How to prove the theorems of algebra using axioms?

Question

Like the vast majority of math students I'm having trouble with proofs. I would like to get some advice on how to approach proofs. What are the steps that need to be done?, how do I know that the proof is correct?. Here is how I approached a proof.

I need to prove that a(b-c)=ab-ac By the distributive field axiom of real numbers x(y+z)=xy+xz Let a=x;b=y;c=-z So a(b+(-c))=ab+a(-c)=a(b-c). I don't know what's next and when I can say with confidence that I have proven the theorem.

Thanks a lot in advance.

Answer 1

Proofs always have an antecedent. This is what you assume before you begin the proof.

Pf.

Suppose our expression is $a(b-c)$, where $a,b,c\in\mathbb{R}$. Always state what the variables belong to. In this case they belong to the real numbers ($\mathbb{R}$)

w.t.s: (want to show), $a(b-c)=ab-ac$

The above line is for us so we don't loose track of our goal.

By the distributive field axiom of real numbers $x(y+z)=xy+xz$. Always state the name of the theorem when necessary, like you have.

Let $a=x;b=y;c=-z$

So we have that $a(b-c)=ab+(-ac)=ab-ac$.

Good, now we have showed what we wanted through the theorem. Now we end the proof.

$\therefore$ by distributive field axiom, $a(b-c)=ab-ac$.

$QED.$

The QED is Latin for 'quod erat demonstrandum' and is used to signify the end of a proof.

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General Idea: Pf. State antecedent.

State where the variables belong to.

State the theorem's involved.

Show how the theorem's apply and how we reached our result.

Finish with a therefore statement, or simply end with QED.

Answer 2

In your case, you should begin with $a(b-c)$ and end up with $ab-ac$. Here is how you should do it:

\begin{align} \ a(b-c) & \ =a(b+(-c))\\ & \ =ab+a(-c)\\ & \ =ab+(-ac) \\ & \ =ab-ac \\ \end{align}

Usually you tackle these kinds of proofs by applying each needed axiom step by step. Sometimes you can also use previous results. For example, use the above to show what $(a+b)(c-d)$ is.

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Now for more complex theorems, you usually use some techniques. While each branch of maths have their own particular tricks, those 3 techniques can be widely applied:

• Case breakdown

Simply break the case into smaller problems.

For example, in set theory, to show that two sets $A$ and $B$ are equal, you usually first prove $A\subseteq B$ and then $A\supseteq B$. Proving "if and only if" theorems also usually use this technique.

• Contradiction

You assume the result you want to prove is false, and then show that it leads to an absurd.

For example, proving that there is no greatest even integer: suppose there is a greatest integer which we call $n$. Then for all even number $m$ we have $m \le n$. Let $m=n+2$. Then $m$ is even and $m\ge n$. But that's an absurd, since we assumed $n$ was the greatest even number. So, our assumption was false, thus there can be no greatest even integer.

• Induction

Used when you have some finitely sequential theorem to prove (the number can be arbitrarily large, but it cannot be infinite). You verify if you hypothesis hold for the first term, then verify that, if it holds for the previous term, it holds for the current one.

The idea is that it holds for $n=1$, thus it holds for $n=2$, thus it holds for $n=3$ and so on.

For example, proving that $1+3+5+...+(2n-3)+(2n-1)=n^2$. It's obvious it holds if $n=1$. So, suppose it holds for $n-1$. Thus by hypothesis:

\begin{align} \ 1+3+...+(2(n-1)-1)\ & \ =1+3+...(2n-3) \\ & \ =(n-1)^2\ \end{align}

Then for $n$: \begin{align} \ 1+3+...+(2n-3)+(2n-1)\ & \ =(n-1)^2+(2n-1) \\ & \ =(n^2-2n+1)+(2n-1) \\ & \ = n^2\\ \end{align}

• Counter-example

You can prove the falsity of an statement when you find a counter-example. If you suspect something may go awry if the conditions are not "regular enough", try to use a pathlogical cases.

Each branch can define what is "regular enough" for what you need to work with: compact sets in Topology, continuous/differentiable functions in Analysis etc. And then those that break our intuitive notions: Cantor sets, non-continuous functions, space-filling curves etc.

Finding a counter-example may be hard, but trying a few of those pathological suspects (particular to each area) first might lead to some insight and the need to strengthen your hypothesis.

For example, if you need to provide a counter-example for all subsets of $\mathbb{R}$, try the rationals or irrationals. They're pathological in the sense they aren't intervals (which can be too regular to prove very general results), and every interval contains a rational and an irrational number.

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Hope it helped to nail down some outlines for proofs.

The easiest proof at first is directly applying the axioms and contradiction proofs, once you get the hang of it. Case breakdown may need some expertise for more complex cases, since you must be sure to cover all possible cases.

Induction can be the strangest by far if not properly explained, but after some practice it becomes natural (even more when you learn that the principle of induction is one of the axioms at the foundation of modern arihmetic)

Counterexamples are hit-or-miss: sometimes you can easily see one, sometimes it taks a whole day, and sometimes they don't exist at all.