Can 2 + 2 ≈ 5 be true?

Question

I was wondering a little, about how to proof that 2+2 =5

And here I am:

2.4 + 2.4 = 4.8

If we approximated numbers in each side individually then :

2 + 2 ≈ 5

I know this may not be right, but I don't know why it's wrong.

Answer 1

About an approximation ("$\approx$") one concern numerical analysts will have is how much error of approximation is involved.

Certainly when we truncate $2.4$ to $2$ we can see the error introduced is less than half of one unit. When we then add the two approximations ($2+2$) it is possible that the error of approximation increases (although it is also possible, if the truncation errors were opposite in sign instead of being the same, that the error would decrease by "cancellation").

Depending on the circumstance we may want to measure or bound the error of approximation in different ways. When addition is involved, the triangle inequality lets us bound the absolute error. That is, if the error of approximation in $a$ is $\varepsilon_a$ and that in $b$ is $\varepsilon_b$, then the error in $a+b$ is $\varepsilon_a + \varepsilon_b$. Since we normally don't know exactly what the errors of approximation of summands are, typically we settle for an upper bound:

$$ |\varepsilon_{a+b}| \le |\varepsilon_a| + |\varepsilon_b| $$

This is going to be helpful when upper bounds for the absolute error in $a,b$ are available to give an upper bound for the absolute error in $a+b$.

At each step new error of approximation can be introduced by rounding, as happens in your example. That is, the initial rounding error of $0.4$ in each summand becomes $0.8$ error in the sum (comparing the exact $4.8$ with the approximate sum $4$), and this error gets conflated with a further rounding of $4.8$ to $5$. The final $0.2$ error of rounding combined with the earlier $0.8$ explains the full unit of error of approximation in $2+2\approx 5$.

Answer 2

When you are talking about approximations you are usually doing so in a real world context. And so whether or not $4 \approx 5$ will depend on what you are measuring.

For example, the distance from the earth to the moon is $239,000$ miles. This is approximately equal to $240,000$ miles. So there $1,000$ is basically equal to zero. (That is $9,000$ is approximately $10,000$.)

But if I am on a road trip then $1,000$ miles is definitely not approximately equal to zero.

Answer 3

You haven't shown "$2+2\approx 5$"; instead, you've shown that "$+$" is not a well-defined operation with respect to $\approx$! Specifically, an operation $*$ is well-defined with respect to a relation $E$ if $a_1Eb_1$ and $a_2Eb_2$ implies $(a_1*a_2)E(b_1*b_2)$. Think of it as $*$ is not ambiguous with respect to $E$.

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Note that usually we talk about well-definedness with respect to equivalence relations. $\approx$ isn't an equivalence relation since it's not transitive ($1\approx 1.4\approx 1.8$ but $1\not\approx 1.8$). This can be fixed by defining $a\approx b$ iff the closest integer to $a$ is also the closest integer to $b$ (where we round up on halves, say). This is now an equivalence relation.

This is a minor point, though; the real issue is that you've implicitly assumed that "$+$" is well-defined with respect to $\approx$, and that's not the case (and indeed you've proved that it is not).