Here is a habit that will save you hours in an exam. Before you set up any induction proof — before you even write "base case" — plug real numbers into the statement and check whether it is actually true. Use n = 1, then n = 2, then n = 3. Do the arithmetic on the calculator or by hand. If the statement fails on any small n, stop. There is nothing to prove, and continuing only buys you the wrong answer delivered with high confidence.
This feels too simple to be worth saying. It isn't. Students burn through entire exam questions proving false statements because the textbook or the teacher or the hastily-written problem set had a typo, and the student never tested the claim with real numbers before committing to the symbolic work.
Why the habit matters
Induction is a machine that converts "the statement holds at one value, and it propagates forward" into "the statement holds for every value." The machine is reliable — provided both inputs are true. If the base case is false, the whole edifice collapses. But the symbolic form of the base case can mislead you. You plug n = 1 into the formula, get a symbolic expression, and pronounce it "obviously equal" without ever comparing actual numbers.
The fix is mechanical: evaluate both sides of the equation numerically, side by side. Write the left-hand side's value. Write the right-hand side's value. Ask: are they the same number? If yes, proceed. If no, stop.
Why symbolic checking is not enough: a symbolic expression like "\frac{1 \cdot 2}{2}" looks like it must equal 1, and your brain will assent before your eyes have actually computed. If the formula on the right was supposed to be \frac{n(n-1)}{2} but a typo made it \frac{n(n+1)}{2}, symbolic pattern-matching will not catch the error — both forms look plausible. Arithmetic catches it. Plug n = 1: left side = 1, right side = 0 (for the typo version). Now you know.
The three-number check
Before attempting any induction, compute the statement for n = 1, 2, 3. Tabulate:
| n | Left side | Right side | Match? |
|---|---|---|---|
| 1 | 1 | 1 | yes |
| 2 | 3 | 3 | yes |
| 3 | 6 | 6 | yes |
This is the check for 1 + 2 + \dots + n = \dfrac{n(n+1)}{2}. Three matches do not prove the statement (you cannot prove "for all n" by checking finite cases), but they tell you two critical things:
- The statement is not false at the low end. You are not wasting your time proving a typo.
- You understand what the two sides are supposed to equal. This understanding is essential when you get to the inductive step and need to manipulate the formula with confidence.
If any of the three checks fails, something is wrong. Maybe the formula is wrong, maybe you misread the problem, maybe the indexing is off by one. Fix it before starting the symbolic proof.
A worked trap
Consider the claim: for every positive integer n, n^2 + n + 41 is prime.
A famous almost-theorem. The formula gives primes for n = 1, 2, 3, \dots, 39. If you check the first three values:
- n = 1: 1 + 1 + 41 = 43. Prime. Good.
- n = 2: 4 + 2 + 41 = 47. Prime. Good.
- n = 3: 9 + 3 + 41 = 53. Prime. Good.
And you might conclude: "this pattern is clearly true, let me try to prove it by induction." But the claim is false — at n = 41, you get 41^2 + 41 + 41 = 41 \cdot 43, which is not prime. Three-number checks do not catch every falsity. But they catch many, and they cost nothing.
The moral is not that three checks are sufficient. The moral is that three checks are a cheap filter that eliminates the most common errors — typos, sign mistakes, off-by-one indexing. More careful verification is still needed, but the filter catches enough problems to be worth doing every single time.
A second trap: forgetting to verify the base case at the actual starting value
Sometimes the claim in a problem is "prove for every n \geq 3" or "prove for every n \geq 5." The base case is not n = 1. You must plug in the actual starting value. A claim about "n \geq 5" that you verify at n = 1 is no verification at all — the statement might be false for n = 1, 2, 3, 4 (which is why the problem specifies n \geq 5), and verifying a false case does not help your proof. Worse, if the statement happens to be true at n = 1 by coincidence, you might think you have done the base case when you have actually done nothing useful.
Always plug in the smallest n that appears in the claim. Not the smallest n you are comfortable with.
The misconception this habit counters
Many students believe that induction is a purely symbolic technique — that once you have the formula, you set up the inductive hypothesis, do some algebra, and the proof writes itself. The arithmetic check sits outside the symbolic machine and feels like it "doesn't count."
It does count. The symbolic machinery of induction requires a correct base case. The base case is, by definition, a numerical fact — not a symbolic one. You cannot establish "the statement holds at n = 1" by waving hands at an expression; you must compute both sides and compare. And if you are going to compute, you might as well compute for n = 2, 3 as well, because the extra seconds are trivial and the confidence you gain is large.
A second related habit: when stuck in the step, try a specific k
The companion to "check the base case numerically" is "when the inductive step gets tangled, plug in a specific value of k and see what the computation looks like."
Suppose you are trying to prove P(k) \Rightarrow P(k+1) and the algebra is turning into a mess. Stop. Set k = 3 and redo the algebra with concrete numbers. You will see the pattern of what is happening, notice the step that should have cancelled, and realise where your general argument went wrong. Then you can go back to symbols and write it cleanly.
Symbolic fluency is not a substitute for numerical intuition. It is built on top of numerical intuition. Start with the numbers; derive the symbols from what the numbers reveal.
The routine, in five lines
- Read the claim.
- Compute both sides at n = 1, 2, 3 (or at the smallest value of n in the claim and the next two).
- If any check fails, stop and investigate.
- If all checks pass, proceed with the inductive step.
- If the step gets stuck, plug in k = 3 and compute.
That is it. Five steps. Three minutes of work at the start of a problem. The payoff is that you never waste an exam on a false claim, and you always have a concrete anchor when the symbols get hairy.
Related: Mathematical Induction · Skipping the Base Case Because It's 'Obvious' — Then Discovering the Statement Was Actually False · Do I Ever Need More Than One Base Case? · Missing Base Case — The Domino Chain That Never Starts · Base Case Index Slider — Watch the Proved Set Shrink From n ≥ 1 to n ≥ 5