# Adding 3000 (an approximate number) and 5 (an exact number)–would it still be 3005?

There is an old anecdote. A group of tourists visits a museum. A tourist asks the museum worker: 'How old is this statue?' The worker responds: 'It is 3005 years old.' The tourist expresses his astonishment: 'Wow, it can't be! It must be some kind of mistake!" The worker explains calmly: "At the moment I was hired, the director told me the statue is 3000 years old. I have been working here for five years already. So, 3000 + 5 = 3005."

It can be a very silly or even a shameful question for an adult, but... was the worker actually right?

I always had problems with approximate calculations. Actually, none of my teachers have ever explained it to me. I live in Moldova, and it is a poor country, although I had been studying for 18 years (lyceum, college and university). I decided to ask my question here; it's better to ask someone and to acquire some knowledge than to remain a dumbass for the rest of the life.

My first step was to assume that 3000 in this problem is an approximate number and 5 is an exact number. Then I watched a video: significant figures. It seems like in case of addition, we must keep as many decimal places as it has the addend with least number of decimal places. For example, adding 2.36 and 12.1. If those were exact numbers, the sum would be equal to 14.46. However, they're approximated numbers; 2.36 has two decimal places and 12.1 has only one. The smaller number is 1. So we must round up 14.46 to one decimal place. The result is 14.5.

Now, back to our problem. I assume 3000 has zero decimal places; 5 also has zero decimal places. 3000 + 5 = 3005. We must round up 3005 to zero decimal places and it remains as is.

So is there an error?

• I've heard the same story with $100$ million year old fossils... Of course, saying something is that old is very, very imprecise. While it would not be any less accurate to say that those fossils were in fact $100$ million plus $1$ years old, it would be odd. It would suggest a degree of precision which is clearly neither intended nor available. So, with that in mind, I'd say the statue stays $3000$ years old for at least the next $100$ years or so.
– lulu
May 28 at 23:10
• Exactly $5$ plus approximately $3000$ equals approximately $3005$, which equals approximately $3000$. May 29 at 0:20
• "Significant figures" are unrelated to the decimal point. In your example, the number 3000 has only one significant figure, which is the 3. So you would round 3005 to one significant figure, which becomes 3000. The number of significant figures is the number of digits we are sure of (up to rounding, starting from the left). But the correct way to do these calculations is to consider the error intervals, as explained in Tony Mathew's answer.
– wimi
May 29 at 7:57
• Your question isn't silly. It's one of the things I learnt as a student, but didn't properly comprehend the full implications. For example, you've helped me realize that the significant figures rule can have an error May 29 at 10:59

Starting with your comment on significant figures, I like to view the decimal precision used as implicitly telling you the possible error/deviation/uncertainty. So if someone says 12.1, unless that's the exact value, what they're saying is that 12.1 is the rounded (to one decimal place) version of the true value. Let's assume they're rounding to the nearest (sometimes people round-down; e.g. omitting the latter decimal places). Then the true value must be between 12.05 and 12.15. So if $$x$$ is the true value, the following are different ways to write an expression for it $$\begin{gather} 12.05 \leq x \leq 12.15 \\ x = 12.1 \pm 0.05 \\ x = 12.1 + \epsilon, \quad -0.05 \leq \epsilon \leq 0.05 \end{gather}$$

So if we consider 2.36 + 12.1, we can write calculation as (I prefer the algebraic approach) \begin{align} y &= 2.36 + \epsilon_y, \; -0.005 \leq \epsilon_y \leq 0.005 \\ x &= 12.1 + \epsilon_x, \; -0.05 \leq \epsilon_x \leq 0.05 \\ z &= y + x = 14.46 + \epsilon_y + \epsilon_x \\ &= 14.46 + \epsilon_z, \quad \epsilon_z \equiv \epsilon_y + \epsilon_x \\ &\Rightarrow \; -0.055 \leq \epsilon_z \leq 0.055 \\ &\Rightarrow \; 14.405 \leq z \leq 14.515 \end{align}

So in terms of decimal precision, 2.36 + 12.1 should be written as 14, since it can lie between 14.405 (2.355 + 12.05) and 14.515 (2.365 + 12.15). But for simplicity of teaching students, and because often a slight error in decimal precision isn't very important, people generally use the significant figures rule you mentioned to arrive at 14.5.

But before we use this approach for your original problem, we need to establish the uncertainties in the given data. The worker said 5 years. It's reasonable to assume that they would have given that number even if they had actually been working for 4 years 6 months, or 5 years 6 months. So there's an uncertainty of 0.5 in their data point. The director's 3000 is bit more subjective. Based on my gut feeling, I'ld give it an uncertainty of 500. My logic is that the director said 3000, but thought of it as "3 thousand". So there's only one significant figure in their answer, but in the thousands place. So using these uncertainties,

\begin{align} y &= 3000 + \epsilon_y, \; -500 \leq \epsilon_y \leq 500 \\ x &= 5 + \epsilon_x, \; -0.5 \leq \epsilon_x \leq 0.5 \\ z &= y + x = 3005 + \epsilon_y + \epsilon_x \\ &= 3005 + \epsilon_z, \quad \epsilon_z \equiv \epsilon_y + \epsilon_x \\ &\Rightarrow \; -500.5 \leq \epsilon_z \leq 500.5 \\ &\Rightarrow \; 2504.5 \leq z \leq 3505.5 \end{align}

So my mathematical answer would be between 2000 and 4000 years. But practically speaking, since the uncertainty in the director's data is much larger than the contribution from the worker's data, I'ld still just say "around 3000 years"

• Interesting answer, but please do not use the big O notation in this way, it is used for asymptotic comparison, so O(0.01) is the same as O(1). Use e.g. ±
– Joce
May 29 at 7:54
• Fair enough. I've editted that out May 29 at 10:40