Answer: 40
Step-by-step explanation:
We are attempting to show that ab=ba, where a and b are non-negative integers. Let’s introduce a new equivalence, b+e=a (i.e. e is defined as the difference between a and b; note that if e=0 then the proof becomes trivial. Also note that we assume a≥b; this does not affect our result as the variables are symmetrical). Now we write:
1) ab=∑ai=ib
This is nothing more than stating the definition of b multiplied by a, ie b summed a times. We can also write:
2) ba=b(b+e)
since b+e=a, by our own definition. We now try to show that equation (2) can be rewritten in the form of equation (1). We expand on equation (2) by writing:
3) b(b+e)=∑bi=1(b+e)
This is very similar to what we did regarding equation (1), ie b(b+e) is just (b+e) summed b times. Using some properties of addition, we can transform the right hand side of (3) to read:
4) ∑bi=1(b+e)=∑bi=1b+∑bi=1e
Now what we are going to do is assume the very thing we set out to prove! That is usually a big no-no unless you are using induction, which is basically where this is going.
5) Assume: ∑bi=1e=∑ei=1b
We now substitute (5) back into (4) and proceed:
6) ∑bi=1(b+e)=∑bi=1b+∑ei=1b
The right hand side of (6) can be simplified:
7) ∑bi=1b+∑ei=1b=∑b+ei=1b
Finally, we make use of the fact that b+e=a:
8) ∑b+ei=1b=∑ai=1b
Comparing (8) to (1) yields what we set out to prove:
9) ab=ba