Order of operations — BODMAS - PEMDAS with nested brackets: Worked examples

First: what the four tiers actually are

Before any example, let us pin down the vocabulary so nothing on this page is used before it is defined.

So whenever a "Why this step?" below says "Orders beat addition" it means Tier 2 (exponents/roots) resolves before Tier 4 (add/subtract) — that is all "Orders" ever means on this page.

The scenario matrix

Every expression in this topic falls into one of these cells. If you can handle all of them, you have handled the whole topic.

Cell	The scenario it tests	Where it bites you
A	Two same-tier operations ( $+$ and $-$ , or $\times$ and $\div$ )	People wrongly do the "later letter" first
B	Mixed tiers, no brackets	Which tier resolves first?
C	Nested brackets, incl. curly ${}$ and a root	Which bracket is "innermost"?
D	A fraction bar as hidden brackets	Flat rewriting changes the meaning
E	Unary minus vs an exponent ( $- 3^{2}$ vs $(- 3)^{2}$ )	The minus sign's "reach"
F	Zero inputs — $\times 0$ , divide-by-zero, and powers $0^{n}$ , $n^{0}$	Degenerate / undefined cases
G	A real-world word problem	Turning English into a correct expression
H	An exam twist — implied multiplication like $48 \div 2 (1 + 2)$	Same-tier ambiguity, resolved by convention
I	Stacked exponents like $2^{3^{2}}$ (which power first?)	Exponent towers group top-down

The prerequisites we lean on: Arithmetic operations — addition, subtraction, multiplication, division, Exponents and powers — laws of indices, Fractions — numerator and denominator as grouping, Negative numbers and the unary minus sign, and Algebraic expressions — evaluating and substitution.

A picture of "tiers" first

Before the examples, hold this picture in your head. Think of the four tiers as four shelves, top shelf resolved first. Within one shelf you sweep left to right like reading a sentence.

Figure — Order of operations — BODMAS - PEMDAS with nested brackets

Figure 1 — four labelled shelves (Brackets, Orders, ×÷, +−) stacked top-to-bottom; a yellow left-to-right sweep arrow sits inside the same-tier shelves, and white downward arrows show you drop to the next shelf only when the one above is empty.

The worked examples

Cell A — same-tier, left to right

Cell B — mixed tiers, no brackets

Cell C — nested brackets, curly braces and a root

Worked example Example 3 (Cell C) —

$2 \times {10 - [3 + (4 - 1)^{2}]} + 1$ Forecast: three layers of grouping and a root sign. Where do you start — the outer curly ${}$ , the square $[]$ , the round $()$ , or the ?

Remember from the definition: the nesting from inside out is round $\to$ square $\to$ curly, and the bar over a root is itself a bracket. So we peel from the deepest point outward.

Innermost round bracket first: $(4 - 1) = 3$ . Why this step? $()$ sits deepest inside everything; nothing outside can use it until it is a single number.
The exponent under the root: $3^{2} = 9$ . Why this step? Everything under the bar is grouped, and inside that group the exponent (Tier 2) resolves before the root can act.
The root: $9 = 3$ . Why this step? A root is a Tier-2 (Orders) operation; once its grouped contents are one number, take the root.
Square bracket next: $3 + 3 = 6$ . Why this step? With the root done, the square bracket $[3 + 3]$ is a plain addition — finish it before the curly wrapper can use it.
Curly bracket: $10 - 6 = 4$ . Why this step? The outermost group ${10 - 6}$ is now a single subtraction; do it before leaving Tier 1.
$2 \times 4 = 8$ . Why this step? $\times$ (Tier 3) beats $+$ (Tier 4).
$8 + 1 = 9$ .

Answer: $9$ .

Verify: rebuild bottom-up: $4 - 1 = 3 \to 3^{2} = 9 \to 9 = 3 \to 3 + 3 = 6 \to 10 - 6 = 4 \to 2 \times 4 = 8 \to 8 + 1 = 9$ . And $9 = 3$ because $3^{2} = 9$ , so the root truly undoes the square. Each layer used only finished numbers. ✓

Cell D — the fraction bar is a hidden bracket

Worked example Example 4 (Cell D) —

$\frac{12 - 2 \times 3}{1 + 2 ^{2}}$ Forecast: the bar looks like it splits things, but it also groups. Guess the value.

A fraction bar acts like invisible brackets around the whole top and the whole bottom (see Fractions — numerator and denominator as grouping): $\frac{12 - 2 \times 3}{1 + 2 ^{2}} = \frac{( 12 - 2 \times 3 )}{( 1 + 2 ^{2} )}$

Top: $2 \times 3 = 6$ , then $12 - 6 = 6$ . Why this step? Inside the top "bracket", $\times$ (Tier 3) beats $-$ (Tier 4).
Bottom: $2^{2} = 4$ , then $1 + 4 = 5$ . Why this step? Inside the bottom "bracket", Orders (Tier 2) beat $+$ (Tier 4).
Now $\frac{6}{5} = 6 \div 5 = 1.2$ . Why this step? Only when both groups are single numbers do we finally divide.

Answer: $\frac{6}{5} = 1.2$ .

Verify (why the bar matters): if you wrongly flattened it to $12 - 2 \times 3 \div 1 + 2^{2} = 12 - 6 + 4 = 10$ — a completely different number. The grouping is doing real work. ✓

Cell E — unary minus vs exponent

Worked example Example 5 (Cell E) — compare

$- 4^{2}$ and $(- 4)^{2}$ Forecast: are these the same? Most people say both are $16$ .

Look at Figure 2 as you read. It shows the same digits read two ways. To make the difference readable without relying on colour, each side is also labelled in words and the grouping is drawn with a labelled underbrace, and the two results carry a $-$ or $+$ sign in plain text.

Figure 2 — same digits, two readings. LEFT panel labelled "power grabs only the 4": in $- 4^{2}$ an underbrace marks $4^{2}$ as the grouped part, the minus is applied last, result $= - 16$ (also written in words "negative sixteen"). RIGHT panel labelled "bracket pulls the sign in": in $(- 4)^{2}$ an underbrace marks $- 4$ as the base, result $= + 16$ ("positive sixteen"). Colour is decorative only; every cue is also stated in text.

Case $- 4^{2}$ :

The exponent binds tighter than the minus, so it means $- (4^{2})$ . Why this step? The unary minus (see Negative numbers and the unary minus sign) is a low-priority "flip sign at the end" operation; the power (Tier 2) grabs only the $4$ next to it — exactly the underbrace over $4^{2}$ in Figure 2.
$4^{2} = 16$ , then apply the minus: $- 16$ .

Case $(- 4)^{2}$ :

The bracket forces the minus into the base first: the base is $- 4$ . Why this step? Brackets (Tier 1) are the override — they make the whole $- 4$ the thing being squared, as the underbrace over $- 4$ on the right of Figure 2 shows.
$(- 4)^{2} = (- 4) \times (- 4) = 16$ .

Answers: $- 4^{2} = - 16$ , but $(- 4)^{2} = 16$ .

Verify: a negative times a negative is positive, so $(- 4) (- 4) = + 16$ . And $- 16$ is just $16$ with the sign flipped once — exactly what "square, then negate" does. Different order, different sign. ✓

Cell F — zero and degenerate inputs

Worked example Example 6 (Cell F) — four zero cases in one sweep

Forecast: does $5 + 0 \times 7$ collapse to $0$ ? Is $0^{3}$ the same as $3^{0}$ ? What happens with a zero on the bottom of a fraction?

Part 1 — $\times 0$ : $5 + 0 \times 7$

$0 \times 7 = 0$ . Why this step? $\times$ (Tier 3) beats $+$ (Tier 4); do the multiply first even though it looks harmless.
$5 + 0 = 5$ . Why this step? Adding $0$ changes nothing.

Answer (part 1): $5$ , not $0$ . The zero only kills the term it multiplies, because BODMAS keeps the $+ 5$ separate.

Part 2 — power $0^{3}$ (zero as the base):

$0^{3} = 0 \times 0 \times 0 = 0$ . Why this step? An exponent (Tier 2) is repeated multiplication; multiplying zeros gives zero. Any positive power of $0$ is $0$ .

Answer (part 2): $0$ .

Part 3 — power $3^{0}$ (zero as the exponent):

$3^{0} = 1$ . Why this step? By the laws of indices (see Exponents and powers — laws of indices), $\frac{3 ^{n}}{3 ^{n}} = 3^{n - n} = 3^{0}$ . This argument only works because the base $3$ is non-zero, so $3^{n} \neq = 0$ and the fraction $\frac{3 ^{n}}{3 ^{n}}$ is a legal "something over itself $= 1$ ". So any non-zero number to the power $0$ is $1$ . (The single oddball $0^{0}$ , where the base is zero, is left undefined / context-dependent — do not rely on it.)

Answer (part 3): $1$ .

Part 4 — divide-by-zero: $\frac{6}{3 - 3}$

Bottom (grouped by the bar): $3 - 3 = 0$ . Why this step? Denominator finishes before the division.
$6 \div 0$ is undefined — no number times $0$ gives $6$ . Why this step? Division asks "what times the bottom gives the top?"; nothing times $0$ gives $6$ , so there is no answer.

Answer (part 4): undefined. The rules got us to the divide-by-zero honestly — recognising it is part of covering the case.

Verify: $0 \times 7 = 0$ and $5 + 0 = 5$ ✓; $0^{3} = 0$ ✓; $3^{0} = 1$ ✓. For part 4, if $6/0 = k$ existed then $k \times 0 = 6$ , but $k \times 0 = 0 \neq = 6$ for every $k$ — contradiction, so it is undefined. ✓

Cell G — a real-world word problem

Worked example Example 7 (Cell G) — the shopping cart

Statement: You buy $3$ notebooks at $\$ 4 $e a c han d$ 2 $p e n s a t$ $5 $e a c h, t h e n u se a$ $6$ coupon. What is the total cost?

Forecast: is it $(3 \times 4 + 2 \times 5) - 6$ , or should you add first? Guess the expression and the number.

Translate the English into one expression: $3 \times 4 + 2 \times 5 - 6$ Why this step? "3 notebooks at $4" is a multiplication $3 \times 4$ ; "2 pens at $5" is $2 \times 5$ ; the coupon subtracts $6. No brackets are needed because BODMAS already does each multiplication (Tier 3) before combining (Tier 4).
Multiplicative shelf first: $3 \times 4 = 12$ and $2 \times 5 = 10$ . Why this step? Each product is a completed sub-total (dollars for notebooks, dollars for pens) before we combine.
Additive shelf, left to right: $12 + 10 = 22$ , then $22 - 6 = 16$ .

Answer: $\$ 16$.

Verify (units): notebooks $\$ 12 $+ p e n s$ $10 $=$ $22 $b e f or e d i sco u n t; min u s t h e$ $6 $co u p o n g i v es$ $16$. Every quantity is dollars and the arithmetic matches the story. ✓

Cell H — the exam twist (implied multiplication)

Worked example Example 8 (Cell H) —

$48 \div 2 (1 + 2)$ Forecast: the famous internet fight. Is it $8$ ? Is it $72$ ? Guess.

Bracket first: $(1 + 2) = 3$ . Now it reads $48 \div 2 \times 3$ (the $2 (\dots)$ means $2 \times (\dots)$ ). Why this step? Brackets (Tier 1) outrank everything; and "a number written against a bracket" is just multiplication — the same tier (Tier 3) as $\div$ , not higher.
Same-tier sweep, left to right: $48 \div 2 = 24$ . Why this step? $\div$ and $\times$ share Tier 3; the leftmost job ( $48 \div 2$ ) goes first.
$24 \times 3 = 72$ .

Answer: $72$ by standard BODMAS convention.

Verify: left-to-right on the shared Tier 3 gives $48 \div 2 = 24$ , then $24 \times 3 = 72$ . The alternate "glued" reading gives $48 \div (2 \times 3) = 8$ . ✓

Cell I — stacked exponents (which power first?)

Worked example Example 9 (Cell I) —

$2^{3^{2}}$ versus $(2^{3})^{2}$ Forecast: an exponent on top of an exponent. Do you square the $3$ first, or the $2^{3}$ first? Guess both values.

Case $2^{3^{2}}$ (a plain exponent tower):

Evaluate the top exponent first: $3^{2} = 9$ . Why this step? Exponent towers group right-to-left (top-down) by convention: $2^{3^{2}}$ means $2^{(3^{2})}$ , because the higher exponent is "attached to" the one just below it. This is the one place inside Tier 2 where we do not go left to right.
Now $2^{9} = 512$ . Why this step? With the exponent reduced to the single number $9$ , raise the base $2$ to it.

Answer (tower): $2^{3^{2}} = 2^{9} = 512$ .

Case $(2^{3})^{2}$ (brackets change the grouping):

Bracket first: $2^{3} = 8$ . Why this step? The bracket (Tier 1) forces the lower power to finish first, overriding the top-down convention.
$8^{2} = 64$ . Why this step? Square the single number $8$ .

Answer (bracketed): $(2^{3})^{2} = 8^{2} = 64$ .

Verify: the two readings genuinely differ — $512 \neq = 64$ — so the grouping is not optional. By the laws of indices $(2^{3})^{2} = 2^{3 \times 2} = 2^{6} = 64$ , matching the second case, while the tower keeps the exponents stacked as $2^{9} = 512$ . This is why $2^{3^{2}}$ must mean the top first: if it meant "bottom first" it would just be $(2^{3})^{2}$ and the tower notation would be pointless. ✓

Active recall

Recall Cover the answers first

What does "Orders" mean in BODMAS? ::: Powers and roots (exponents like $2^{3}$ and radicals like $9$ ) — Tier 2.
Which grouping symbol is usually outermost when you nest three? ::: Curly ${}$ , with square $[]$ inside it and round $()$ deepest.
Cell A: $18 - 7 + 4 = ?$ ::: $15$ (left to right, subtract then add).
Cell B: $5 + 6 \times 2 - 4 \div 2 = ?$ ::: $15$ .
Cell C: $2 \times {10 - [3 + (4 - 1)^{2}]} + 1 = ?$ ::: $9$ .
Cell D: $\frac{12 - 2 \times 3}{1 + 2 ^{2}} = ?$ ::: $\frac{6}{5} = 1.2$ .
Cell E: $- 4^{2}$ vs $(- 4)^{2}$ ? ::: $- 16$ vs $16$ .
Cell F: $0^{3} = ?$ , $3^{0} = ?$ , $6 \div (3 - 3) = ?$ ::: $0$ , $1$ , undefined.
Cell G: $3 \times 4 + 2 \times 5 - 6 = ?$ ::: $\$ 16$.
Cell H: $48 \div 2 (1 + 2) = ?$ ::: $72$ by convention.
Cell I: $2^{3^{2}} = ?$ vs $(2^{3})^{2} = ?$ ::: $512$ vs $64$ .

Connections

Parent: BODMAS / PEMDAS
Arithmetic operations — addition, subtraction, multiplication, division
Exponents and powers — laws of indices
Fractions — numerator and denominator as grouping
Negative numbers and the unary minus sign
Algebraic expressions — evaluating and substitution
Calculator vs mental arithmetic — parsing expressions