[Phys-l] google +- precision

[John Denker <jsd@av8n.com>]

On 12/28/2009 02:40 PM, curtis osterhoudt wrote:
> Note that even on the first question, I was a bit confused by the
> answer that google gave me. I thought, "Well, it's obviously 2 for
> the remainder, since if I subtract two from the dividend, that'll be
> all nines and divide evenly." Then, when I plugged 1000000001/9 into
> google, expecting to get 0.2222... for the fractional part of the
> remainder, I instead got what was explicitly an integer: 111111111.
> Boooo, google!

That's amusing.  Yes, it's true that
  http://www.google.com/search?q=1000000001/9
gives a result that "looks like" an exact integer, namely 
 111111111
On the other hand,
  http://www.google.com/search?q=1000000001/9*9
gives 1000000001 not 999999999.  Hmmmmmm.  Evidently google
knows more than it is letting on.

Apparently I hit google right on the edge of its roundoff
decision.  A shorter dividend would have given a result that
"looks" more like a rounded-off number:
   http://www.google.com/search?q=100000001/9
gives
  11111111.2

And if we take a step in the other direction, toward larger
dividends, we find that
  http://www.google.com/search?q=10000000001/9
gives an answer in scientific notation, which is again
less misleading than the "apparent" integer.

All of this fits with my longstanding opinion that all 
of us are addicted to the notion that "sig figs" are an 
acceptable way of indicating precision.  This addiction 
is not good for us.

Constructive suggestions:
 1) When reading a number, never assume that the number of
  digits tells you anything about the uncertainty.  The
  uncertainty could be much more or much less than what
  you would guess based on the number of digits.
 2) When writing numbers, if they are not exact, always
  express the uncertainty separately and explicitly.
  In the case of a rounded number, where the uncertainty
  is roughly half a count in the last place, you can
  express it as 111111111(½) or 111111111(⁄) for short.

For details on this, see
  http://www.av8n.com/physics/uncertainty.htm    

=================

On the other hand, if you really want google to compute the 
answer to
   http://www.av8n.com/physics/algebra-trig-quiz.htm#main-remainder
then you can ask for it directly and succinctly:
  http://www.google.com/search?q=1000000001+mod+9

=================

On the third hand, do you really need google to solve this
problem, or even to check it?

First of all, there is the old bookkeeper's trick of casting
out nines.  Add up all the digits without regard to place
value.  If the sum is q mod 9, then the original number
(including place value) is also q mod 9.  There's a similar
rule for casting out threes.

Also, as others have pointed out, it seems likely that 999999999
is evenly divisible by 9.  This embodies the general problem-
solving skill that sometimes, given a seemingly-hard problem,
there is a similar problem that is very easy, and gives you a
lever for cracking the original problem.

And then there is the plodding "Hufflepuff" solution.  Just
do the long division.  It very quickly settles into a pattern,
and unless you are very very dim you will notice the pattern,
so you can do the entire long division in less time than it
takes to talk about it.

And if you know anything about modular arithmetic, there are
yet other ways of doing it.

========

Most of the problems on my little quiz have this property, that
there are multiple ways of solving the problem.  This is partly
intentional but mostly inevitable.  Most real-world problems
have this property.  Most of the problems I consider interesting
have this property.

Students should learn that any important problem is worth
solving _twice_.  If two independent methods of solution give
the same answer, it greatly increases your confidence in the
answer.