# Paper Folding: Locating the square root of a number on the number line

Paper folding (origami) has been used by many teachers, particularly in Japan, to teach mathematics.   In this post, we are going to use paper folding to locate the square root of any number on the number line. A GeoGebra simulation of the paper folding is shown below.

Paper Folding Instructions

1. Get a rectangular piece of paper and fold it in the middle, horizontally and vertically, and let the creases (see green segments in the applet) represent the coordinate axes.
2. Construct point M at (0,1), and fold and make a crease at y = -1 (see blue segment above).
3. Let P denote a point with coordinates (0,n) where n < 0 (drag P along the negative y-axis).
4. Make a fold through P (by dragging Q in the applet) that places M(0,1) on the line y = – 1.
5. The x intercept of the fold is $\sqrt{n}$.
The Mathematics Behind

Theorem:  The x intercept of the fold above is $\sqrt{n}$.
Before we proceed with the proof, in the applet check the Show/Hide Triangles check box to display the triangles MNO and OPN. You can also uncheck the Show/Hide Fold check box to get a better view of the triangles.

Proof:
Let be x-intercept of the fold.  It is clear that $\Delta NOM \sim \Delta PON$. This implies that
$\displaystyle\frac{ON}{OM} = \frac{OP}{ON}$.
But $ON=m$, $OM=1$, and $OP = n$.  Subsituting the values, we have
$\displaystyle\frac{m}{1} = \frac{n}{m}$.
Therefore $m^2 = n$ and  $m = \sqrt{n}$.
This means that for every value for $n < 0$,  the intersection of the fold is always $\sqrt{n}$.  In effect,  using this method, we can locate the number or the length  $\sqrt{n}$ on the number line.
***
Reference:  My old dilapidated geometry notebook, so I don’t really know to whom I shall attribute this. Our teacher in Geometry never used any reference. He could talk all day without looking at any book.

## 11 comments on “Paper Folding: Locating the square root of a number on the number line”

1. s. johnsey on said:

It is not clear to me in your diagram above that PON and NOM are similar as you stated in your proof. Can you explain. thanks.

2. Guillermo Bautista on said:

Hi Susan. Remember that before the proof, the condition is to fold the paper such that M (0,1) is on the line y = -1. This is represented by M’ which is actually a reflection of M along the fold. This means that MM’ is perpendicular to PQ. That makes POM, MON, and PMN all right triangles which implies that they are similar.

For the details:

Angle PON and angle MON are congruent (they are right angles). Angle M is congruent to angle PNO. So, by AA similarity, the two triangles are similar.

Please tell me if you have more questions.

3. Sneha on said:

Great activity. I made an accompanying set of questions to get the students thinking and one of them was to change the y-coordinate of P to positive values and look at whether the reflection of M cou old be placed on the line y = -1.
An effective way to visualise that the square roots of negative numbers were not on the number line.
P.S. I tried to make the geogebra sheet myself – is there any way that we can make the reflected point snap to the line y = -1?
When I move the point P I have to go through the whole rotating thing again, a button to move the image to the line y = -1 would be useful.

• Guillermo Bautista on said:

@Sneha:
Thanks. I don’t think there’s a way to snap the point to the line. The point only snaps to grid intersections. Since we can choose an arbitrary fold, it is impossible to create such grid.

4. Guillermo Bautista on said:

@Sneha: If you have improved the applet, I would love to post it here to your credit.

5. Sneha on said:

The geogebra sketch is improved to me because I created it and understand where it is coming from. However, it does not look as good as yours:-)

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