Proving is a process - an example proof on a property of logarithms
Many students fear writing proofs and just cannot do them. Remember that writing proofs is just like solving any other problem: it is a process. Often, it is a process of trial and error. The proofs you see in math books are polished and finished and read oh-so smoothly from step to step. It makes you wonder how in the world did they suddenly know to use this or that as the next step?
In reality, solving math problems or writing proofs is similar to when people solve problems in their chosen trade and use their reasoning powers naturally without any anxiety over it. For example, a doctor has a problem every time a patient comes in with mysterious symptoms. An auto mechanic has to solve the problem as to why the car does not run smoothly. A housewife may be solving the problem of how to make the special cookies she tasted a week ago somewhere.
Yet these people don't lose hope or give up when facing problems. They start somewhere, see if it works, and go on from there. Trial and error; another trial.
In times past, children used to take on their father's trade. They would observe the father work, and no doubt they would see dad face problems he didn't know how to solve at first. The children saw a problem solving process first hand and learned from that!
Letting students be like apprentices that observe skilled math problem solving can be very useful to them! Try it sometimes: use a challenging problem and think through the problem aloud as you solve it. Make plain to the listeners the entire process of mistakes, different ideas, and maybe even checking back to reference books. It can be eye-opening to youngsters that math problems are usually NOT solved by some mysterious "AHA!" moment where you know the answer at first sight.
Here you will see an example of a problem solving process... What follows was my true thinking process in solving a particular problem. This is exactly how I thought through it.
Somebody asked me to prove that
n logaM = logaMn.
I tackled the problem as a math teacher that has been away from logarithms for 5-6 years. I knew that antilogarithms, or raising the number a to some power would probably be helpful since I know that in general, logarithms often get solved by the exponentiation process.
So as an attempt to make things simpler, I decided to call the complex looking thing on the right side x. I wrote x = logaMn, from which using the definition of logarithm I got ax = Mn.
Okay, that looked simpler, so I did the same for the other side of the original equation, calling it y:
y = n logaM
from which dividing both sides by n, I got y/n = logaM, and now using the definition of logarithm I got ay/n = M.
So thus farm using only the definition, I had these two equations: ax = Mn and ay/n = M. They looked kind of similar. I saw quickly that taking the nth root of the first equation, it would become ax/n = M (because nth root is the same as number to the 1/n th power). And if you have ay/n = M and ax/n = M, then obviously ay/n = ax/n. It looked good, because I do know the fact that in this situation, the exponents must be the same, or y/n = x/n and so y = x.
And that's exactly what I was supposed to prove, because I chose x and y to be the two sides in the original problem.
BUT, my proof certainly didn't look like one of those textbook "dandy" 2-3 steps proofs. Could I arrange it to look like that? Well, looking back, if I take antilogarithms (or raise a to these powers) on both sides of this equation, I get
|n logaM ? logaMn|
|a n logaM ? alogaMn|
Okay, it looks kind of complicated, BUT it also looks like it could proceed somewhere, because I can try use the property of exponents that cdz = (cd)z :
|a n logaM ? alogaMn|
|(a logaM)n ? alogaMn|
Well, the left side is now simply Mn (since a logaM is M) — and the right side simplifies too, because alogaMn is simply Mn. (a raised to a power and logarithm base a are opposite operations).
But this still wasn't a "textbook" polished proof, because I was using a question mark instead of equal sign to mark that I don't yet know if the two things are equal. To write a proof, I'd start from something that is known and proceed from there to whatever I needed to prove, whereas above I took what I was supposed to prove and started playing around, hoping to get to a result that is true. I did get to something that is true, so I can now rewrite the process in the opposite order, and that should look like a finished proof.
So starting from the known fact that
Mn = Mn
we can write M = a logaM and Mn = a loga(Mn), and get:
(a logaM)n = aloga(Mn)
Next, on the left side we can use the property of exponents that says cdz = (cd)z:
a logaM n = aloga(Mn)
Now, in the above equation we have two powers that are equal. Since the bases are equal (a and a), the exponents also must be equal:
logaM n = loga(Mn)
Let's reverse the order of multiplication on the left side...
n logaM = loga(Mn)
...and that's it!
Now, I assume textbooks might present a different proof, and probably simpler... Don't get me wrong, I have proved it! But I would need to try again, I guess, to find the ultimate "simple" proof.
However, I hope you got the point of how proving or any other problem solving is a process. Sometimes you may find a solution that is a true solution, just not the most aesthetic one. But if your quest is to recreate those cookies you tasted at a friend's house, even if you don't quite achieve the same exact taste, you can still enjoy the cookies you made - if they still turn out to be cookies, that is.
Also remember, it's OK to go astray for a while when solving a problem... as long as you keep "monitoring" your own progress and take some other route if things don't look too good - but at the same time don't erase all your erroneous work. Instead, analyze it. Analyzing your own work will so often give you a clue for a new way forward.