Why PEMDAS, and the internet's favourite argument
Tonight's idea
Write \(2 + 3 \times 4\) on a board and ask a room what it equals. Everyone who says 14 is following a convention: multiply before you add. Someone who says 20 is reading a different sentence. The convention exists so that everyone reads the same sentence.
Why this convention and not the other? Because multiplication is repeated addition. \(2 + 3 \times 4\) means "two, plus three fours". Written out, \(2 + 4 + 4 + 4\), and nobody is tempted to get 20. Exponents bind tighter still, because they are repeated multiplication: \(2 \times 3^2\) means \(2 \times (3 \times 3)\), two times nine.
So the ranking is: brackets (do this first), then exponents, then multiplication and division together, then addition and subtraction together. Within a level, go left to right: \(8 - 3 - 2 = 3\), not 7. A fraction bar and a square root sign are brackets in disguise: everything above the bar is grouped, everything under the root is grouped.
Now the famous one: \(8 \div 2(2+2)\). Strict left-to-right says \(8 \div 2 = 4\), then \(4 \times 4 = 16\). But many textbooks and most scientists treat a number written right next to a bracket, like \(2(2+2)\), as one glued-together unit: \(2(2+2) = 8\) comes first, so the answer is 1. Both readings follow a real convention. The expression is ambiguous, and the only correct response is to refuse to write it that way. Write \(\frac{8}{2}(2+2)\) or \(\frac{8}{2(2+2)}\), and the argument ends.
Watch one: brackets change the sentence
\(2 + 3 \times 4 - 1\). Multiplication first: \(2 + 12 - 1 = 13\).
\((2 + 3) \times 4 - 1\). Bracket first: \(5 \times 4 - 1 = 19\).
\(2 + 3 \times (4 - 1)\). Bracket first: \(2 + 3 \times 3 = 11\).
\((2 + 3) \times (4 - 1) = 5 \times 3 = 15\).
Same four numbers, same three symbols, four answers. The brackets are not decoration.
Your turn
- Evaluate: \(7 + 2 \times 5\); \(20 - 8 \div 2\); \(3 \times 2^3\); \(24 \div 4 \div 2\); \(2 + 3^2 \times 2\); \(\frac{6 + 9}{3} - 1\).
- Insert brackets into \(2 + 3 \times 4 - 1\) to make it equal 19, then 11, then 15. Which value needs no brackets?
- Tickets cost $8 for adults and $5 for children. A family buys 2 adult and 3 child tickets and pays with a $50 note. Write one expression for the change, with brackets. Then show what happens to the meaning if you leave the brackets out.
- Four fours: using exactly four 4s and any of \(+, -, \times, \div\) and brackets, make as many of the numbers 0 to 9 as you can. (Hint: 44 counts as using two 4s.)
- Someone writes \(10 - 2 + 3\) and gets 5. What did they do, and what rule did they break?
- Rewrite \(8 \div 2(2+2)\) twice using fraction bars, once so it clearly equals 16 and once so it clearly equals 1.
Stretch: towers
What is \(2^{3^2}\)? One reading is \((2^3)^2 = 8^2 = 64\). The other is \(2^{(3^2)} = 2^9 = 512\). Mathematicians agree on 512: exponent towers are read from the top down. Why is that the sensible choice? (Hint: \((2^3)^2\) can already be written more simply as \(2^6\), so the bottom-up reading would never need a tower at all.) Then decide: \(6 \div 2(1+2)\), 9 or 1, and write a one-sentence verdict.
Answers
- 17; 16; 24; 3; 20; 4.
- \((2+3) \times 4 - 1 = 19\); \(2 + 3 \times (4-1) = 11\); \((2+3) \times (4-1) = 15\). Thirteen needs no brackets.
- \(50 - (2 \times 8 + 3 \times 5) = 50 - 31 = 19\) dollars change. Without the brackets, \(50 - 2 \times 8 + 3 \times 5 = 50 - 16 + 15 = 49\), which adds the children's tickets to the change instead of subtracting them.
- One solution each: \(0 = 44 - 44\); \(1 = 44 \div 44\); \(2 = 4\div4 + 4\div4\); \(3 = (4+4+4)\div4\); \(4 = 4 + (4-4)\times4\); \(5 = (4\times4+4)\div4\); \(6 = (4+4)\div4 + 4\); \(7 = 44\div4 - 4\); \(8 = 4+4+4-4\); \(9 = 4+4+4\div4\). Many others work.
- They did \(2 + 3 = 5\) first and then \(10 - 5\). Addition and subtraction are the same level and go left to right: \(10 - 2 = 8\), then \(8 + 3 = 11\).
- \(\frac{8}{2}(2+2) = 16\) and \(\frac{8}{2(2+2)} = 1\).
Stretch. \(2^{3^2} = 2^9 = 512\). Reading bottom-up would make the tower redundant, since \((a^b)^c = a^{bc}\) is already a single power; top-down gives towers a job of their own. \(6 \div 2(1+2)\): left-to-right gives 9, glued-bracket gives 1. Verdict: the expression is badly written, and anyone who wants a specific answer should write a fraction.