What follows is an example of a direct proof, the most basic form of proof. This post won’t be very interesting for people who already write proofs.

*The photographs are of M.C. Escher’s Metamorphosis II, a woodcut print made during 1939-1940. *

★You don’t need to understand every detail. The ideal response is a spark of interest in mathematics. At minimum, you should know in advance what an integer is.

The integers: numbers that can be written without fractions or decimals.

symbol denoting the set of integers

The set of integers: {…, -3, -2, -1, 0, 1, 2, 3, … }

First, the proof is presented with zero explanation. Just note how short it is and how most of it consists of elementary arithmetic operations.

Afterwards, the proof is re-presented with excessive explanation.

**Result: If x is an even integer, then 5x – 3 is an odd integer. **

**Proof: Assume that x is an even integer. Since x is even, we can write x = 2k for some integer k. Then **

**5x – 3 = 5(2k) – 3 = 10k – 3 = 10k – 4 + 1 = 2(5k – 2) + 1 **

**Since 5k – 2 is an integer, 5x – 3 is odd. **

**QED **

Now, taking it step by step:

**Result: If x is an even integer, then 5x – 3 is an odd integer. **

**Proof: Assume that x is an even integer. **

Direct proofs demonstrate that when the if-statement is true, then the then-statement is forced to be true. So the first step of a direct proof always assumes the if-statement is true.

**Since x is even, we can write x = 2k for some integer k.**

We don’t know what x is specifically, but we do know that x is a number divisible by 2 since x is even. k is an unknown, nonspecific integer, which is why we say in a vague way ‘for some integer k’. Side note: if x is an odd integer, we represent it as x = 2k + 1 for some integer k, since an odd integer has remainder 1 when divided by 2.

**5x – 3 = 5(2k) – 3 = 10k – 3 = 10k – 4 + 1 = 2(5k – 2) + 1**

5x – 3 is the expression whose oddness/evenness we’re ultimately concerned about. So we take the information from our last step, that we can write x = 2k, and we replace the x in 5x – 3 with our new representation of x, that is 2k. So

5x – 3 = 5*(2k) – 3

Now we simplify our new expression with arithmetic.

5x – 3 = 10k – 3

We change 10k – 3 to 10k – 4 + 1 since these expressions are equal, but 10k – 4 + 1 divided by 2 will have a remainder of 0 or 1 whereas 10k – 3 has a remainder of -1. According to our definition of odd numbers (represented with 2k + 1, where k is an integer) and even numbers (2k where k is an integer), we want a remainder 1 or 0.

Now we want to know if our expression (10k – 4 + 1) is odd or even, so we divide it by 2. If there is no remainder in dividing by 2, then our expression is even. Observe that there is a remainder:

5x – 3 = 10k – 4 + 1 = 2(5k – 2) + 1

**Now, since (5k – 2) is an integer, 5x – 3 is odd. **

In the statement 5x – 3 = 2(5k – 2) + 1, the expression (5k – 2) is just some integer. We know this because k, 5, and -2 are integers and multiplying or adding integers results in an integer (see algebraic properties).

Since 5k – 2 is just some integer, we can reduce it to the unknown integer symbol k. In the equation, this reduces to 5x – 3 = 2(5k – 2) + 1 = 2k + 1, which is an odd integer. So we know that 5x – 3 is always equal to some odd integer when x is an even integer. So the result is true.

**QED**

QED just means the proof is done. It stands for the Latin phrase ‘quod erat demonstrandum’ which means ‘which had to be demonstrated’.

You can also indicate that your proof is done with this symbol ∎, which is called a tombstone.

Note that we started the proof by assuming the first part of the if-then statement is true. We assumed x is an even integer. Then we try to prove that the second part (5x – 3 is an odd integer) is true. Why this proves the result is true (even in cases when the first part is false) relies on simple, unintimidating logic.

Let P be a statement and Q be another statement. P ⟹ Q means that ‘P implies Q’ or ‘if P, then Q’.

This is a truth table for implication:

In the proof above, P is ‘x is an even integer’ and Q is ‘5x – 3 is an odd integer’. P ⟹ Q is:** **If x is an even integer, then 5x – 3 is an odd integer. In the truth table above, if P is false, then regardless of whether Q is true or false, P ⟹ Q is true. If P is true and Q is true, then again P ⟹ Q is true. Consequently, we only need to prove that it never happens that P is true and Q is false. Otherwise, our result P ⟹ Q isn’t true.

I learned most of what I know about proof writing from Mathematical Proofs by Chartrand et al. It was helpful in the transition from calculus/linear algebra to higher level math classes. However, it certainly doesn’t require calculus or linear algebra. With this book, you can start from scratch as long as you know some algebra.

I assure you that the other types of proofs (especially contradiction and induction) are more fun and interesting than direct proofs. Furthermore, there are much more elegant, relevant, beautiful, significant, interesting results and theorems than this one. The main point of using this proof as an example was to demonstrate that it isn’t out of anyone’s grasp to understand and to write proofs.

I used to think math was dry, uncreative, lifeless, sterilized, and monotonous. I only started studying it (beyond a high school level) regularly and seriously a little over a year ago. Since then I’ve found that mathematics can be fascinating, beautiful, meaningful, even romantic. I’ve felt some of the same captivation and passion while writing proofs that I feel while writing poems.