# Sinificant figures

imagine having a square box with 4 meters of length and hieght its area obviosly is 16 meters square but according to significant figure the number of digits in a result of a calculation cannot be greater than the number of digits in the observations (length and hieght both have one digit ) so isnt that a bit contradictory?????????

16 meters squared would be correct if the dimensions were measured to be 4.0 meters. So in this case the correct answer is 2 x 10^1 meters squared (if you mean the area of one of the square sides on the box). If you mean the volume of the box then we calculate 4 x 4 x 4 = 64 meters cubed. So then the correct answer would be 6 x 10^1 meters cubed or 6 x 10^7 centimeters cubed (as you already learned it has to be rounded to the same amount of significant figures as the measurement with the smallest amount of significant figures). Don't forget you must use the unrounded result if any further calculations have to be done with it. The final answer has to be rounded the same way again.

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imagine having a square box with 4 meters of length and hieght its area obviosly is 16 meters square but according to significant figure the number of digits in a result of a calculation cannot be greater than the number of digits in the observations (length and hieght both have one digit ) so isnt that a bit contradictory?????????
There are 2 different considerations here.
Normally in problem calculations in physics, we are given numbers to put into equations and calculate answers. The general advice given is as you say. Give the answer to the least number of significant figures in the data supplied. The data supplied, on the other hand, should be consistent. The value in your example should (would?) be stated as 4.0m

On the other hand, we also use values measured in experiments to calculate something else using a formula, possibly. Here the treatment is different.
It's contradictory if the value of 4m you "obtained" for the length was only certain to ±1m
As it is written it's a bit ambiguous. Having said that, if you want to carry through your calculation, you could say that the value of 4±1 has a 25% uncertainty. [± 1 in 4]
When you square it you increase the uncertainty to 50%
Your final value of 16 could be quoted as having a 50% uncertainty, that is, ±8
16±8m²
You can look at it another way:
If the initial value is 4±1m then it could possibly be 3m or 5m
3x3 is 9 and 5x5 is 25
So your result could be anywhere between 9m² and 25m². This is almost the same result as using the percentages.
The lesson is, if you want your answer of 16, you need to express the value of 4 to a suitable degree of certainty.

PS On the other hand, if you're a mathematician and dealing only with integers, there is no problem. 4²=16 every time!

There are 2 different considerations here.
Normally in problem calculations in physics, we are given numbers to put into equations and calculate answers. The general advice given is as you say. Give the answer to the least number of significant figures in the data supplied. The data supplied, on the other hand, should be consistent. The value in your example should (would?) be stated as 4.0m

On the other hand, we also use values measured in experiments to calculate something else using a formula, possibly. Here the treatment is different.
It's contradictory if the value of 4m you "obtained" for the length was only certain to ±1m
As it is written it's a bit ambiguous. Having said that, if you want to carry through your calculation, you could say that the value of 4±1 has a 25% uncertainty. [± 1 in 4]
When you square it you increase the uncertainty to 50%
Your final value of 16 could be quoted as having a 50% uncertainty, that is, ±8
16±8m²
You can look at it another way:
If the initial value is 4±1m then it could possibly be 3m or 5m
3x3 is 9 and 5x5 is 25
So your result could be anywhere between 9m² and 25m². This is almost the same result as using the percentages.
The lesson is, if you want your answer of 16, you need to express the value of 4 to a suitable degree of certainty.

PS On the other hand, if you're a mathematician and dealing only with integers, there is no problem. 4²=16 every time!

why would you consider that the values measured in an controlled enviorment have so much uncertainty ?

why would you consider that the values measured in an controlled enviorment have so much uncertainty ?

I wouldn't. It's just an example.
In an experiment, if the value of 4 was obtained by measurement, there would have be a corresponding value of the uncertainty quoted. 4±something. That something would, in turn, determine how confident you are of the value of 4 squared.
To confidently give the result 16±something, you would need to have measured the value of the length of the side to be at least 4.0 ±something
As I said, there is a difference between working with physical measurements in calculations, and using integers in pure mathematics. As an integer, 4 is 4±nothing. It's known to absolute precision.

I wouldn't. It's just an example.
In an experiment, if the value of 4 was obtained by measurement, there would have be a corresponding value of the uncertainty quoted. 4±something. That something would, in turn, determine how confident you are of the value of 4 squared.
To confidently give the result 16±something, you would need to have measured the value of the length of the side to be at least 4.0 ±something
As I said, there is a difference between working with physical measurements in calculations, and using integers in pure mathematics. As an integer, 4 is 4±nothing. It's known to absolute precision.

so what you are saying is that the measurements always have some amount of uncertainty related to them if we add or subtract that uncertainty for example for the length of a wire measured 4m we will 4+or- the uncertainty which will give another digit meaning that we can afford to have more than one digit in the result of an experiment in which we may use this measurement right ?????????

Yes, all measurements have some amount of uncertainty associated with them.
If you use such a measurement to calculate something else, then that final calculation has a corresponding uncertainty associated with it.
The more accurate your original measurement, the more accurate your final calculation.
As a general rule, the more (legitimate) significant figures you have in the original value, the more you can have in the final calculation.
The point I was making is that you shouldn't get this mixed up with purely mathematical calculations such as what is 4 X 4? The answer is exactly 16 because the original value of 4 is not subject to any uncertainty. It is an exact integer; and so is 16.
If you measure the side of a square and use it to calculate the area, then the uncertainty in the area is determined by the uncertainty in the original measurement or the side.
In any physics experiment, you would always estimate the uncertainty in that original measurement, and express it as, for example, 4.0 ±0.1m; or even 4.000 ±0.001m if you can measure to 1mm.

alrigth let us suppose that the uncertaity is very low let say +0.0005 and the original measure ment is 5m so the after correction we will have 4.0095 and let suppose that the measurement is of the side of a squared box now that we have 5 significant figures considering the box is a square when the area of one side of the box will be calculated we can afford to have 5 sinificant figures in the result assuming the other side has same uncertainty but we dont see that in an actual experiment why ????

Sorry, I don't understand what you are now asking.
What do you mean "we dont see that in an actual experiment"?

In my first post I wrote
There are 2 different considerations here.
Normally in problem calculations in physics, we are given numbers to put into equations and calculate answers. The general advice given is as you say. Give the answer to the least number of significant figures in the data supplied. The data supplied, on the other hand, should be consistent. The value in your example should (would?) be stated as 4.0m

I then wrote
On the other hand, we also use values measured in experiments to calculate something else using a formula, possibly. Here the treatment is different.

Sorry, I don't understand what you are now asking.
What do you mean "we dont see that in an actual experiment"?

In my first post I wrote

I then wrote