# Page:Popular Science Monthly Volume 83.djvu/391

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THE FOURTH DIMENSION

distances perpendicular to the ${\displaystyle xy}$ plane are positive if measured above, negative if measured below. This notation enables us to locate any point in our space.

Now we know of ${\displaystyle 2}$-space only as a section of ${\displaystyle 3}$-space, and a duodim is purely an imaginary being to us; and we know of ${\displaystyle 1}$-space only as a section of ${\displaystyle 2}$-space (and therefore of ${\displaystyle 3}$-space), and the unodim is imaginary. We have seen that a duodim might interfere with life in ${\displaystyle 1}$-space, but the unodim would not know at all what had caused the

Fig. 8.

interference. We have also seen that a tridim might in a similar way interfere with life in ${\displaystyle 2}$-space. The important point to observe is that in either case the inhabitant of the lower space would not understand what had caused the change.

A duodim could lock up his treasure in circular or polygonal vaults, such as "${\displaystyle a}$" or "${\displaystyle b}$," safe from ${\displaystyle 2}$-space intruders, but a tridim could help himself to anything he pleased without breaking the sides of the vault. By analogy, a ${\displaystyle 4}$-space being could do many things in ${\displaystyle 3}$-space impossible to man and entirely inexplicable to him. No ${\displaystyle 3}$-space safe or vault would be secure from a ${\displaystyle 4}$-space burglar. He could get a ball out of a hollow shell without breaking the surface, he could get out the

Fig. 9.

contents of an egg without cracking the shell and enjoy the kernel of a nut without the use of a nut-cracker.

A geometrical illustration similar to those already given is found in Fig. 9. Here "${\displaystyle a}$" and "${\displaystyle b}$" are symmetrical tetrahedrons,[1] in length

1. A model of "${\displaystyle a}$" and "${\displaystyle b}$" can be readily constructed as follows:

Cut out the figure (Fig. 10) from a piece of cardboard, perforated along the lines ${\displaystyle AB,BC,CA}$, and having ${\displaystyle AF=AE,CE=CD}$ and ${\displaystyle BD=BF}$. Fold over the triangle ${\displaystyle ABF,ACE,CBD}$ till the points ${\displaystyle F,E}$ and ${\displaystyle D}$ meet in a point, thus making one tetrahedron: fold the triangles in the opposite direction and the symmetrical tetrahedron will be formed. The one corresponds to the image of the other in a mirror.