Why 8 Decimal Places? The Surprising Reason Behind Finite Numbers in Roots

[trini]

Ok so I'm playing around with my claculator the other day and i realize that sometimes when you play around with roots(for example root 2 +root 3) you get finite numbers, and these numbers are always 8 dp long. Is there some reason for this? here are a few examples:

root 2 + root 3 = 3.14626437
root 6 x root 8 = 6.92820323
root 3 + root 6 = 4.18154055
root 2 + root 5 = 3.65028154

just to name a few. considering that individually all these roots are infinite, how do they always condense to 8 dp finite numbers?

 

[Borek]

http://en.wikipedia.org/wiki/Floating_point

It all depends on the internal format used.

 

[trini]

ok so they're not actually finite, just a result of internal approximations?

 

[vociferous]

Ok so I'm playing around with my claculator the other day and i realize that sometimes when you play around with roots(for example root 2 +root 3) you get finite numbers, and these numbers are always 8 dp long. Is there some reason for this? here are a few examples:

root 2 + root 3 = 3.14626437
root 6 x root 8 = 6.92820323
root 3 + root 6 = 4.18154055
root 2 + root 5 = 3.65028154

just to name a few. considering that individually all these roots are infinite, how do they always condense to 8 dp finite numbers?

There could be a couple of reasons for that.

Reason 1: The calculator stores results at higher precision but only displays an 8 digit mantissa.

Reason 2: That is the precision that the calculator uses.

To understand precision, you have to understand that integers have a certain range. For instance, an unsigned 8 bit integer has a range of 0 to 2^8. An eight bit signed float has a greater range, but it must sacrifice precision in the mantissa to store values in the exponent.

For instance, if you have a 16 bit number, the range of 0- 10^99 in the floating point digit takes a maximum of seven bits, one bit needs to be used for the sign bit, and the other eight bits can be used to store for the precision of the value, which would limit you to 2^8=256, or effectively, two digit precision. So then your answer to the square root of two might be 1.4.

 

[Borek]

ok so they're not actually finite, just a result of internal approximations?

Bullseye.

 

[Algr]

It would take the calculator an infinite amount of time to calculate an infinite amount of digits - and what practical purpose would that serve?

Calculators aren't designed for deep theoretical stuff. They are for real world operations - carpenters measuring wood to cut, architects, navigators calculating an airplane's course, and so on. Those tasks are unlikely to have measurements or observations that are accurate to eight digits, so such accuracy isn't necessary.

And it is assumed that any good craftsman understands the limits of his tools.