lvc@cbscd5.UUCP (09/09/83)
Here is 'proof' that -1 = 1 : -1 = -1 => -1/1 = 1/-1 => sqrt(-1/1) = sqrt(1/-1) => (sqrt is the square root operation here) sqrt(-1)/sqrt(1) = sqrt(1)/sqrt(-1) => sqrt(-1)*sqrt(-1) = sqrt(1)*sqrt(1) => -1 = 1 And one for you calculus fans : x^2 = x + x + ... + x (x times) differentiate both sides with respect to x 2*x = 1 + 1 + ... + 1 (x times) 2*x = x (divide by x) 2 = 1 Larry Cipriani cbosgd!cbscd5!lvc
leimkuhl@uiuccsb.UUCP (09/12/83)
#R:cbscd5:-55100:uiuccsb:9700005:000:613
uiuccsb!leimkuhl Sep 11 19:27:00 1983
The proof that -1=1 is interesting since it points up the fact that the sqrt
function for positive reals is NOT like the complex square root. In
general, sqrt(x/y)=-sqrt(x)/sqrt(y) if x/y<0, x,y real. One should use caution
when applying results derived in first year algebra courses before being
acquainted with the complex domain.
In this case, (1**(1/2))/((-1)**(1/2)) = 1/i = -i,
while (1/(-1))**(1/2) =(-1)**(1/2) = i
so the statement sqrt(1/-1)=sqrt(1)/sqrt(-1) is false.
The second "paradox" is really dumb. First you assume x is an integer, then
try to use real calculus. Brilliant.john@hp-pcd.UUCP (09/14/83)
#R:cbscd5:-55100:hp-pcd:6100003:000:314 hp-pcd!john Sep 13 09:43:00 1983 Regarding the Calculus Proof: The equality x^2 = x+x+x...+x ( X times ) is only true for one value of x. Once you pick x you get a parabola that intersects a line at that value of x. To imply that because they intersect that they must have the same slope is wrong. John Eaton hplabs!hp-pcd!john