# How do you write a mathematical proof?

1. Sep 25, 2006

### Blahness

I've figured out a faster way of doing simple mixture problems; how do I mathematically prove it consistantly works?

Edit: Is my question badly worded somehow? 30 people have looked at this; 0 have replied. Kind of awkward, but if you don't know what I'm trying to ask, or I need to give more information, please say so.

Last edited: Sep 25, 2006
2. Sep 25, 2006

### chroot

Staff Emeritus
"How do you write a mathematical proof?" is as vague a question as "how do you perform brain surgery?"

Learning how to write rigorous proofs takes a complete education, and many years. It's just not something that someone can answer in a forum post.

For starters, however, you should note that every step in a proof is based either on an axiom of the mathematical system you're using, or a theorem which itself is based on those axioms. Just because something "seems logical" doesn't mean it's really proven or provable.

- Warren

3. Sep 25, 2006

### CRGreathouse

Likewise, I find the question too vague. You don't say much about the problem, you say nothing about what you've tried to do so far, and you haven't even mentioned what level you're at in math (though the fact that you are working on a mixture problem gives an impression).

4. Sep 25, 2006

### pivoxa15

I'd start looking at a beginners logic book than beginners maths proof book. The logic book will give you very good foundations for not just maths but thinking in general.

5. Sep 25, 2006

### 0rthodontist

Logic is good to know but it seems that most math proofs involve a lot of intuition--things that are clearly true and that could be verified messily but the author doesn't bother. As a comment on the intuitive abstraction involved in such informal proofs, one my books from last year stated that many paragraph-long theorems it contains, if reduced to the level of pure symbolic logical deduction, would fill the entire book.

6. Sep 25, 2006

### chroot

Staff Emeritus
Any decent proof provides references to any theorems it uses for brevity.

- Warren

7. Sep 25, 2006

### 0rthodontist

Sure, but it's not just a matter of invoking a theorem on a couple statements, deriving the next statement, invoking another theorem, etc. until you find that you have derived the goal. For example, here's a theorem from the same book:

(if L1 and L2 are languages, then L1 / L2 is defined as the quotient of the two languages, the set of all strings x such that there exists some y in L2 such that xy is in L1)

Theorem: If R is regular, then so is R / L for any language L.
Let M be a finite automaton for R. We define the new finite automaton M' to accept R/L as follows. M' is an exact copy of M, with one exception: we define the accepting states of M' differently--thus M' has the same states, transitions, and start state as M, but possibly different accepting states. A state q of M is an accepting state of M' if and only if there exists a string y in L that takes M from state q to one of its accepting states.
(The Theory of Computation, Bernhard Moret, p. 78-79)

This is very easy to understand if you understand the jargon involved. However, symbolizing it into a series of elementary logical deductions looks like a difficult, tedious task, even if you can invoke a lot of other theorems without re-proving them. The natural language version above doesn't explicitly invoke any other theorems--well, maybe the definition of regular language, but nothing else.

Last edited: Sep 25, 2006
8. Sep 26, 2006

### Blahness

Hmm, I see. Would it be easier if I explained it?

Let's take an arbitary situation, where Mr. Doctor needs 12% and 26% of two solutions to make 100mL of a 15% solution.

First, you take 26%, subtract 15%, that gives you 11, which is the first half of the ratio, and how much you need of the 12%.

Then, you take the 15%, subtract 12%, and that gives you 3, which is the amount of 26% you need compared to 11 of the 12%.

So, the ratio's 11:3, which totals 14. If you take the mL needed, 100, and divide it by 14, that's how much each "unit" is. 100/14 = 7.142(rounded), so the ratio is 11(7.142) : 3(7.142).
Final amounts needed are 78.562mL of the 12% and 21.426mL.

Make sense? I just wish to know how to prove it. :yuck:

9. Sep 26, 2006

### TrevE

I'm confused. Is it a criterion to mathematically prove that mixture problem? Although I'm still new to proofs, I always thought that mixture and other similar problems fell into the calculatory portion, or what is more commonly referred to as "plug-n-chug", of mathematics.

10. Sep 26, 2006

### Data

You can certainly prove that an algorithm for solving problems like this works in general.

Okay, so first let's reformulate your problem:

If $m_1, \, m_2, \, r \in (0, 1)$, $m_1 < m_2, \ m_1 \leq r \leq m_2$, and $a > 0$ then we need to find $k_1, \, k_2 \geq 0$ such that $k_1 + k_2 = a$ and $k_1m_1 + k_2m_2 = ar$.

Here's the interpretation of what I just said when you're doing a mixture problem: $m_1$ and $m_2$ are concentrations of the starting solutions, with $m_1$ being the smaller of the two. $r$ is the concentration of the resulting solution, and $a$ is the amount that we want of resulting solution. We want to find the amounts $k_1$ of the first solution and $k_2$ of the second solution that we should combine in order to have $a$ total solution (ie. $k_1 + k_2 = a$), and in order to make the resulting concentration $r$. This last condition is equvalent to wanting an amount $ar$ of solute in the resulting solution, so it is equivalent to needing $k_1m_1 + k_2m_2 = ar$ (since $k_1m_1$ is the amount that you get from the first starting solution and $k_2m_2$ is the amount that you get from the second starting solution).

You can see that if you are going to describe things mathematically you want to eliminate all of the extraneous circumstantial information. Mathematics isn't concerned with mixtures!

So now you have the problem, stated mathematically. The next step, naturally, is to formulate your conjecture mathematically. In the terms of the mathematical statement of the problem I've written above, your algorithm looks like this:

Step 1: Let $M = m_2 - r$.
Step 2: Let $m = r - m_1$.
Step 3: Let $D = m + M = m_2 - m_1$.
Step 4: Let $U = 1/D$.
Step 5: Let $k_1 = aMU$, $k_2 = amU$.

You can work backwards to get $k_1$ and $k_2$ in terms of $a, m_1, m_2,$ and $r$:

$$k_1 = aMU = \frac{a(m_2 - r)}{D} = \frac{a(m_2 - r)}{m_2 - m_1}$$

and

$$k_2 = amU = \frac{a(r - m_1)}{D} = \frac{a(r - m_1)}{m_2 - m_1}$$.

You are now in a position to state your conjecture:

If $m_1, \, m_2, \, r \in (0, 1)$, $m_1 < m_2, \ m_1 \leq r \leq m_2$, and $a > 0$ then setting $k_1 = a(m_2 - r)/(m_2 - m_1)$ and $k_2 = a(r - m_1)/(m_2 - m_1)$ gives $k_1 + k_2 = a$ and $k_1m_1 + k_2m_2 = ar$.

We will see later that some of the assumptions made in the statement of the conjecture are not at all necessary, so in fact we can prove more general statements with the same argument as we will use for this one.

Now all that's left is the proof:

We have

$$k_1 = \frac{a(m_2 - r)}{m_2 - m_1}, \ k_2 = \frac{a(r- m_1)}{m_2 - m_1},$$

so indeed,

$$k_1 + k_2 = \frac{a(m_2 - r)}{m_2 - m_1} + \frac{a(r-m_1)}{m_2 - m_1}$$

$$= \frac{a(m_2 - r) + a(r - m_1)}{m_2 - m_1} = \frac{a(m_2 - m_1 + r - r)}{m_2 - m_1} = \frac{a(m_2 - m_1)}{m_2 - m_1}$$

$$= a$$

and

$$k_1m_1 + k_2m_2 = \frac{a(m_2 - r)}{m_2 - m_1}m_1 + \frac{a(r - m_1)}{m_2 - m_1}m_2$$

$$= \frac{a}{m_2 - m_1}\left[ (m_2 - r)m_1 + (r - m_1)m_2\right] = \frac{a}{m_2 - m_1}\left[ m_2m_1 - rm_1 + rm_2 - m_1m_2\right]$$

$$=\frac{a}{m_2-m_1}\left[(rm_2 - rm_1) + (m_2m_1 - m_1m_2)\right] = \frac{a(rm_2 - rm_1)}{m_2 - m_1} = \frac{ar(m_2 - m_1)}{m_2-m_1}$$

$$=ar.$$

QED. (I left in most of the arithmetic, but I might have fallen asleep and skipped some steps! :tongue2:)

As I said before, in fact this is a proof for the more general result that the system of equations $\{k_1m_1 + k_2m_2 = ar, \ k_1 + k_2 = a \}$ has the solution $k_1 = a(m_2-r)/(m_2-m_1)$, $k_2 = a(r-m_1)/(m_2-m_1)$ whenever $m_2 \neq m_1$ (ie. all the other assumptions I made when I stated the earlier conjecture were unnecessary, and just there to make it look more like a mixture problem). You can see this because we never used any of the other assumptions in the proof (you need $m_1 \neq m_2$ because otherwise you can't divide by $m_2-m_1$ in the first place!).

Last edited: Sep 26, 2006
11. Sep 27, 2006

### Blahness

Well-explained and perfectly followable!

Thank you! (Seems to be proven by the normally done method of doing these problems, $\{k_1m_1 + k_2m_2 = ar, \ k_1 + k_2 = a \}$), but it proves it works in all situations as shown.

Thank you!