Is it faster to count to the infinite going one by one or two by two? [closed]
YES and NO.
Galileo made the "discovery", the so-called Galileo's paradox, that if you start with the succession of natural numbers:
$1, 2, 3, ...$
and you map it into the succession of even numbers :
$2, 4, 6, ...$
you may map (i.e.associate) every number into its double (today, we call it one-to-one mapping).
So, you have the same "number" of numbers and of even numbers.
Modern set theory (from Cantor on) solved the paradox extending the "counting" process to infinite sets, but proving that the euclidean common notion that "The whole is greater than the part" [see Euclid, The Elements, trans T.L.Heath, Dover, Common notions 5] will not hold for an infinite set.
According to modern set theory, the two above sets can be put in one-to-one correspondence, so they have the same cardinal number, and their "type" of infinity is called denumerable (a set is called denumerable exactly when it can be put in one-to-one correspondence with the set of natural numbers).
But, again by a result of Cantor, not all infinite sets can be put in one-to-one correspondence: there are infinite sets "more infinite" than another. A set of "a larger kind" of infinity is the set of real numbers; it is not denumerable, in the meaning defined above.
As for counting "faster": of course, if you count two by two, after a while you will be way "ahead" of your friend that is counting by one. The only problem is that you will not "end" before him, because there is no final goal to be reached!
Fast or slow has to do with speed. So, if you properly define speed, and if you assume additional hypothesis, like "one count per second", then you can say that counting two by two is faster.
In fact, if you count one by one, then you will count one number per second. If you count two by two, it will be two counts per second. And since $2 > 1$, two by two is faster.
After you two agree that two by two is faster, the kid should be able to realize that this isn't exactly what s/he wanted to ask. As the kid rephrases the question, s/he should get closer and closer to being able to answer by perself. My conclusion is:
Two by two is faster, but it is kind of "useless", since you will not be able to finish anyway.
Asking questions is more important then answering them.
Contrary to the common layman assumption, there is no unique context in which mathematics is being done. Words, especially coming from natural language, can be interpreted in different ways when changing the context.
In this case, the word "faster" can be interpreted in different ways, and it will affect the answer.
"Faster" as a real valued limit. We can think about this as two functions, $f(n)=n$ and $g(n)=2n$. These can be thought as sequences of real numbers, and we can ask whether or not one sequence dominates the other, and what is the limit of their ratios? $$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\frac12$$ Since $\frac12<1$ we have that indeed $g$ is the faster sequence.
"Faster" as a computational process. This is similar to the above case, we define two sequences and calculate the limit of their ratios. However this time we say that if the ratio is a positive (but finite!) constant, then they are more or less the same speed. In this case both sequences have the same "speed".
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"Faster" as a set theoretic process (cardinality). This is not similar to the two cases above, in this case we only measure the cardinality of the output of the sequence. We can ask whether or not the cardinality, which is the rawest notion of "size" for sets, of the two sets $\Bbb N$ and $\{2n\mid n\in\Bbb N\}$ are the same, despite one being a proper subset of the other.
The answer here is that the cardinality is the same, because there is a bijection between the two sets, indeed $g(n)=2n$ is this bijection. So the two methods of counting end up having the same "speed" again.
There are definitely more ways to define which sets go "faster" to infinity. And the result will definitely change from one context to another. The point is that there are different ways we can measure how fast a particular set of integers, or a sequence if you will, thins out as it grows, and it is useful to consider different ways in different contexts.
If you want to explain to a child about the importance of context in which a term is interpreted, ask them what it is easier to carry a $5\ \rm kg$ of iron cast into one block, or a balloon containing $1\ \rm kg$ of air which is huge ($0.85\rm m^3$ in volume). While the iron is definitely heavier, the balloon is definitely much larger and more difficult to handle.
So carrying it by hand makes the iron easier to carry. But if you have a cart on which you can put both objects, then the balloon is much easier to carry because you are pulling a lighter object and you have to put less effort into that.
Therefore there are two ways to interpret "easier to carry" and they depend on the the tools that you have, and so "faster" in mathematics depends on the context in which you are asking the question.
Show your kid this table: $$\begin{matrix} 1&2&3&4&5&6&7\\ \hline 2&4&6&8&10&12&14 \end{matrix}$$ Now let him decide which counting takes longer ...
To answer this, lets try to think like a child and find:
Ironically, counting two by two should be actually slower!
If you count up to the $N$th element of the corresponding sequences ($\mathbb{N}$ and $2\mathbb{N}$), we observe the following pattern: the more digits the number has the longer it takes to spell it.
Take for example $N=8$, the time you need to spell
"one, two, three, four, five, six, seven, eight"
is considerably shorter than the time you need to spell
"two, four, six, eight, ten, twelve, fourteen, sixteen".
One can imagine that this holds true for "most" of the $N$, so skipping the odd numbers will be slower.
Of course, this is far away from a mathematical proof, but probably a child would like the way of thinking. One could bring up the whole Cantor/countability problem at a later point in time, for example after clarifying what we mean by "count".
I did not attempt to prove this "claim". :-) Maybe one can find a subsequence that, after fixing the counting times for the digits, has longer counting times in the first version, who knows...
EDIT: If we represent the numbers with their binary coefficients (for example $(0,1,0,\ldots)$ for 2) and if we assume that one neets equal amounts of time for each syllable we have that the time $\ell((0,1,0,\ldots))$ it needs spell the binary number is $3$: $1$ for "one" and $2$ for "ze-ro". So if we spell $a_n = n$ in its binary representation it takes $\ell(a_n)$ time to do so. Since multiplication by two corresponds to a shift in binary space (add a zero) we have $\ell(b_n)=\ell(2a_n)=\ell(a_n)+2 > \ell(a_n)$. Therefore it always takes longer to spell the even sequence. Asymptotically, of course, the speed is the same: $\frac{\ell(b_n)}{\ell(a_n)}=1+\frac{2}{\ell(a_n)} \rightarrow 1$ as $n \rightarrow \infty.$ A fun fact is, that this is independent of the language. Just replace the number $2$ by the number of syllables used for $0$. :-)