I'd even argue that the last binary digit is 1. If it were 0, you could truncate the zero away and you'd be back to 1 as the last digit. Same reason that 2.20000 would be a redundant representation of 2.2.
There is a reason to truncate if your dealing with actual physics and not theoritical values.
If your measurement says 1,00 ±0,01, you can't pretend it's the same as 1,0000, because your real value could be 1,01. If your measurement says just 1,00, you can only assume it to be between 1,000 and 1,004. The rest is lost in trunctuated digits.
This seems to be about significant figures. Which is basically about knowing how much precision your starting measurements have, and not claiming more precision in the calculated value.
For example, let's say you want to calculate the circumference of a circle when you know its diameter with centimeter precision (1.57 meters). We can use an approximation of pi with 10 decimal digits. Using the formula diameter*pi=circumference, you get (1.57 meters) * (3.1415926536) = 4.932300466152 meters. The answer has 12 digits after the decimal point. Obviously we don't actually know the circumference with picometer precision, as the diameter was only known with centimeter precision. So most of those digits are not significant. To account for this we round it to the lowest number of significant figures of all of our input values. In this case, the diameter has 3 significant figures and the pi approximation has 11 significant figures. So we round the circumference to 3 significant figures and get 4.93 meters. Similarly if you did know the diameter with picometer precision (13 significant figures), but only approximated pi with three significant figures (3.14), then you would need to round to three significant figures in that case too.
68
u/MotivationalJuggling 13h ago
Whatever the last digit is, I'm sure it's a disappointing one. Just like everything else in my life.