# Sum and product of integers conserving order

I have the feeling this is true, but can't prove it: $$\sum_n^An\lt\sum_n^Bn\implies\prod_n^An\lt\prod_n^Bn$$ Where $A\subset\mathbb N-\{0, 1\},B\subset\mathbb N-\{0, 1\}$

Example: $$3+4\lt5+6\implies3\cdot4\lt5\cdot6$$

I'm still in high school, and tried with the tools I have, but couldn't figure anything out.

How would I prove this?

-
So far I'm the only one who's up-voted the question. –  Michael Hardy Dec 17 '13 at 19:44

Counterexample: $4 + 8 < 3 + 10$ but $4\cdot8 > 3\cdot10$

-
Well, I guess this is enough! Thanks (answer will be accepted as soon as SE lets me) –  throwaway Dec 17 '13 at 19:14
I'd like to add the following: If $m_1 < m_2$ and $n_1 < n_2$ and all are positive, then $m_1 + n_1 < m_2 + n_2$ and $m_1n_1 < m_2n_2$. The example you chose was like that... –  Gautam Shenoy Dec 17 '13 at 19:21

Here's another way to see that the statement can't be true. If it were true that $$\sum_{n\in A} n < \sum_{n\in B} n \implies \prod_{n\in A} n < \prod_{n\in B} n$$ then by symmetry it would also have to be true that $$\sum_{n\in B} n < \sum_{n\in A} n \implies \prod_{n\in B} n < \prod_{n\in A} n$$ Combined, these two statements would yield the weaker statement $$\sum_{n\in A} n \ne \sum_{n\in B} n \implies \prod_{n\in A} n \ne \prod_{n\in B} n$$ which in turn would yield its contrapositive, $$\prod_{n\in A} n = \prod_{n\in B} n \implies \sum_{n\in A} n = \sum_{n\in B} n$$ But this says that the product of a set of numbers determines the sum, which is outrageous.

Edit to try to clarify in response to Michael Hardy's answer: The conjecture as stated in the question might or might not immediately elicit any feeling of truth or falsity; the OP thought it was probably true. But I believe that most people, if set the task of determining the sum of some secret numbers, having been told only their product, would very quickly come to feel that the task was impossible, whether or not they were immediately able to furnish a counterexample. My principal goal in presenting this line of reasoning is to ground the falseness of the conjecture in the kind of everyday number sense that's at work when people reach such conclusions without having explicit counterexamples in mind. Of course a concrete numerical counterexample is more conclusive, but in isolation it gives little or no insight. Also of course, any particular reader might have a different "everyday number sense" than what I imagine, whether through their own idiosyncrasy or through my misjudgement, and this explanation might thus fail. Finally of course, whatever number sense a reader has does indeed ultimately come from concrete numerical experience, so this "argument" in a sense conceals an appeal to concrete numerical examples — but people generalize from their experience, and it's just good communication to refer to those generalizations, even if they're informal, even if doing so leaves some formal work yet to be done.

-

This will express what I take to be Steven Taschuk's point far more concretely. $$60 = 2\times30=3\times 20=4\times15=5\times12$$

But $$5+12\quad<\quad4+15\quad<\quad3+20\quad<\quad2+30.$$ The sums get bigger but the product does not.

(And we would have no way to conclude that this "is outrageous", as Steven Taschuk puts it, without knowing that there are concrete examples.)

-

The easiest to find counterexample would be $$1.2+1.2 < 1+1+1 \not\rightarrow 1.44 < 1$$

You can add infinitly many 1's to be greater than any other sum, but the product of 1's will always be 1.

-