Proof & binomial 1 · The language of proof

Module 6 of 46Lesson 1 of 7 · Proof & binomial 1 · The language of proof

Simple proof & the binomial theorem · Essential understanding 1

The language of proof

IB AA SL 1.6with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: A proof is a chain of correct steps that shows a statement is true for every case, not just the ones you tried. In the exam you will mostly show that two expressions are equal, by starting on the left-hand side (LHS) and changing it, line by line, until it becomes the right-hand side (RHS).

By the end of this lesson you can:

explain why testing a few cases is not a proof
use the symbols $\equiv$ , $\Rightarrow$ and $\therefore$ correctly
prove a numerical statement (with surds or fractions) by exact working
prove an identity by working from LHS to RHS
answer “show that … hence …” questions

🧑🏾 Imagine this…

Thato says: “Every time I add two odd numbers I get an even number: 3 + 5 = 8, 7 + 9 = 16, 11 + 1 = 12.” Is that a proof?
No. Thato has tried three cases. There are infinitely many odd numbers! A proof must work for all of them at once, and the way to do that is to use letters. (You will do this one properly in Lesson 2.)

Equation

true only for certain values, e.g.

x + 3 = 7

is true only when

x = 4

Identity (≡)

true for all values of the letters, e.g.

(x + 1)^2 \equiv x^2 + 2x + 1

. We write ≡ (“is identically equal to”), although = is also accepted in exams.

Proof

a logical argument that starts from known facts and ends at the statement you want, with every step justified.

Counterexample

one single case that makes a statement false. One counterexample is enough to disprove a statement.

LHS, RHS

the left-hand side and right-hand side of an equation or identity.

★ The language of proof

Show that / prove that: the answer is given, so you must show every step. Never skip the key line.
Hence: you must use the result you have just shown. Using a calculator to get the answer another way earns no marks.
LHS → RHS layout: write “LHS” once, then $= \ldots = \ldots$ down the page, ending on exactly the RHS. Do not touch the RHS while you work.
Symbols: $\Rightarrow$ means “so” or “implies”, and $\therefore$ means “therefore”. $\blacksquare$ or “QED” marks the end.
In mark schemes the final line carries AG (answer given): you only get it if the working before it is complete and correct.

⚠️ A pattern is not a proof

Checking

n = 1, 2, 3, 4, 5

never proves a statement. The expression

n^2 + n + 41

gives primes (41, 43, 47, 53, 61, 71, …) for every

n

from 0 to 39, yet it fails at

n = 40

(the value is

1681 = 41^2

). To prove, use algebra. To disprove, one counterexample is enough.

⚠️ Do not work on both sides at once

Writing

\text{LHS} = \text{RHS}

at the top and then changing both sides assumes the thing you are trying to prove. Start with one side only, and finish on the other.

💡 Which side first?

Start with the more complicated side (more brackets, more terms). It is easier to simplify than to build up.

🧪 Try it yourselfProof or not a proof?

Tap the best answer. These use the ideas from this lesson.

Worked examples

Example 1.1Testing is not proving

Jabu claims: “

n^2 + n + 41

is a prime number for every positive integer

n

.” Show that Jabu is wrong.

🧠 Think it through

To disprove a statement you need only one counterexample.
The 41s are suspicious: when $n = 41$ , every term has a factor of 41.
Substitute $n = 41$ and factorise.

✍️ By hand

n = 41:\; 41^2 + 41 + 41 = 41(41 + 1 + 1)

= 41 \times 43 = 1763

1763 has factors 41 and 43, so it is not prime.

🔢 On your TI-Nspire CX II

Check on the calculator (this checks the arithmetic, it is not the proof).

✅ Answer When n = 41 the value is 41 × 43 = 1763, which is not prime. One counterexample is enough, so the claim is false.

The values for n = 1 to 39 all happen to be prime. That is why a long list of successes still proves nothing.

Example 1.2A numerical proof with surds

Show that

\dfrac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}

, without using a calculator.

🧠 Think it through

Start on the LHS and rationalise the denominator.
Multiply top and bottom by the conjugate $2 - \sqrt{3}$ .
The bottom becomes a difference of two squares: $(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$ .

✍️ By hand

\text{LHS} = \dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}}

(M1)

= \dfrac{2 - \sqrt{3}}{4 - 3}

(A1)

= 2 - \sqrt{3} = \text{RHS}

(AG)

✅ Answer LHS

= \dfrac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} = \dfrac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}

, as required.

Example 1.3Proving an identity, LHS to RHS

Show that

(2n + 1)^2 \equiv 4n(n + 1) + 1

🧠 Think it through

The LHS has a bracket squared, so expand it.
$(2n+1)^2 = (2n+1)(2n+1)$ . Multiply carefully: four terms.
Then look for a common factor of $4n$ in the first two terms.

✍️ By hand

\text{LHS} = (2n + 1)(2n + 1)

= 4n^2 + 2n + 2n + 1

(M1)

= 4n^2 + 4n + 1

(A1)

= 4n(n + 1) + 1 = \text{RHS}

(AG)

✅ Answer LHS

= 4n^2 + 4n + 1 = 4n(n+1) + 1

, which is the RHS.

This identity is useful: it says the square of any odd number is “4 × something + 1”. You will use it in Lesson 2.

Example 1.4Show that … hence …

(a) Show that

(n + 1)^2 - (n - 1)^2 \equiv 4n

.
(b) Hence find the value of

1001^2 - 999^2

, without a calculator.

🧠 Think it through

(a) Expand both brackets. The $n^2$ terms should cancel. Watch the minus sign in front of the second bracket!
(b) “Hence” means: spot $n + 1 = 1001$ and $n - 1 = 999$ . What is $n$ ?

✍️ By hand
(a)    (M1)
   (A1)
   (AG)
(b)    (and  ✓)
   (A1)

🔢 On your TI-Nspire CX II

Optional check after finishing by hand.

✅ Answer (a) shown. (b) 4000.

Using the identity with n = 1000 is “hence”. Typing 1001² − 999² into a calculator is not.

Example 1.5A proof that makes a sum collapse

(a) Show that

\dfrac{1}{n} - \dfrac{1}{n + 1} \equiv \dfrac{1}{n(n + 1)}

.
(b) Hence find the sum of the five fractions

\dfrac{1}{1 \times 2},\; \dfrac{1}{2 \times 3},\; \dfrac{1}{3 \times 4},\; \dfrac{1}{4 \times 5},\; \dfrac{1}{5 \times 6}

, as a single fraction.

🧠 Think it through

(a) Put the LHS over a common denominator $n(n+1)$ .
(b) Use (a) on every term: each fraction $\dfrac{1}{k(k+1)}$ becomes $\dfrac{1}{k} - \dfrac{1}{k+1}$ . Write them all out and see what cancels.

✍️ By hand
(a)    (M1)
   (AG)
(b) Sum 
   (M1)
Everything cancels except the first and the last:    (A1)

🔢 On your TI-Nspire CX II

Check by adding the five fractions directly.

✅ Answer (a) shown. (b)

\dfrac{5}{6}

Spotting that a proof from part (a) makes part (b) easy is the whole point of “hence”.

Challenge questions

⭐ Challenge 1.6Show thatPaper 1 style

Show that

(x + 3)^2 + (x - 3)^2 \equiv 2(x^2 + 9)

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Expand each bracket separately, then add.
Cross terms $+6x$ and $-6x$ cancel.

✍️ By hand

\text{LHS} = (x^2 + 6x + 9) + (x^2 - 6x + 9)

= 2x^2 + 18 = 2(x^2 + 9) = \text{RHS}

✅ Answer LHS

= 2x^2 + 18 = 2(x^2+9)

, as required.

⭐ Challenge 1.7A trick for squaresPaper 2 style

(a) Show that

(a + b)^2 - (a - b)^2 \equiv 4ab

.
(b) Hence, by choosing suitable values of

a

and

b

, find

53^2 - 47^2

.
(c) Use your GDC to check your answer to (b). Then write down a different pair of numbers

(p, q)

such that

p^2 - q^2 = 4 \times 50 \times 3

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

(a) Expand both brackets; the $a^2$ and $b^2$ terms cancel.
(b) You need $a + b = 53$ and $a - b = 47$ . Add and halve to get $a$ , subtract and halve to get $b$ .
(c) Type 53² − 47² into the calculator. For the last bit think about what (a) says for any a and b.

✍️ By hand

(a)

\text{LHS} = a^2 + 2ab + b^2 - a^2 + 2ab - b^2 = 4ab

(AG)

(b)

a + b = 53,\; a - b = 47 \Rightarrow a = 50,\; b = 3

53^2 - 47^2 = 4 \times 50 \times 3 = 600

🔢 On your TI-Nspire CX II

Check (b).

✅ Answer (a) shown. (b) 600. (c) 600 ✓. Any p = a + b, q = a − b with ab = 150 works, e.g. a = 15, b = 10 gives (p, q) = (25, 5): 625 − 25 = 600.

Next: Lesson 2 · Algebraic proof with integers

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL Proof & binomial 1 · The language of proof

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

The language of proof

Worked examples

Challenge questions