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Given

  1. A straight line of arbitrary length
  2. The ability to construct a straight line in any direction from any starting point with the "unit length", or the length whose square root of its magnitude yields its own magnitude.

Is there a way to geometrically construct (using only a compass and straightedge) the a line with the length of the square root of the arbitrary-lengthed line? What is the mathematical basis?

Also, why can't this be done without the unit line length?

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    $\begingroup$ I cannot understand your last remark. If the lenght of a segment is 4 units, its square root is 2 units, that is its half; but if it is 16 units, its square root is 4 units, that is a quarter. Thus to make sense of the notion of a square root you must specify which is the unit you use, which is tantamount to have a unit length segment. $\endgroup$
    – mau
    Commented Jul 26, 2010 at 7:32
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    $\begingroup$ Maybe because a square root does not have a geometric meaning in one dimension (and maybe not an inherent geometric meaning at all), whereas bisection has to do with ratios, which have geometric interpretation. Square roots are related to the geometric mean, which does have geometric meaning. $\endgroup$
    – Isaac
    Commented Jul 26, 2010 at 7:40
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    $\begingroup$ So probably the best answer is that square root is not e dimensional invariant; you are mixing lengths (the square root itself) and areas (the original number, that you must see as an area). Greek geometry was very strict in it; only with the rise of algebra such distinctions were lost. $\endgroup$
    – mau
    Commented Jul 26, 2010 at 7:43
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    $\begingroup$ This is also a proposition of Euclid. Isaac's and mau's constructions are similar. $\endgroup$
    – Larry Wang
    Commented Jul 26, 2010 at 8:00
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    $\begingroup$ "Also, why can't this be done without the unit line length?" Consider an initial line segment 25 cm long. Should its "square root" be longer or shorter? Sounds simple - you might say it is obvious that $\sqrt{25}=5$ so we should have a rather shorter segment 5 cm long... but we might just as well say that our initial segment was 0.25 metres long, and $\sqrt{0.25}=0.5$ suggesting we need a new segment 50 cm long, double the initial length! Unless we declare a "unit" to be metres or centimetres or inches or whatever, then our initial length is arbitrary and there's no way to square root it. $\endgroup$
    – Silverfish
    Commented Jun 10, 2016 at 15:02

5 Answers 5

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If you have a segment $AB$, place the unit length segment on the line where $AB$ lies, starting with $A$ and in the direction opposite to $B$; let $C$ be the other point of the segment. Now draw a semicircle with diameter $BC$ and the perpendicular to $A$; this line crosses the semicircle in a point $D$. Now $AD$ is the square root of $AB$.

$\triangle BCD$ is a right triangle, like $\triangle ACD$ and $\triangle ABD$; all of these are similar, so you find out that $AC/AD = AD/AB$. But $AC=1$, so $AD = \sqrt{AB}$.

See the drawing below:

constructing square root of a line segment

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    $\begingroup$ yup, I transliterate from Italian (where we talk about "triangolo rettangolo"). Thanks for pointing it out. $\endgroup$
    – mau
    Commented Jul 26, 2010 at 8:52
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    $\begingroup$ "Now draw a semicircle with diameter AC and the perpendicular to B;" I believe you mixed up A and B here. $\endgroup$
    – rollover
    Commented May 1, 2014 at 3:09
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    $\begingroup$ How to prove that BCD is a right triangle? $\endgroup$
    – Jam
    Commented Mar 1, 2016 at 12:20
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    $\begingroup$ It is inscribed in a semicircle $\endgroup$
    – mau
    Commented Mar 1, 2016 at 12:31
  • $\begingroup$ Maybe it should be noted that this is essentially Euclid's proposition II.14 - not so much Euclid's proposition III.35 - which might be equivalent? $\endgroup$ Commented Aug 31, 2018 at 8:25
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Without the unit-length segment--that is, without something to compare the first segment to--its length is entirely arbitrary, so can't be valued, so there's no value of which to take the square root.

Let the given segment (with length x) be AB and let point C be on ray AB such that BC = 1. Construct the midpoint M of segment AC, construct the circle with center M passing through A, construct the line perpendicular to AB through B, and let D be one of the intersections of that line with the circle centered at M (call the other intersection E). BD = sqrt(x).

AC and DE are chords of the circle intersecting at B, so by the power of a point theorem, AB * BC = DB * BE, so x * 1 = x = DB * BE. Since DE is perpendicular to AC and AC is a diameter of the circle, AC bisects DE and DB = BE, so x = DB^2 or DB = sqrt(x).

edit: this is a special case of the more general geometric-mean construction. Given two lengths AB and BC (arranged as above), the above construction produces the length BD = sqrt(AB * BC), which is the geometric mean of AB and BC.

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    $\begingroup$ FYI: The theorem relied on here is Euclid's proposition 35 (Book III). $\endgroup$
    – Larry Wang
    Commented Jul 26, 2010 at 8:04
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    $\begingroup$ @Kaestur: Ahh, yes, that's what I've called the "power of a point" theorem (or, at least, that's the relevant form where the point is inside the circle). $\endgroup$
    – Isaac
    Commented Jul 26, 2010 at 8:07
  • $\begingroup$ Given a concrete segment with some length, the unit length sometimes can be constructed like 3、sqrt(2), etc. $\endgroup$ Commented Aug 21, 2022 at 6:54
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take a line AB of 1 unit. draw a line segment BC perpendicular to AB and join CA. take the radius of CA and with centre of compass on A draw an arc cutting the extension of line AB. that point is square root 2.(this is in grade 9 syllabus for us). what MAU answered is also another way.

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This answer focuses on OP's last question about the unit-length line.

I would answer the rest of the question exactly as mau. It can be seen that in mau's drawing, the unit length segment is a necessary element, and without that segment this construction is not possible. Now I add a small point here:

If we don't have the unit length segment, we may still be able to complete the construction if we know the length whose square root we are seeking. The rationale is that the value of the length contains information. It implies the relative size of a segment with respect to the unit length. So it is useful!

In particular, in the case where the said length is an integer, we are able to use that information to construct a unit length segment (how?). Then we can apply mau's method.

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  • $\begingroup$ Given a segment with length π you cannot construct the unit length right? $\endgroup$ Commented Aug 21, 2022 at 6:52
  • $\begingroup$ @user1206899 That's right: not with straightedge and compass. $\endgroup$
    – Saeed
    Commented Aug 22, 2022 at 9:13
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Distances are reckoned in terms of some unit, so in that sense, a unit length segment is always necessary. That said, you don't necessary have to append on a segment with the number who's square root you want to find, at least in special cases.

For example, suppose you want to find the square root of 5. Construct a right triangle with side lengths 1 and 2. This can be done with straight edge and compass. Then the hypotenuse has length $\sqrt{5} $(times the unit).

The procedure can get more complicated. Since 7 is a prime which is not expressible as the sum of two perfect squares, you need the right triangle to have one side of length 2 and the other side of length $\sqrt{3}$ where $\sqrt{3}$ had been constructed before

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