A Conversation for Mathematics

The calculator's inherent inability to count

Post 1

Sean

Take an ordinary pocket calculator

key in 1 / 3

write down the answer

clear the memory

type in your answer the machine gave you and multiply by 3

any self respecting numerate person knows that if you divide anything by 3 and then muliply by 3 you get what you started with - not a pocket calculator;

according to this piece of bad technology

1 = 0.9999999

WHICH IS A LIE!!!!!!!!!!



The calculator's inherent inability to count

Post 2

the Shee

Wait! I know the answer to this! smiley - biggrin

Take a look at a fraction like 1/9: in decimal form, it is .1111111 (repeating).

2/9: .2222222 (repeating)
3/9: .3333333 (repeating)
4/9: .4444444 (repeating)
5/9: .5555555 (repeating)
6/9: .6666666 (repeating)
7/9: .7777777 (repeating)
8/9: .8888888 (repeating)

The decimal form of a number divided by 9 then, is the number infinitely repeating behind a decimal point.

So, .9999999 (it only goes so far because the calculator only has so large a window) is the same thing as 9/9, which is the same thing as 1.

smiley - winkeye
Shee


The calculator's inherent inability to count

Post 3

Calculator Nerd 256

also, it's possible to prove that .2499999(repeating) = 1/4
.24999999*10 = 2.49999999
2.4999999999999 etc.
-.2499999999999 etc.
---------------
2.25 = (10 - 1) * 2.499999999
2.25 * 4 = 9
smiley - geek>8^B


The calculator's inherent inability to count

Post 4

U198950

Well, actually, it's just the calculator's inability to show more than eight characters, or as many characters you set it to show.

Remember, 1/3 is acually a very big number, with an endless display of .3333333

Your calculator doesn't really want to show you infinite threes, so it just displays the last eight digits.

What YOU did was take that ordinary pocket calculator and only worked with what it showed you. Any self-respecting numerate person knows that pocket calculators aren't very smart.

So, in any case, what you put in the calculator was not the answer to 1/3, but "0.3333333" or something to that affect. If you just left the answer alone, you would have gotten 1, because the calculator would remember that the answer to 1/3 was a repeting decimal.

So, in the end, it was the human that made the flaw, not the calculator. Which is always the case.


The calculator's inherent inability to count

Post 5

Calculator Nerd 256

absolutely right
don't you feel sorry for ignoramouses-is that the right plural?- like that?
smiley - geek>8^B


The calculator's inherent inability to count

Post 6

26199

Not always smiley - smiley... our numerical analysis lecturer told us about a calculator he once bought second hand that came with an 'errata'... there was one particular number (not especially large or anything) that the calculator couldn't display, it'd just show 'error' instead smiley - silly


The calculator's inherent inability to count

Post 7

Calculator Nerd 256

yeah, the overflow?
it has only so many bits of memory and the largest number is 2^n - 1 where n is the bit depth
you probably got something like 1.8446744e+21
smiley - geek>8^B


The calculator's inherent inability to count

Post 8

26199

Nope -- not the overflow... that's the point smiley - smiley

It just happened to be a quirk of the particular floating-point representation the calculator used...


The calculator's inherent inability to count

Post 9

26199

(hmm, and that would work for integers -- but calculators don't use integers... the real max is 0.11...111*2^(max exponent) -- the max exponent is probably close to a power of two, but they're not a straight integer representation either... check this out if you're interested: http://www.psc.edu/general/software/packages/ieee/ieee.html)


The calculator's inherent inability to count

Post 10

Calculator Nerd 256

oh, yeah
ur right
swy
smiley - geek>8^B


The calculator's inherent inability to count

Post 11

ITIWBS

Also try 1.111111111/9 on your calculator, giving 0.1234567901...
(usually).

Work it out by hand.

Now its 0.123456789..., which is what its supposed to be.

A microchip older than Pentium 4 usually gets it right.

A problem of a bad algorithm predisposing to cumulative errors originating with Pentium 4.


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The calculator's inherent inability to count

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