/sci/ - Science & Math

[Anonymous No.6822561]

So I'm doing my intro to proofs class and I thought I run something very easy by you guys. We were told to find something easy out and then prove it. I chose to show that when multiplying to numbers that add up to some amount, you get the largest number by having them equal each other. (e.g. let's take 6 for example, the largest number there is 3*3 = 9, as opposed to 2 * 4 = 8, or 1*5 = 5)

Here's my proof, I want to know if it qualifies as a proof.

Theorem:
Let N = a + b, where a, b > 0
then ab is the largest when a = b

Proof:
<div class="math">a^2 > ab</div> (To be proved)
where a != b
Let b < a
<div class="math">a^2 > ab</div>
<div class="math">a > b</div>
Q.E.D

The reason I'm worried and the reason I think this works is because I'm allowed to say "let b < a", right? It's just an arbitrary variable, so I can choose any of them to be greater or smaller right?

[Anonymous No.6822576]

>>6822561

should be claim: N^2/4 >= ab

[Anonymous No.6822594]

why not just use the fact that the function
f(a) = a*(N- a) for fixed N
has a maximum at a = N/2?

[Anonymous No.6822612]

>>6822594
Well this was just the one I thought of, but that one works too.

[Anonymous No.6822649]

>>6822561
The idea is right but the proof has a structural flaw.

You said, let a < b. That suggests that there is another case where b < a that you didn't cover. So either you have to prove that it has to be the case that b < a or you have to restructure your proof or you have to add in some text justifying why you can choose b < a. Another red flag is that you never actually used the assumption 'a + b = N'. So here's a question, is it necessary that a + a < N?

I think what you're trying to say, in your proof, is.

Let a, b be positive integers. Let N = a + b.
>Case 1:
Suppose b < a
then ab < a^2 (multiply by a on both sides)
>Case 2:
Suppose a < b
then ab < b^2
...

The problem with the proof is more clear now. As an example, suppose you're talking about N=6. Then a + b are both positive integers such that 6 = a + b. So for example, 2 + 4. Then your proof is saying.
if 2 < 4 then (2)(4) < (4)(4)
in other words 8 < 16.
But that's not what you want to say because while 2 + 4 = 6, it is not the case that 4 + 4 = 6.

What you're actually trying to prove is that (2)(4) < (3)(3).

[Anonymous No.6822719]

>>6822649
OK, that's helpful. I don't need to cover the two cases because they're identical, except with different letters for the same variable, a or b. I guess I could say "let the smaller one be a" but I said "let a < b" which is the same. So no, I don't think that first part is accurate.

The second part yes, what I was supposed to do was to say (a^2)/4 or (b^2)/4

[Anonymous No.6822724]

>>6822719
Sorry, N^2/4 or what:
>>6822576
said

[Anonymous No.6822736]

>>6822724
You could also write everything in terms of the "center number".

In other words.
Let c = N/2
Your other two numbers are c plus/minus some radius.
Let r be an integer such that 0 <= r <= c (<= means less than or equal).
Then let a = (c + r) and b = (c - r)
Note that a + b = N.

Now all that's left to show is that c^2 is greater than or equal to ab.
ab = (c + r)(c - r) = c^2 - r^2
and since r^2 is greater than or equal to zero it follows that c^2 is greater than or equal to ab.

There's a lot of different ways to do this proof. Play around with it.