PotatoSauce <email@example.com> wrote: > On Monday, May 7, 2012 6:05:38 PM UTC-4, Ross wrote: <snip> > > The second is noting that: > > > > x^4 = x * x * x * x > > x^3 = x * x * x > > x^2 = x * x > > x^1 = x > > x^0 = ? > > How about > > x^4 = 1*1 * x * x * x * x > x^3 = 1*1 * x * x * x > x^2 = 1*1 * x * x > x^1 = 1*1 * x > x^0 = 1*1 > > I put two "1"s instead of just 1, in case you feel like multiplication > can only be done when there are at least two numbers.
I had thought of posting essentially the same thing, together with an explanation:
Algebra students know that a term's numerical coefficient, when not otherwise specified, is _understood_ to be 1. But let's _explicitly_ show that numerical coefficient on the right side of the equations in the pattern that Ross gave.
x^4 = 1 * x * x * x * x x^3 = 1 * x * x * x x^2 = 1 * x * x x^1 = 1 * x x^0 = 1
Here's what's going on. In _any_ multiplication, there is a factor of 1 "in the background", so to speak. Of course, since 1 is the multiplicative identity, having it in the background is perfectly innocuous. And if our product is in "simplified form", that 1 in the background will _never_ be seen _unless there is no other factor of any sort_.
In the pattern above, going from the top line down, we erase a factor of x on the right each time until, when no factors of x remain, the background 1 is finally revealed!