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How to write proofs: a quick guide
Eugenia Cheng
Department of Mathematics, University of Chicago
Web: http://www.math.uchicago.edu/∼eugenia
October 2004
A proof is like a poem, or a painting, or a building, or a bridge, or a novel, or a symphony.
“Help! I don’t know how to write a proof!” Well, did anyone ever tell you what a proof is, and how to go about writing one? Maybe not. In which case it’s no wonder you’re perplexed. Writing a good proof is not supposed to be something we can just sit down and do. It’s like writing a poem in a foreign language. First you have to learn the language. And then you have to know it well enough to write poetry in it, not just say “Which way is it to the train station please?” Even when you know how to do it, writing a proof takes planning, effort and inspira- tion. Great artists do make sketches before starting a painting for real; great architects make plans before building a building; great engineers make plans before building a bridge; great authors plan their novels before writing them; great musicians plan their symphonies before composing them. And yes, great mathematicians plan their proofs in advance as well.
Contents
- 1 What does a proof look like?
- 2 Why is writing a proof hard?
- 3 What sort of things do we try and prove?
- 4 The general shape of a proof
- 5 What doesn’t a proof look like?
- 6 Practicalities: how to think up a proof
- 7 Some more specific shapes of proofs
- 8 Proof by contradiction
- 9 Exercises: What is wrong with the following “proofs”?
3 What sort of things do we try and prove?
Here is a classification of the sorts of things we prove (this list is not exhaustive, and it’s also not clear cut – there is some overlap, depending on how you look at it):
- x = y i.e. “something equals something else”
- x =⇒ y
- x ⇐⇒ y
- x is purple (or has some other interesting and relevant property)
- ∀x p(x) is true i.e. “all animals of a certain kind x behave in a certain way p(x)”
- ∃x s.t. p(x) is true i.e. “there is an animal that behaves in a certain way p(x)”
- Suppose that a, b, c and d are true. Then e is true. [Note that this is just a version of 2 in disguise.]
4 The general shape of a proof
Let’s now have a look at the general shape of a proof, before taking a closer look at what it might look like for each of the cases above. We must always remember that there is a beginning, a middle and an end.
Example 1. Using the field axioms, prove that a(b − c) = ab − ac for any real numbers a, b, c. You may use the fact that x.0 = 0 for any real number x.
beginning field axioms definition x − y = x + (−y) given x. 0 = 0
middle a(b − c) = a(b + (−c)) definition = ab + a(−c) distributive law
ac + a(−c) = a(c + (−c)) distributive law = a. 0 additive inverse = 0 given ∴ a(−c) = −(ac) definition of additive inverse
∴ ab + a(−c) = ab − ac
end ∴ by line 2, a(b − c) = ab − ac as required
Example 2. Let f and g be functions A f −→ B g −→ C. Show that if f and g are injective then g ◦ f is injective
beginning definition of injective definition (g ◦ f)(a) = g(f(a)) assumption that f and g are injective i.e. ∀a, a′^ ∈ A f(a) = f(a′) =⇒ a = a′ ∀b, b′^ ∈ B g(b) = g(b′) =⇒ b = b′
middle (g ◦ f)(a) = (g ◦ f)(a′) =⇒ g(f(a)) = g(f(a′)) by definition =⇒ f(a) = f(a′) since g is injective =⇒ a = a′^ since f is injective
∴ (g ◦ f)(a) = (g ◦ f)(a′) =⇒ a = a′
end i.e. g ◦ f is injective, as required
Example 3. Prove by induction that ∀n ∈ N, 1 + · · · + n =
n(n + 1) 2
beginning Principle of Induction
middle for n = 1, LHS = 1
RHS =
∴ result is true for n = 1
If result is true for n = k then
1 + · · · + k + (k + 1 ) =
k(k + 1 ) 2
= k(k + 1 ) + 2 (k + 1 ) 2 = (k + 1 )(k + 2 ) 2 i.e. result true for n = k + 1
∴ result true for k =⇒ result true for k + 1
end ∴ by the Principle of Induction, the result is true for all n ∈ N
Sense any make doesn’t it backwards but things right the write you if.
This is a terrible thing to do but not a terminal catastrophe – if you have all the right ideas but in the wrong order, all you need to do is work out how to put them in the right order...
2. Take flying leaps instead of earthbound steps.
This category includes leaping from one statement to another
- without justifying the leap
- leaving out too many steps in between
- using a profound theorem without proving it
- (worse) using a profound theorem without even mentioning it
For example, spot the flying leap in the following “proof”:
a(b − c) = ab + a(−c) = ab − ac
3. Take flying leaps and land flat on your face in the mud
By which I mean making steps that are actually wrong. The end may well justify the means in some worlds, but in mathematics if you use the wrong means to get to the right end, you haven’t actually got to the end at all. You just think you have. But it’s a figment of your imagination. Here’s an example of a very imaginitive “proof” that is definitely flat on its face in the mud:
a(b − c) = ab + a(−c) = ab + a(−c) + a. 1 = ab + a( 1 − c) = ab − ac Of course, it’s even worse if you do something illegal and thereby reach a conclusion that isn’t even true. Like
x^2 = y^2 =⇒ x = y
or x^2 < y^2 =⇒ x < y.
What is wrong with these two “deductions”?
4. Handwaving
Handwaving is when you arrive at a statement by some not-very-mathematical means. The step isn’t necessarily wrong, but you haven’t arrived at it in a good logical manner. Perhaps you had to resort to writing a few sentences of prose in English rather than Mathematics-speak. This is often a sign that you’ve got the right idea but you haven’t worked out how to express it. Spot the handwaving here – you can see it from a mile off:
a(b − c) = ab + a(−c) a(−c) = −ac because if you add ac to both sides then both sides vanish which means they′re inverse
∴ ab + a(−c) = ab − ac Handwaving is bad but is not ultimately catastrophic – you just need to learn how to translate from English into Mathematics. This is probably easier to learn than the problem of coming up with the right idea in the first place.
5. Incorrect logic
This includes the two great classics
- negating a statement incorrectly
- proving the converse of something instead of the thing itself
What is the negation of the following statement:
∀ε > 0 ∃δ > 0 s.t. ∀x satisfying 0 < |x − a| < δ, |f (x) − l| < ε
The correct answer is at the bottom of the page^1. If you get it wrong, you go directly to Jail. Do not pass Go. Do not collect $200.
6. Incorrect assumption
You could have all your logic right, you could make a series of perfectly good and sensible steps, but if you start in the wrong place then you’re not going to have a good proof. Or, if you use any assumption along the way that simply isn’t true, then it’s all going to go horribly pear-shaped...
7. Incorrect use of definitions – or use of incorrect definitions
This is a very, very avoidable error. Especially if it’s not a test and so you have all your notes and all the books in the world to consult: getting the definitions wrong is a really
(^1) ∃ε > 0 s.t. ∀δ > 0 ∃x satisfying 0 < |x − a| < δ s.t. |f (x) − l| ≥ ε
- Try and manipulate both the beginning and the end to try and make them look like one another. This is like building from both ends of the bridge until they meet in the middle, and it’s okay as long as you write the whole thing out properly in the right order afterwards.
- Take big leaps to see what happens, and then make the big leaps into smaller leaps afterwards.
- See if the situation reminds you of any situations you’ve ever seen before. If so, perhaps you can copy the method.
You should always read over your proof after you’ve written it to make sure every single step makes sense. When you’re writing a proof the first time through, you might get carried away in a frenzy of inspiration and become blind to the world around you – by which I mean that you might do something wrong without noticing it. It’s important to go back in a calm state and pretend to be more stupid than you are. Or more sceptical. Or untrusting. When you finish a proof you should feel like you understand what’s going on, but you should go back over it pretending that you don’t understand, and see if your proof explains it to you.
7 Some more specific shapes of proofs
Now let’s look at the various types of things that we try to prove (as listed in Section 3), and think about how we can prove them.
1. x = y or “something equals something else”
The proof might take the following general shape:
x = a = b = c = d = y
Or:
x = a = b = c
y = e = d = c
∴ x = y
Note that this is very different from:
x = y a = e b = d c = c
2. x =⇒ y
Now the proof might look like this:
x =⇒ a =⇒ b =⇒ c =⇒ d =⇒ y
Or:
We know that a =⇒ b
Also a ⇐⇒ x and b ⇐⇒ y
∴ x =⇒ y
3. x ⇐⇒ y
Now the proof might look like this:
x =⇒ a =⇒ b =⇒ c =⇒ d =⇒ y
Conversely y =⇒ p =⇒ q =⇒ r =⇒ x
Hence x ⇐⇒ y
Note that we have picked an arbitrary x, and then we just can just prove that this x has the desired property, and we’re done. You might say, “But we’ve only proved it for this x, and not every x”. But the point is that this is a random x not one specific one, it’s a sort of generic x that shows the proof will work for any specific one that we substituted in. It’s not like proving the result for one particular number, say, 23.
6. ∃x s.t. p(x) is true
Here, all we have to do is find one x for which p(x) is true. So we can just say “let x = 23” and then show that 23 has the desired property p. For example, prove that:
∃ δ > 0 s.t. |x| < δ =⇒ |x^2 | <
Put δ = 101. Now |x^2 | = |x|^2 so we have
|x| <
=⇒ |x^2 | <
This is fine; of course we could also have picked
δ =
or δ =
The latter especially would be a little eccentric but would still be a perfectly valid (if violent) choice of δ to finish the problem off. Of course, sometimes it’s a bit hard to just pluck a valid x out of thin air. It’s a bit like pulling a rabbit out of a hat – it looks like magic, but of course you are the one who put the rabbit there in the first place. So if it’s a complicated example we probably work out (on a rough piece of paper) which x is going to do the trick, and once we have it all worked out we can pull it out of the hat.
7. If a, b, c, d are true then e is true
When you have a whole lot of things (a, b, c, d,.. .) you’re allowed to assume, it gets more complicated. You might have to develop several parts of it sort of at the same time before proceeding to the end, like a novel where there are several strands of plot happening at the same time before they all come together at the end for the final d´enouement. The proof might look like this:
a =⇒ z b and z =⇒ y c =⇒ x x and d =⇒ w y and w =⇒ e
In fact if we draw a little picture of what happened, it’s much easier to see what’s going on (and see where the beginning, the middle and the end have got to):
a b c d
z x
y w
e
?
?
?
?
?
Here’s an example of this phenomenon at work:
Prove that if a > 0 ∈ R then a^2 > 0. You may assume that for all x, y ∈ R, (−x)(y) = −(xy) and −(−x) = x.
Now a < 0 means 0 − a ∈ P
i.e. −a ∈ P.
Therefore (−a)(−a) ∈ P
by closure of P under multiplication. Now
(−a)(−a) = −a(−a) by the first given assumption = −((−a)(a)) commutativity of multiplication = −(−(a^2 )) first given = a^2 second given
∴ (−a)(−a) > 0 =⇒ a^2 > 0 as required
Here’s an example:
Using the field axioms, prove that 0 has no multiplicative inverse in R.
Suppose that 0 does have a multiplicative inverse. This means
∃x s.t. 0.x = 1
But we know that 0x = 0 ∀x ∈ R
and
∴ 0 .x has no multiplicative inverse.
9 Exercises: What is wrong with the following “proofs”?
Example 1. Prove that ∀x 6 = 0 ∈ R, ∣ ∣ ∣∣^1 x
|x|
x
∣ =^
|x|
x ≥ 0 ⇐⇒
x
so
x
x
x < 0 ⇐⇒
x
so −
x
−x
= −
x
Example 2. Let f and g be functions A f −→ B g −→ C. Show that if f and g are injective then g ◦ f is injective
f(a) = f(a′) =⇒ a = a′ g(a) = g(a′) =⇒ a = a′ (g ◦ f)(a) = (g ◦ f)(a′) =⇒ g(f(a)) = g(f(a′))
But g injective =⇒ a = a′.
∴ g ◦ f is injective as required.
Example 3. Prove that ∀a > 0 ∈ R, ∃x ∈ R s.t. x^2 > a
(2a)^2 = 4a^2 > a ∴ put x = 2a
Example 4. Using only the field axioms, prove that ∀x, y ∈ R
x^2 − y^2 = (x + y)(x − y).
(x + y)(x − y) = x(x − y) + y(x − y) distributive law = x^2 + x(−y) + yx + y(−y) distributive law = x^2 + x(−y) + xy + y(−y) commutativity of multiplication = x^2 + x.(− 1 ).(y) + xy + y.(− 1 ).(y) additive inverse = x^2 + xy.(− 1 ) + xy + (− 1 ).(y^2 ) commutativity of multiplication = x^2 + xy(− 1 + 1 ) − y^2 distributivity, additive inverse = x^2 + xy( 0 ) − y^2 additive inverse = x^2 + 0 − y^2 definition of 0 = x^2 − y^2 additive identity
