a + b + c = 1, what are the requirements for numbers to satisfy this?

Question

This is a very informal "proof", I think, which shows what sorts of number could satisfy the equation a + b + c = 1. I'm not actually sure whether this is even a proof, but I'll refer to it from now on as a proof anyway.

I started off by thinking about whether all the numbers could be positive, all the numbers could be negative, and then continue from there. My questions are:

• What am I actually doing here? What would you call this piece of maths? Is it a proof? It doesn't seem to be proving anything.
• what could I do to improve my proof?
• For example, are there scenarios I haven't considered? What alternative ways of doing this are there?

And here it is.

Could a, b and c all be positive? NO

The smallest possible total in that case is 1 + 1 + 1 = 3 and all other positive numbers are going to produce larger numbers than that, so no (note: I'm assuming 0 is not positive.)

Could a, b, and c all be negative? NO

Let's write this as -a + -b + -c = 1 The easiest way for this to equal 1 is if the total of two of the numbers, let's pick a and b, equal o, and subtracting c from -a + -b results in 1. This means that c would actually be +1, not a negative number. So this scenario is impossible.

Could two of the numbers be negative, and the final one positive? YES

Let's change the previous scenario it slightly so that a + b now has to be less than 0. For a difference of +1 between (- a + - b) and c, c would have to be |- a + -b| + 1. So in this case, it's possible.

Could a and b be positive while c is negative? YES

(a + b) - c = 1

can be rearranged into:

(a + b) - 1 = c

Answer 1

This is Linear Algebra.

The rule of thumb in Linear Algebra is that you will need as much (independent) equations as you have variables to come up with a unique solution. In your case there is one equation, but three variables. This means there are two degrees of freedom and no unique solution exists.

One solution to your equation would be (could you write another solution like this?)

\begin{align} a&=a^* \\ b&=b^* \\ c&=1-a^*-b^* \end{align}

where you choose $a^*$ and $b^*$, to be an integer in you case. But this is not restricted to the integers.

Answer 2

You've nearly proved all the possibilities, provided your numbers are integers. Your final scenario "$a$ and $b$ positive, $c$ negative" needs to be "two of them positive, the other one negative". Otherwise you're leaving out the cases where $a$ or $b$ is the negative one.

You've proved all the cases you list, but left those two out.

As mentioned in comments, you also need to say right at the start that they're integers, to define what you're proving.

But as far as you went, yes it's a proof.

And don't worry that what you've proved seems obvious: obvious things need proving too, since something can look obvious but be wrong.

Also it's a good way of practising proving things!

Afterthought

To check for missed cases: you've got three integers, each of which can be either positive or negative. So there are $2×2×2=8$ cases. You've covered

• all positive: $1$ case
• all negative: $1$ case
• one positive, two negative: $3$ cases depending which is the positive number
• $a$ and $b$ positive, $c$ negative: $1$ case.

Adding these together, we've only covered $6$ cases, so two are missing. Looking at the list, the last item seems to spoil the pattern, so maybe the missing cases belong there . . . Should it be like the one before but the other way round? . . . Yes it should, and now we've got all $8$ cases.