Practice Test

8.EE.1 integer exponent properties

Question 1 of 2

2026 Practice Test C Math, question 18

Question

I. $s - t$
II. $st^{2}$
III. $s^{t}$

If s is a positive integer and t is a negative integer, which of the above expressions must be a positive number?

AII only
BI and II only
CII and III only
DI, II, and III

Show the answer

Correct answer: D

Each of the three expressions must be considered:

Expression I: $s - t$ , where s is a positive integer and t is a negative integer:

Subtracting a negative integer is the equivalent of adding a positive integer, so $s - t$ is equivalent to finding the value of $s + (-t)$ , which must result in a positive number.

Expression II: $st^{2}$ , where s is a positive integer and t is a negative integer:

Squaring a negative integer results in a positive integer, so $s(-t)^{2}$ is equivalent to finding the value of $st^{2}$ , which must result in a positive number.

Expression III: $s^{t}$ , where s is a positive integer and t is a negative integer:

A negative exponent results in the reciprocal of the same exponential expression, so $s^{t}$ is equivalent to finding the value of $\frac{1}{s^{(-t)}}$ , which must result in a positive number.

Therefore, Expressions I, II, and III must all result in a positive number.

Question analysis

Three expressions, s − t, s times t squared, and s to the power t, are listed, and students decide which must be positive when s is a positive integer and t is a negative integer. It tests sign rules for subtraction and multiplication and the meaning of a negative exponent. Each expression has to be judged for every possible pair, not just one example.

NYS standards: 8.EE.1 integer exponent properties
7.NS.1c rational number subtraction
7.NS.2a rational number multiplication
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 students learn sign rules for subtracting and multiplying signed numbers. Here they reason about letters rather than numbers, decide what must always hold, and handle a negative exponent, a grade 8 idea.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly analyze
What makes it hard: Variables instead of numbers moderate
Something you must already know moderate
Every choice must be checked minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 42 words.
Corrected from the DOE handbook: Two steps of the explanation that treated t as a positive number were corrected.
A punctuation error in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The three expressions test three different sign facts: subtracting a negative, squaring a negative, and raising to a negative power. The last is the trap, since a negative exponent seems to promise a negative result when it really gives a fraction, and that belief alone leads to a listed choice. Trying s = 2 and t = −3 gives 5, 18 and 1/8, all positive, and the rules show why no other pair can change that.

Variables instead of numbers moderate Understanding the question
No values are given for s and t, only their signs, and must means each expression has to be positive for every such pair, not just for one example. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Something you must already know moderate Working it out
Expression III needs the meaning of a negative exponent, s to the power t equals 1 over s to the power −t, a grade 8 rule the question does not state. (Anderson & Krathwohl, 2001)
Every choice must be checked minor Choosing the answer
Each of the three expressions is judged on its own and the choices combine the verdicts, so one wrong judgment anywhere points to a wrong combination. (Embretson & Wetzel, 1987)

Strategy: Try a pair such as s = 2 and t = −3 in each expression, then explain why its sign could or could not change for any other positive s and negative t.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Trusts only the squared expression, believing that subtracting t must make s smaller and that a negative exponent makes a negative number.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Handles the subtraction and the square correctly but believes a negative exponent makes the result negative, though s to the power t is a positive fraction such as 1/8.
C Negatives and direction (Brown & Burton, 1978)
Reveals a misconception Expected pull: moderate
Reads s − t as taking a number away from s, so thinks it could be negative, missing that subtracting the negative t adds to s.

What was corrected from the DOE handbook, and why

Explanation: The question makes t the negative integer, but two steps of the DOE’s explanation write the negative integer as −t, as if t were positive. For Expression I it says “s − (−t) is equivalent to finding the value of s + t, which must result in a positive number”: s − (−t) is not Expression I, and with the question’s t, s + t need not be positive (s = 1 and t = −2 give −1). For Expression III it says “s^(−t) is equivalent to finding the value of 1/s^t”, about s^(−t) rather than Expression III. The site now says that s − t is equivalent to s + (−t), and that s^t is equivalent to 1/s^(−t). Since t is negative, −t is a positive integer, so s + (−t) is the sum of two positive integers and 1/s^(−t) is the reciprocal of a positive integer: both positive, and the answer, I, II and III, is unchanged. The step for Expression II, s(−t)² = st², is true for any t and is unchanged.
Explanation: The DOE’s explanation introduces Expression III as “s^t. where s is a positive integer and t is a negative integer”, with a period where Expressions I and II (and the 2022–23 guide, which first printed this explanation) have a comma. The site has the comma.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Subtracting a number: $s - t = s + (-t)$; s and t are integers, and −t is the opposite of t. Subtracting a number is the same as adding its opposite. When t is negative, −t is positive. So subtracting t adds a positive amount.
A negative exponent: $s^{t} = \frac{1}{s^{-t}}$; s is the base and t is the exponent. When t is negative, −t is positive. So a negative exponent means 1 over the base to a positive exponent. For example, 2 to the power −3 is 1 over 2³, which is 1/8.

Sort out the question
An integer is a whole number or its opposite, such as −2, 0 or 5. Here s is a positive integer, like 1, 2 or 3. And t is a negative integer, like −1, −2 or −3. Must means always. An expression counts only if it is positive for every such pair.
Try one pair of numbers
Try s = 2 and t = −3 in each expression:
- I: $2 - (-3) = 2 + 3 = 5$
- II: $2 \times (-3)^{2} = 2 \times 9 = 18$
- III: $2^{-3} = \frac{1}{2^{3}} = \frac{1}{8}$
All three are positive this time. Now see why that always happens.
Explain expression I
Use the rule for subtracting:
$s - t = s + (-t)$
t is negative, so its opposite, −t, is positive. So s − t is a positive plus a positive. That is always positive.
Watch out: subtracting does not always make a number smaller. Subtracting a negative makes it bigger. Missing this leads to choice C.
Explain expression II
$t^{2}$ means t × t. A negative times a negative is positive, so $t^{2}$ is always positive. Then $s \times t^{2}$ is a positive times a positive. That is always positive too.
Explain expression III
Use the rule for a negative exponent:
$s^{t} = \frac{1}{s^{-t}}$
−t is a positive integer, so $s^{-t}$ is a positive integer. So $s^{t}$ is 1 over a positive number. That is a positive fraction.
Watch out: a negative exponent does not make the answer negative. It makes a fraction. Thinking it does leads to choice B, I and II only.
Choose the answer
Each of the three expressions is positive for every positive s and negative t. None of them can ever be zero or negative.
So the answer is D, I, II, and III.

Check your answer

Try the values closest to 0: s = 1 and t = −1. I gives $1 - (-1) = 2$ . II gives $1 \times (-1)^{2} = 1$ . III gives $1^{-1} = \frac{1}{1} = 1$ . All three are still positive.

Question 2 of 2

2023 Test B Math, question 46

Question

$0^{3}+4^{0}+2^{-1}+(-1)^{2}$

What is the value of the expression above, in decimal form?

A0
B0.5
C1.5
D2.5

Show the answer

Correct answer: D

$0^{3}+4^{0}+2^{-1}+(-1)^{2}$

$=0+1+\frac{1}{2}+1$

$=2\frac{1}{2}=2.5$

Question analysis

An exponents question: students find the decimal value of a sum of four powers, zero cubed, four to the zero power, two to the negative first power and negative one squared. Each of the four terms follows a different rule, and a slip on any one of them changes the total.

NYS standards: 8.EE.1 integer exponent properties
7.NS.2a rational number multiplication
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 work uses whole-number exponents and signed numbers. Zero and negative exponents are grade 8 content, and this expression tests both alongside a negative base, each with its own common misconception.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know major
Error-prone arithmetic minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 12 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every term tests a different exponent rule, and each wrong choice comes from breaking exactly one of them. The zero exponent is the most common casualty, since 4⁰ looks as if it should be 0 like 0³; a negative exponent read as a negative number is close behind. Writing each term’s value above it before adding, 0, 1, 1/2 and 1, keeps the four rules separate.

Something you must already know major Working it out
Three rules the test does not give decide the sum: any nonzero number to the 0 power is 1, 2 to the −1 power is 1/2, and (−1)² is positive 1. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
The four terms carry a zero base, a zero exponent, a negative exponent and a negative base, so a sign or a stray 0 can slip in at each one. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Evaluate each power on its own, using the rules for a zero exponent and a negative exponent and the sign of a squared negative number, then add the four values.

Traps

A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

A A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Treats 2 to the power −1 as −2 instead of one half, so the four terms 0 + 1 − 2 + 1 add to 0.
B Order of operations (Brown & Burton, 1978)
Reveals a misconception Expected pull: weak
Squares as if the minus sat outside the parentheses, taking (−1)² as −1, so the terms add to 0 + 1 + 0.5 − 1 = 0.5.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Takes 4 to the power 0 as 0 rather than 1, so the expression becomes 0 + 0 + 0.5 + 1 = 1.5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Zero exponent: $a^0 = 1$; a is any number except 0. An exponent of 0 always gives 1. It works because each step down in the exponent divides by a. For example, 4 to the first power is 4, and 4 ÷ 4 = 1.
Negative exponent: $a^{-n} = \frac{1}{a^n}$; a is the base, and n is a positive whole number. A negative exponent does not make the number negative. It means 1 divided by the base to the positive power. For example, 2 to the −1 power is one half.

Sort out the question
The expression adds four powers. A power is a base with an exponent, a small raised number. The exponent tells how many times to use the base as a factor, a number you multiply. Find each power, then add.
Find the first two powers
Zero times itself is still zero.
$0^3 = 0 \times 0 \times 0 = 0$
Next, use the zero exponent rule.
$4^0 = 1$
Watch out: $4^0$ is 1, not 0. Taking it as 0 gives choice C.
Find the negative power
Use the negative exponent rule.
$2^{-1} = \frac{1}{2^1} = \frac{1}{2} = 0.5$
Watch out: $2^{-1}$ is not −2. That mistake gives choice A.
Find the last power
The base is −1, inside the parentheses. Multiply it by itself. A negative times a negative is positive.
$(-1)^2 = (-1) \times (-1) = 1$
Watch out: The answer is not −1. That mistake gives choice B.
Add the four values
Now add them. The question wants a decimal, so write one half as 0.5.
$0 + 1 + 0.5 + 1 = 2.5$
So the answer is D, 2.5.

Check your answer

See why $4^0 = 1$ with a pattern. $4^2 = 16$ and $4^1 = 4$ . Each step down divides by 4, so $4^0 = 4 \div 4 = 1$ . In the same way, $2^0 = 1$ and $2^{-1} = 1 \div 2 = \frac{1}{2}$ .

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