Re: number line twisted!!!!!!!!

On Tue, 6 May 2008 04:41:08 -0700 (PDT), dhritimant@xxxxxxxxx wrote:

Consider the relation y= 1000/x. We vary x in the following way:- we
move through the number line from right to left (i.e from infinity to -
infinity) and assign the numbers we encounter to x.
we find that the value of y first increases(till x=o) but again
starts decreasing. WHY!!!!!!!!!!!!!!!!!!!!!!!!!!

Consider the circle of radius 1 centred on the origin. Let M be the
point (-1, 0) on the x-axis. Consider two points P and Q, at opposite
ends of a diagonal, free to move around the circle. If t is the slope
of MP, and u is the slope of MQ, then tu = -1 (angle in a semicircle).
If we allow oo ('infinity' sign) as the value of a slope (it's not a
number: you can't add it to a number, compare it with a number, etc.),
then either P or Q may coincide with M, the chord MP or MQ becoming a
tangent. When t = 0, u = oo; and vice versa. So, you can think of
the real numbers, extended by the adjunction of a single point oo
at 'infinity', as being mapped 1-1 onto the circle: with -1/0 = oo,
-1/oo = 0, and (t, u) with tu = -1 at the opposite ends of a diagonal.
Points on a circle don't 'increase' or 'decrease', but they can move
clockwise or anticlockwise; if you delete the point M from the circle,
the 'increasing' direction is anticlockwise, from M back to M again.
Of course, slope 0 corresponds to the point N = (1, 0) opposite to M,
slope 1 is the point (0, 1), and slope -1 the opposite point (0, -1).
Finally, changing the sign of the slope corresponds to reflection in
the x-axis (slope oo is of point M on the x-axis, and has no 'sign'),
and simply inverting the slope (with no reversal of sign) corresponds
to reflection in the y-axis. So if you want to visualise the relation
between x and y (first rescaling to get rid of the factor 1000), think
of a point moving around a circle, and its reflection in the vertical
diagonal of the circle.

--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril
.

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