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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.

mohamedrashad
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    Because you did an approximation, you should technically use $\approx$ instead of $=$. Note that $2+2\approx 5$ isn't wrong (although very few would write that without at least some context). – Arthur Sep 11 '16 at 18:39
  • If you don't know why something is wrong, which happens suprisingly often in mathematics, a good first step is to try and phrase very precisely why you think it should be true. – user2520938 Sep 11 '16 at 18:41
  • @Arthur I edited it, thank you – mohamedrashad Sep 11 '16 at 18:43
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    It simply cannot be proved that $2+2=5$. Do you have any reasons to think that it can be proved? – drhab Sep 11 '16 at 18:46
  • If you allow $\approx$ to behave as follows: If $a \approx a'$ and $b \approx b'$ then $a + b \approx a' + b'$ ($\approx$ is an additive homomorpism) or if $a\approx b$ and $b \approx c$ then $a \approx c$ ($\approx$ is transitive), then you can get any equation that you want. – user357980 Sep 11 '16 at 18:46
  • I know it can't, But I want to know why this proof is wrong :) – mohamedrashad Sep 11 '16 at 18:47
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    @mohamedrashad Your argument doesn't show $2+2=5$. – Wojowu Sep 11 '16 at 18:49

3 Answers3

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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 $E$.

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).

Noah Schweber
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  • In fact one might even argue that the OP doesn't have a well defined relation. – Thomas Sep 11 '16 at 19:03
  • @Thomas That's pretty clearly the case; however, I think it's worth making the point that even if the OP gives a precise definition of "$\approx$," they won't have shown what they think. – Noah Schweber Sep 11 '16 at 19:04
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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 $2+2\approx 5$.

hardmath
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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.

Thomas
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