Mathematical philosophy, a study of fate and freedom; lectures for educated laymen, by Cassius J. Keyser.

NON-GEOMETRIC INTERPRETATION 95
given system s; it is clear that there is in s at least one
dyad d"(x", y") such that d\d2 is congruent with d'd",
i.e., such that
'V/(Xl - X2)2 +(yl -y2)2 = /(Xtl -Xit)2 + (yt _ytt)2
for x" is at our disposal and y" is a function of it. But
is there in s more than one such d"? We know that s is
y=mx+b or else x=m'y+b'; let us use the former,
for the reasoning will be the same as for the latter. If
there be a second d", denote it by d"'(x"', y"'), where
x"' =x" + ô; then since d1d2, d'd", d'd"' are congruent, we
/ (X' -X")2 + (y' -y")2 = V(X- _-~"'2 + (y _-)y ")2;
note that y' = mx'+b, y" = -mx" +b, y"' = m(x" + ô) +b; substituting these values in the last radical equation, and
simplifying, we get ô2+2(x"-x') =0; whence =0 or
=2(X' -x"); the former value of ô gives x" =x"', and so
does not give a second d"; the latter value of ô gives
X"' =x" -z(x" -x'), and so there is one and but one other
d"; now note that x"' -x' = - (x" -x'); hence if one d"
is on one side of d', the other d" is on the other side. And
so the postulate is verified.
Postulate (I5).-This postulate is so manifestly satisfied that we need not tarry to prove the fact.
/Postulate (I6).-That this postulate is verified may be
readily proved as follows: Let di(xi, yi), d2(x2, y2) and
d3(x3, y3), three dyads of any given systems s, be such
that the segments d\d2 and d2d3 have in common no dyad
save d2; let di', d2', d3', three dyads of any given system s',
be such that d2' is the only dyad common to the segments
d1'd2' and d2'd'. Let d1d2 be congruent with d'd2',
and d2d3 with d2'd3'; we are to prove that d1d3 and d1'd3'
are congruent. We may take s to be y =mx+b, and s'