[dento]

I case it helps, this is how it was taught to me when introducing proofs.

For your example case, it's simply the matter of seeing multiplication as repeated addition. So, 2x is x+x, and from this follows:

2(x+y) = (x+y)+(x+y) = x+y+x+y = x+x+y+y = 2x+2y

However, for the other case, when we convert the multiplication by 2 to an addition and back:

2(xy) = (xy)+(xy) = xy+xy = 2xy

Now we can show that this works for n instead of two:

n(x+y) = (x+y)+..[n times]..+(x+y) = x+..[n times]..+x + y+..[n times]..+y = nx+ny

And for multiplication:

n(xy) = xy+..[n times]..+xy = nxy

[JadeNB]

Thanks for the suggestion, which I think is exactly the right way to address this particular issue.

Indeed, now that I've learned from a few times teaching the course that it always comes up, I will address it. But, if I spend my time building up basic-algebra proficiency at this level, then I'm never going to get into the meat of proving things. Perhaps the answer is to view the very act of writing your argument as a proof—and I like that perspective (because it genuinely is!); but, if I were to break that down to its full details, then I would have to write 2(x + y) = (x + y) + (x + y) = x + (y + (x + y)) = x + (y + (y + x)) = x + ((y + y) + x) = x + (x + (y + y)) = (x + x) + (y + y) = 2x + 2y, and writing down proofs of that sort risks alienating students who are eager for a conceptual picture, and guarantees giving the wrong idea of what sort of activity proving is. (And it also risks confusing why later I will say that, for a silly example, π(x + y) = πx + πy "by the distributive law"—I can't think of multiplication by pi as repeated addition, so I must cite the distributive law; and then why didn't I just do that before?)