
Re: NonEuclidean Arithmetic
Posted:
Sep 3, 2012 12:38 PM


He is pointing out that there is a problem with the usual way of talking about multiplication as repeated addition.
For example: The phrase "5 added to itself once" can be taken to mean 5 + 5. Likewise the phrases "5 added to itself twice and "5 added to itself three times" respectively can be taken to mean 5 + 5 + 5 and 5 + 5 + 5 + 5. And so on.
The usual way of putting putting forth 5 times 3 as repeated addition is to say that it is 5 added to itself three times. But by the above, this can be taken to mean that 5 times 3 is 5 + 5 + 5 + 5, which is wrong.
That is, to say it right  to say it so that there is a lower probability of such confusion  we need to make it so that what we say implies that the number of times one of the factors appears as an addend is equal to the other factor.
He did this by giving an alternative way of saying it, which would result in 5 times 3 being equal to 0 + 5 + 5 + 5. The number of times that 5 appears as an addend is equal to the other factor 3.
As I said before, I've given an alternative way of saying it many times here at mathteach, which would be along the line of "x times y" is "x instances of y added up" or "y instances of x added up". (Or we could say something like "x times y" is "the sum of x instances of y" or "the sum of y instances of x".) Or we could use some other term than "instance" for this, as long as the language is clearer than the usual way in terms of implying that the number of times one of the factors appears as an addend is equal to the other factor.
When applied to this example above, saying that 5 times 3 is 5 instances of 3 added up or 3 instances of 5 added up would be saying that 5 times 3 is equal to 3 + 3 + 3 + 3 + 3 or 5 + 5 + 5. The number of times one of the factors appears as an addend is equal to the other factor.
Like I said before, I prefer something along the line of this latter approach because I prefer that the sum written out as a model for multiplication have all the same addends. (With the former approach, not all the addends are the same  one of the addends is 0 while all the other addends are the same nonzero number.)
On Sun, Sep 2, 2012 at 12:17 PM, Robert Hansen <bob@rsccore.com> wrote: > So, are you saying that 3 x 1 = 4? > > When a teacher says "added x number of times" they write the multiplicand x > number of times, not the addition symbol. When I ask you what is the sum of > 12, 34, 16 and 7 (4 addends) I am asking what is SUM(12,34,16,7) and the > algorithm to do that is to start with 12, then add 34, then add 16, then add > 7. > > There has to be a word for when people pivot on semantics like this and > create problems that aren't even there. I have been calling them > "semanticists" but that isn't accurate. > > The photocopy example is actually a solution to the problem "1 + x = 4", not > "x * 1 = 4". > > Bob Hansen > > > On Sep 1, 2012, at 1:58 AM, Jonathan Crabtree <sendtojonathan@yahoo.com.au> > wrote: > > P.S. Think of a photocopying machine. You have one letter and you need four > altogether. So what button do you press on the multiplication machine? > Three! ie k1 Press the four button and you end up with five.

