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hardmath
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My book says:
The bit I don't get is
Why? Apply this logic to one of the numbers at random for arguments sake, let's say the first number .234673, it couldn't be .2346735-.2346739 or else it would have been rounded to .234674. It's true value (if it's greater) has to lie somewhere between .2346731-2346734 or else it would have been rounded up to .234674 .
So if it can't be .0000005 greater as that would get it rounded up to .234674, and you can't count .2346730 when calculating error as .2346730 is the same as .234673 which would be correct and wouldn't be a possible error you have to factor in, then knowing the figures are accurate to 6 significant figures it would appear to me that logically the possible error range is
.2346731-.2346734 = .0000004 greater
.2346725-.2346729 = .0000005 lesser
And this would apply to all 4 figures, so the sum could be (4 X .0000005) -.000002 lesser or (4X.0000004) .0000016 greater.
Now it's in the book so obviously there's a flaw in my reasoning and since this is very basic math it's probably a very basic error so that's my question, thanks for reading and I apologize for the longwindedness.
Approximation in Addition and Subtraction. As an illus-
tration of approximation in addition, let us add the decimal fractions
.234673, .322135, .114342, .563217, each being known to be correct to
six figures. The addition gives 1.234367, but it is not certain that this
result is correct because, in accordance with the discussion of the pre-
ceding article, each of the given decimals may be either greater or
less than the true value by as much as .0000005 (due to rejection of
the seventh figure). Since there are four numbers the total error in
the sum may be as much as 4 X .0000005, or .000002. That is, the true sum may be as much as 1.234369 or as little as 1.234365. In
either case the result correct to six figures would be written 1.23437.
The bit I don't get is
each of the given decimals may be either greater or
less than the true value by as much as .0000005 (due to rejection of
the seventh figure).
Why? Apply this logic to one of the numbers at random for arguments sake, let's say the first number .234673, it couldn't be .2346735-.2346739 or else it would have been rounded to .234674. It's true value (if it's greater) has to lie somewhere between .2346731-2346734 or else it would have been rounded up to .234674 .
So if it can't be .0000005 greater as that would get it rounded up to .234674, and you can't count .2346730 when calculating error as .2346730 is the same as .234673 which would be correct and wouldn't be a possible error you have to factor in, then knowing the figures are accurate to 6 significant figures it would appear to me that logically the possible error range is
.2346731-.2346734 = .0000004 greater
.2346725-.2346729 = .0000005 lesser
And this would apply to all 4 figures, so the sum could be (4 X .0000005) -.000002 lesser or (4X.0000004) .0000016 greater.
Now it's in the book so obviously there's a flaw in my reasoning and since this is very basic math it's probably a very basic error so that's my question, thanks for reading and I apologize for the longwindedness.
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