by Axel Arturo Barceló Aspeitia
The issue of whether deduction can gives us new knowledge is one of extreme philosophical awkwardness. On the one hand, almost anyone with some knowledge of logic or mathematics would find the question triffling in so far as it is obvious that at least some genuine knowledge must be obtained through competent deduction. However, finding a good example has proved an ellusive endeavour. The holly grial would be to find a simple, straightforward deduction that would certainly deliver new knowledge. But this seems almost impossible, in so far as if the deduction is simple, then the consequence must be obviously contained in the premises and as such, it would be difficult for anyone to have conscious knowledge of the premises without also knowing the conclusion. Nevertheless, in a recent talk, profrs. Héctor Hernández Ortíz and Víctor Cantero just gave what I think is the best example in this regards: a straightforward simple deduction from very intuitive premises – or, at least, premises that anyone with even the most rudimentary knowledge of arithmetic would accept – to a very unintuitive but true conclusion – in this case, a conclusion that only people with some sophisticated knowledge of arithmetic would accept. The arguments is as follows:
Controversial conclusion to prove: 0.999999999… = 1
Uncontroversial premise: 1/3 = 0.3333333333…
We multiply both sides by 3: 3(1/3) = 3(0.3333333333…)
An we immediately get the desired conclusion: 1 = 0.999999999…
According to Hernández and Cantero, this is a case of a non-vicious circular agument. But where exactly is the circularity, and is it actually non-vicious? Or is it question begging instead? Here is my take on where to find the ciruclarity in the argument. The key step is:
3(0.3333333333…) = 0.999999999
Presumably, the reason why we find this step uncontroversial is because it is based on the very elemental arithmetic fact that 3 times 3 is 9. However, it is not part of our basic arithmetical knowledge how to multiply infinite decimal expressions, and since this is a multiplication of an infinite decimal series a finite number of times, the issue in the background is how do we extrapolate from what we know about the multiplication of finite expressions to infinite expressions. One way of solving this would be to reason by some kind of informal induction thus:
3 x 0.3 = 0.9
3 x 0.33 = 0.99
3 x 0.333 = 0.999
3 x 0.3333 = 0.9999
3 x 0.33333333… = 0.9999999999…
Presumably, this, or something relevantly similar, is behind why we find the claim that 3 x 0.33333333… = 0.9999999999… very intuitive. However, one could question the validity of this induction by appealing to something also very basic, i.e., something we learn in elementary arithmetic lessons about how to multiply numbers in decimal system. As you all might recall, we do not start multiplying large numbers from the left to the right, but from the right to the left. However, this is what seems to be happenning in this case: since we cannot start multiplying from the right, because the series in infinite, we start from the left. Nevertheless, there is a reason why we do not start from the left: because numbers might “carry over” from the right and this might mess up our result. In other words, the numbers to the right might add up to more than ten (or a hundred or a thousand, depending on the size of the number) and this would require us to revise our result. For example, 371 times 2 does not start with 6, but with 7 because 71 times 2 is more than a hundred, so the one from the one hundred carries over, so to speak, to the left. In contrast, 312 times 2 starts with 6 because 12 times 2 is less than a hundred and so nothing gets carried over to the left. In a little more formal terms, the general rule when multiplying in decimal notation is that,
For any digit D occupying the nth place (from right to left) of number C expressed in decimal notation, the product of C times N expressed in decimal notation has digit (D times N) [or the rightmost digit of D times N expressed in decimal notation if D times N is larger than te] if and only if the decimal expression to the right of D (that is, the decimal expression composed of the first n-1 digits of C counting from right to left) times N is strictly less than the number corresponding to the numeral in decimal notation of a one followed by n-1 zeroes.
This sounds complex, but the idea is simple. Consider our previous sunple example: if we multiply by 2 a number starting with 3 (from left to right) and followed by, say, other two digits whatsoever (to its right), the result will start with 6 (from left to right) only if 2 times the number to the right of the 3 is less than 100 (that is, a one followed by two zeroes). That is, if we multiply 312 times 2, the result – 624 – starts with 6 – 3 times 2 – because the numer to the right of the 3 –12 – times 2 – 24 – is less than a hundred – a one followed by two zeroes, because there are two numerals to the right of 3 in 312; but if we multiply 371 times 2, the result – 742 – does not start with 6, because 71 times 2 – 142 – is more than a hundred. That is why we do not start multiplying from the left.
The rule is simple and basic. It is one of the first things we learn when we learn how to multiply expressions in decimal notation. However, when we apply it to this infinite case, we do not get the result that Hernández and Cantero wanted.
According to this rule, 0.33333…. times 3 would start with 0.9…. if and only if the number to the right of the leftmost 3, times 3, were strictly less that 0.1. This means that if 3(0.33333….) = 0.99999…., then 3(0.033333…) must be strictly less than 0.1. But if 3(0.33333….) = 0.99999…., then 3(0.033333…) is 0.0999999…. and thus 0.099999… would have to be strictly less than 0.1. Multiplying both sides of the inequality by ten, we get that 0.999999…. is strictly less than 1 which is in direct contradiction with our starting assumption – and Hernández and González desired result – that 0.999999…. = 1.
Thus, we have two different ways of extrapolating from the case of finite multiplication – that is, the multiplication of finite expressions in decimal notation – to infinite multiplication. Each one gives us different and inconsistent results. There is nothing wrong in adopting the first way and not the second, but there is nothing particulary right about it either. We must take other stuff in consideration to make the decision. And this is what should be meant when we say that this argument is circular but not viciously so. It is circular in so far as it is based on a step that is valid only if we reject other, equally suited ways of performing the relevant operation, inconsistent with it. It is not vicious in so far as it does establish a rational inferential link between the premises and the conclusion. Furthermore, it gives us new knowledge in so far as, as Hernández and Cantero correctly point out, the premises and operations involved in the argument are substantially more intuitive and uncontroversial than the final conclusion.