QuanticaAlgebra IOpen in the courseAn expression is only useful if every reader gets the same value from it. But some expressions can be read in more than one reasonable way, and different readings give different numbers. Before doing any algebra, we have to settle how expressions are read.
Problem
Two students evaluate \(100-10\times 9\). The first works strictly left to right, subtracting before multiplying. The second does the multiplication first, then subtracts. Neither makes a single arithmetic slip, yet they land on different numbers. How far apart are their two answers?
Show a hint
- Work each reading all the way to the end before comparing.
- Left to right means \((100-10)\times 9\). Multiplying first means \(100-(10\times 9)\).
Show the full solution
Reading left to right, \(100-10=90\), then \(90\times 9=810\). Multiplying first, \(10\times 9=90\), then \(100-90=10\). Two reasonable readings give \(810\) and \(10\), and the gap between them is \(810-10=\boxed{800}\).
Neither student made an arithmetic mistake, so checking the arithmetic cannot settle this. We need a rule that says which operation comes first, and mathematicians agreed on one long ago. From here on we also write multiplication as \(3\cdot 9\) or with parentheses, because algebra uses the letter x constantly and \(\times\) is too easy to confuse with it.
Problem
All four levels of the order show up in this one expression. Evaluate $$5+(18-6)\div 2^2$$ doing the steps in the order the rule lists them.
Show a hint
- Resolve \(18-6\) and \(2^2\) before any dividing happens.
Show the full solution
Parentheses first, \(18-6=12\). Exponent next, \(2^2=4\). Then the division, \(12\div 4=3\), and finally the addition, \(5+3=\boxed{8}\).
Problem
Another disagreement. In \(48\div 6\cdot 2\), one student divides first and gets \(8\cdot 2=16\). The other multiplies \(6\cdot 2\) first and gets \(48\div 12=4\). Only one of them is right. Which value is correct?
Show a hint
- Division and multiplication have equal rank, so rank cannot break the tie.
- When rank ties, reading order decides.
Show the full solution
\(48\div 6=8\), then \(8\cdot 2=\boxed{16}\). The first student is right, because division and multiplication have equal rank and a tie is broken left to right, not in favor of multiplication.
Problem
One multiplication, one subtraction, one addition. Evaluate $$19-8+2\cdot 3$$ keeping careful track of which of the three moves first, which second, and which last.
Show a hint
- Only the multiplication outranks the rest, so it goes first.
- Then subtraction and addition have equal rank. Read left to right.
Show the full solution
The multiplication goes first, \(2\cdot 3=6\), leaving \(19-8+6\). Subtraction and addition have equal rank, so work left to right, \(19-8=11\), then \(11+6=\boxed{17}\). Doing the \(8+6\) first would give \(19-14=5\), which answers a different expression.
Problem
In $$200-3\cdot 4^2$$ the \(4\) sits between a multiplication and an exponent. Decide which one applies to it first, then evaluate.
Show a hint
- The exponent applies only to the \(4\), not to \(3\cdot 4\).
Show the full solution
\(4^2=16\), then \(3\cdot 16=48\), and \(200-48=\boxed{152}\). The exponent applies only to the \(4\). Squaring \(3\cdot 4\) instead would give \(200-144=56\).
Problem
Evaluate \((-7)^2+\left(-7^2\right)\). The two pieces look identical, so compute each one separately before adding.
Show a hint
- The parentheses in \((-7)^2\) put the minus sign inside the base.
- In \(-7^2\) the exponent applies only to the \(7\). Square first, then negate.
Show the full solution
The first piece squares the negative itself, \((-7)^2=(-7)(-7)=49\). The second squares first and negates after, \(-7^2=-(7^2)=-49\). The sum is \(49+(-49)=\boxed{0}\).
Problem
Not one parenthesis appears in $$\dfrac{22+2\cdot 14}{5^2-15}$$ yet two invisible pairs of them are at work. Evaluate the expression.
Show a hint
- The bar wraps the whole top and the whole bottom. Finish each completely before dividing.
- In the denominator, the exponent goes before the subtraction.
Show the full solution
On top, \(2\cdot 14=28\) and \(22+28=50\). On the bottom, \(5^2=25\) and \(25-15=10\). Divide last, \(50\div 10=\boxed{5}\). The bar groups the whole top and the whole bottom, so each side has to finish before any dividing happens.
Problem
Both \(625\) and \(49\) are perfect squares, which makes a certain shortcut very tempting. Evaluate $$\sqrt{625-49}$$ and resist it.
Show a hint
- The radical bar groups everything under it. Subtract before any root is taken.
Show the full solution
The radical bar groups everything under it, so \(625-49=576\) comes first, and \(\sqrt{576}=\boxed{24}\). Taking each root separately gives \(25-7=18\), which answers a different expression.
When groups nest, work from the innermost group outward, passing each answer to the next layer. Square brackets are just parentheses that are easier to spot. They mean the same thing and follow the same rules, they simply help the eye keep the layers apart.
Problem
Two layers of grouping and one exponent. Evaluate $$3\left[\,47-(11-5)^2\,\right]$$ working from the innermost group outward.
Show a hint
- Do the parentheses inside before the brackets outside.
- Inside the brackets, the exponent still goes before the subtraction.
Show the full solution
Innermost first, \(11-5=6\). The exponent comes next, \(6^2=36\), then inside the brackets \(47-36=11\), and finally \(3\cdot 11=\boxed{33}\).
Problem
One expression carries every trap in the lesson, a tight exponent, a radical bar, and a fraction bar. Evaluate $$\dfrac{-3^4+117}{\sqrt{45-36}}$$
Show a hint
- The exponent applies only to the \(3\), so \(-3^4\) means \(-(3^4)\).
- Both bars group. Finish the whole top and the whole bottom before dividing.
Show the full solution
On top, \(-3^4=-(3^4)=-81\), so the numerator is \(-81+117=36\). On the bottom, the radical bar groups \(45-36=9\), and \(\sqrt{9}=3\). The fraction bar divides last, \(36\div 3=\boxed{12}\).
The whole rule is short. Groups first, worked inside out, then exponents, then multiplication and division left to right, then addition and subtraction left to right. It only tells you how to read what is written. It does not tell you which changes to an expression are legal, which swaps and regroupings leave the value alone. That is the next lesson.
Practice these ideas
Practice
Start with a straight computation. Evaluate \(46-5\cdot3^2\), doing the operations in the order the rule assigns them.
Show the solution
The power goes first, \(3^2=9\), then the multiplication, \(5\cdot9=45\), and last the subtraction, \(46-45=\boxed{1}\).
Practice
In \(60\div6\cdot5\), a student multiplies \(6\cdot5\) first because PEMDAS lists multiplication before division. Evaluate the expression correctly.
Show the solution
\(60\div6=10\), then \(10\cdot5=\boxed{50}\). Division and multiplication have equal rank, so the tie breaks left to right. Multiplying first would give \(60\div30=2\), and PEMDAS listing M before D does not change that.
Practice
Evaluate \(52-37+8\). Subtraction and addition tie the same way division and multiplication do, and the same rule breaks the tie.
Show the solution
\(52-37=15\), then \(15+8=\boxed{23}\). Subtraction and addition have equal rank, so read left to right. Doing \(37+8\) first would give \(52-45=7\).
Practice
Evaluate \(-2^6\). Before you compute anything, decide exactly what the exponent applies to and what the minus sign is doing.
Show the solution
The exponent applies only to the \(2\), so \(-2^6\) means \(-(2^6)=\boxed{-64}\). Only \((-2)^6\), with written parentheses, puts the whole \(-2\) under the exponent, and that one comes out at positive \(64\).
Practice
Evaluate \(34-(-3)^3\). This time the parentheses are written in, so the exponent applies to the whole \(-3\).
Show the solution
\((-3)^3=-27\), then \(34-(-27)=34+27=\boxed{61}\). The written parentheses put the whole \(-3\) under the exponent, and an odd power keeps the negative sign.
Practice
Evaluate \(\dfrac{75-3}{2\cdot4}\). Written carelessly on a single line this expression would mean something different, so keep the bar's grouping in mind.
Show the solution
Finish the top and the bottom before dividing. The top is \(75-3=72\) and the bottom is \(2\cdot4=8\), so the value is \(72\div8=\boxed{9}\).
Practice
Evaluate \(\sqrt{144+25}\). Both numbers under the bar happen to be perfect squares, which makes the wrong shortcut look extra inviting.
Show the solution
The bar covers the whole sum, so add first, \(144+25=169\), and then \(\sqrt{169}=\boxed{13}\). Taking each root separately gives \(12+5=17\), which answers a different expression.
Practice
Evaluate \(84\div[2\cdot(9-2)]\). Two layers of grouping, so work from the inside out.
Show the solution
Inside out, \(9-2=7\), then \(2\cdot7=14\), and finally \(84\div14=\boxed{6}\). Ignoring the brackets and reading straight through would give \(84\div2\cdot7=294\).
Practice
Evaluate \(9+54\div3^2\cdot2\). One short line that uses every level below grouping, including a left-to-right tie.
Show the solution
The power goes first, \(3^2=9\). Division and multiplication tie, so work left to right, \(54\div9=6\), then \(6\cdot2=12\), and adding the \(9\) in front gives \(\boxed{21}\).
Practice
Parentheses outrank everything, so one well-placed pair can transform a value. Insert a single pair of parentheses into \(2\cdot6+3^2\) to make the result as large as possible, then enter that largest value.
Show the solution
Wrapping everything before the exponent gives \((2\cdot6+3)^2=15^2=\boxed{225}\). Every other placement is smaller, since \(2\cdot(6+3)^2=162\) and \(2\cdot(6+3^2)=30\).