmetic on segments:

We present the Cartesian works with in the plane. V the advent of coordinates, we deserve to now talk around the distance between two point out in the plane.

Let united state assume that the point out (0, 0) and also (1, 0) are initially given. Hence, we have actually a segment of length 1 to begin with.

Suppose that two segments have currently been constructed such the their lengths room a and b. Us shall display how one can further construct segment which have lengths a + b, a-b, ab, a/b and √a. (Here we assume,without ns of generality, the b Adding segments

Figure 2 shows exactly how to include and subtract segments. Right here EG and EH are the sum and also difference dong of the segment AB and CD.

Multiplying segments

The building and construction of segment of lengths ab and a/b depend top top the following construction.

Constructing a perpendicular

Given a line and a suggest P ~ above it, build a perpendicular segment to the line at ns . The building is presented in number 3. In this construction, every circle has actually radius 1.


Figure 3: construction of the perpendicular come a line at a offered point.

Constructing ab and a/b

Using the construction of best angles as shown in number 3, one constructs segment of lengths ab and a/b. View the adhering to figures.


Figure 4: constructing multiples that segments


Figure 5: constructing ratios of segments

In order to prove the CE = ab in number 4 and also BE = a/b in figure 5, one have the right to use similarity that triangles.

Square roots of segments

Next, we attain a segment that has actually length √a as follows.


Figure 6: Square root of a segment

We construct the circle through diameter CD = a + 1. Climate the altitude the the appropriate angled triangle CFD has size √a.

Constructible numbers:

Now the we have learnt just how to perform arithmetic v segments, a natural question come ask would certainly be the following: What are those number which can be acquired as the size of a segment built from the initially offered unit segment involvement (0, 0) and (1, 0)?

In view of this question, it renders sense to have the following definition:

A actual number d is stated to be constructible if d = 0 or there exists a constructible segment that has actually length |d|.

We have learnt exactly how to add, subtract, multiply and divide segments. Therefore, provided any non-zero reasonable number q, we can acquire a segment of length │q│, starting from the segment that unit size that joins (0, 0) and (1, 0). Or in other words, a segment of size │q│ can be presented to be constructible. This is because the rational number │q│ can be acquired from 1 through the using the operations of addition, subtraction, multiplication and department finitely numerous times. Performing the corresponding operations on the unit segment authorized (0, 0) and also (1, 0), us can achieve a segment that has actually length |q|. Therefore, every reasonable number is constructible.

Examples the constructible number that room not rational include numbers such together √2. Take a=2 in number 6. Now that we have created √2, us can also construct √√2. We can go upto √√√√2 and even more!

Summarizing what we have actually learnt it spins now, we view that any number obtained from 1 by transferring out the to work of addition, subtraction, multiplication, division and square root, a finite variety of times, is constructible.

The main result of this article, which us shall soon see, asserts the these are the only constructible numbers! In order to make points clear, we introduce the adhering to definitions.

Let us call the to work of addition, subtraction, multiplication and division of genuine numbers as rational operations. A real number is claimed to lie in real quadratic closure of Q if and only if it have the right to be derived from 1 by a limited sequence of reasonable operations and square source extractions.

Reframing the result that us are around to prove in regards to this brand-new definition, we obtain: A genuine number is constructible if and only if the lies in the real quadratic closure the Q.

Let the statement sink in! This opens up the door to proving that the three classic problems pointed out in the start are unsolvable. Together a specific example, think about the trouble of trisecting 60 by ruler and compass.

We say that an edge θ is constructible if we have the right to construct two lines conference at an angle θ. If we mark off a unit length on one of those lines and also drop a perpendicular come the other, we will certainly have built the actual number cos θ. Now, expect that the angle 60/3 = 20 were constructible. Climate this would suggest that the actual number cos 20 is constructible. See figure 7. In that case, cos 20 would lie in the actual quadratic closure that Q.

Thus, in order to display that the angle 20 is not constructible, we simply need come prove the cos 20 does not lie in the real quadratic closure that Q! Now, this is a problem of algebra and we have powerful tools in our arsenal to handle this problem! we shall placed off the proof of this last action for one more occasion. In spite of that, we have hopefully acquired a flavour of just how to prove that specific geometric constructions have the right to never be performed using just a ruler and a compass. It is yes, really remarkable exactly how we converted a trouble of geometry come a difficulty of summary algebra.


Figure 7: If an edge of 20 were constructible then the very same would have been true the cos 20.

It would be a installation conclusion to finish this conversation with the evidence of the result which develops the structure to handle the difficulties of unsolvability of certain geometric constructions.

Theorem 1. A genuine number d is constructible if and also only if d lies in the actual quadratic closure the Q.

Proof. We have currently seen that any type of number in the genuine quadratic closure of Q is constructible. Let united state prove the converse.

Suppose d ∈ R is constructible. Therefore, over there exists constructible points (x1, y1) and (x2, y2) R2 such the square source of(x1 – x2)2 + (y1 – y2)2 =d.

It suffices to display that the collaborates of any kind of constructible allude P lies in the genuine quadratic closure that Q. Then x1, x2, y1 and y2 and also hence d will also lie in the same.

Recall that constructible points are acquired from intersections of present or one that have actually been constructed.



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<3>Stillwell J. (1996). Elements the Algebra: Geometry, Numbers, Equations, Springer.