NDA Maths · Binomial Theorem

Integer & Fractional Parts of Binomial Expressions

When a surd like (a+√b)ⁿ is expanded, pairing it with its conjugate (a−√b)ⁿ makes the irrational parts cancel — turning a messy surd power into a clean integer plus a small fractional remainder.

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Why this matters

8 PYQs, and they look intimidating until you know the one trick: add the conjugate. Then 'find the fractional part', 'is it an integer', 'find the inverse' all fall out of two facts — the conjugate product is small, and the conjugate sum is an integer.

Test yourself — a quick 18-question Binomial Theorem recall quiz, instant score.

Concept 1 of 2: The Conjugate Trick — Irrational Parts Cancel

Expanding (a+√b)ⁿ gives some rational terms and some terms carrying √b. Expanding the conjugate (a−√b)ⁿ gives the SAME terms but with the √b ones flipped in sign. Add them and every √b term cancels — so the sum is a pure integer. Their product is even simpler.

Definition

For integers $a, b$ with $b$ not a perfect square:

Sum is an integer: $(a+\sqrt{b})^n + (a-\sqrt{b})^n = 2\!\!\sum_{r\text{ even}}\!\!\binom{n}{r} a^{\,n-r} b^{\,r/2}$ — all the odd (irrational) terms cancel, leaving an even integer.
Product is small: $(a+\sqrt{b})(a-\sqrt{b}) = a^2 - b$ , so $(a+\sqrt{b})^n (a-\sqrt{b})^n = (a^2-b)^n$ . When $a^2 - b = 1$ the product is exactly $1$ .
Since $0 < a-\sqrt{b} < 1$ (when $a^2-b=1$ and $a>1$ ), the conjugate power $(a-\sqrt{b})^n$ is a small positive number less than 1.

Conjugate sum & product

(a+\sqrt{b})^n + (a-\sqrt{b})^n \in \mathbb{Z}, \qquad (a+\sqrt{b})^n(a-\sqrt{b})^n = (a^2-b)^n

Worked example

k < (\sqrt{2}+1)^3 < k+2

for a natural number

k

, find

k

The same idea in a real exam question:

NDA · Apr 2025 · Q22Moderate

Example 1 · Binomial Theorem · Integer and Fractional Parts of Binomial Expressions

k < (\sqrt{2}+1)^3 < k+2

, where

k

is a natural number, then what is the value of

k

Add the conjugate — don't expand the whole thing

Trying to expand

(a+\sqrt b)^{20}

term by term is hopeless. The intended move is always to bring in

(a-\sqrt b)^n

: its sum is an integer and its product is

(a^2-b)^n

Drill 2 more on the conjugate trick — irrational parts cancel

Concept 2 of 2: Integer Part + Fractional Part

Write the big surd power as its integer part I plus a fractional part f (with 0 ≤ f < 1). The small conjugate power f′ = (a−√b)ⁿ is itself between 0 and 1. Because I + f + f′ is a whole integer and f + f′ is squeezed strictly between 0 and 2, f + f′ must equal exactly 1.

Definition

Let $N = (a+\sqrt{b})^n = I + f$ with integer part $I$ and $0 \le f < 1$ , and let $f' = (a-\sqrt{b})^n$ with $0 < f' < 1$ (taking $a^2-b=1,\ a>1$ ). From the conjugate trick $N + f' = I + f + f'$ is an integer, so:

$f + f' = 1$ (since it is an integer strictly between 0 and 2).
The integer part is $I = N + f' - 1$ ; the fractional part of $N$ is $f = 1 - f'$ .
Products like $N \cdot f' = (a^2-b)^n$ give relations such as $I\cdot f' = (a^2-b)^n - f f'$ , pinning quantities like $uv$ into $(0,1)$ .

Fractional parts add to 1

f + f' = 1, \qquad f' = (a-\sqrt{b})^n

Worked example

For

(\sqrt{2}+1)^{10} = u + f

(

u

integer,

0\le f<1

) and

v = (\sqrt{2}-1)^{10}

, show

f + v = 1

The same idea in a real exam question:

NDA · Apr 2026 · Q46Moderate

Example 2 · Binomial Theorem · Integer and Fractional Parts of Binomial Expressions

for the items that follow: Let u be a positive integer and f be a real number lying between 0 and 1. Further,

(\sqrt{2}+1)^{10}=u+f

and

(\sqrt{2}-1)^{10}=v

Consider the following statements: I.

(u+v+f)

is an integer. II.

(f+v)

is an integer. Which of the statements given above is/are correct?

The conjugate power IS the missing fractional part

f' = (a-\sqrt b)^n

is not just "small" — it is exactly

1 - f

. Treating

f'

as negligible or zero loses the relation

f + f' = 1

the question is built on.

Drill 4 more on integer part + fractional part

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The Conjugate Trick — Irrational Parts Cancel
Conjugate sum & product
$(a+\sqrt{b})^n + (a-\sqrt{b})^n \in \mathbb{Z}, \qquad (a+\sqrt{b})^n(a-\sqrt{b})^n = (a^2-b)^n$
Integer Part + Fractional Part
Fractional parts add to 1
$f + f' = 1, \qquad f' = (a-\sqrt{b})^n$

Watch out for (2)

Add the conjugate — don't expand the whole thing
→ The Conjugate Trick — Irrational Parts Cancel
The conjugate power IS the missing fractional part
→ Integer Part + Fractional Part

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