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A prime integer greater than 1 has exactly two positive divisors, itself and 1. Testing 61 against the primes 2, 3, 5, and 7 gives no divisor, and because 7² = 49 < 61 < 121 = 11², no prime larger than 7 need be tested. Therefore 61 is prime.
A real number is rational exactly when its decimal expansion terminates or eventually repeats one fixed block forever. This expansion never ends and never settles into a repeating block, so it cannot equal a ratio of two integers. Hence the number is irrational.
The intersection consists of exactly those elements that belong to both sets. The numbers 3 and 4 appear in A and in B, while 1 and 2 belong only to A and 5 and 6 belong only to B. Therefore A ∩ B = {3, 4}.
Expand with the distributive law: 3(1) + 3(−4i) + 2i(1) + 2i(−4i) = 3 − 12i + 2i − 8i². Substituting i² = −1 turns −8i² into +8, so the product is 3 − 10i + 8 = 11 − 10i.
A conditional statement is logically equivalent to its contrapositive, which is formed by negating both parts and reversing the direction: ¬q → ¬p. The converse q → p and the inverse ¬p → ¬q have different truth conditions, and p ∧ ¬q is actually the negation of the original conditional.
The first term is 5 and the common difference is 4. The nth term of an arithmetic sequence is a₁ + (n − 1)d, and nineteen differences separate the first term from the twentieth. Thus a₂₀ = 5 + 19(4) = 5 + 76 = 81.
Factor both numbers: 84 = 2² · 3 · 7 and 126 = 2 · 3² · 7. The greatest common divisor takes the smallest power of each shared prime, giving 2¹ · 3¹ · 7¹ = 42. Every common divisor of the two numbers must divide 42.
The magnitude of a vector (a, b) is √(a² + b²), the Euclidean length of the arrow from the origin. Here a² + b² = 36 + 64 = 100, and the principal square root of 100 is 10. The magnitude is therefore 10.
Scientific notation writes a number as a × 10ⁿ with 1 ≤ |a| < 10. Moving the decimal point from 0.0000425 to 4.25 requires shifting it five places to the right, so the exponent must be −5. Therefore 0.0000425 = 4.25 × 10⁻⁵.
For a 2 × 2 matrix with rows (a, b) and (c, d), the determinant equals ad − bc. Here ad − bc = 4(5) − (−3)(2) = 20 + 6 = 26. The subtraction of a negative product makes the two contributions add.
Reduce the base modulo 5: since 7 ≡ 2 (mod 5), we get 7¹⁰⁰ ≡ 2¹⁰⁰ (mod 5). Because 2⁴ = 16 ≡ 1 (mod 5) and 100 is a multiple of 4, we have 2¹⁰⁰ = (2⁴)²⁵ ≡ 1²⁵ = 1 (mod 5). The remainder is therefore 1.
The standard proof is by contradiction: write √2 in lowest terms as a/b, so that a² = 2b². This forces a to be even, say a = 2k, whence 4k² = 2b² and b² = 2k², so b is even too. That contradicts the assumption that a/b was already in lowest terms.
Each term is 0.4 times the preceding term, so the common ratio satisfies |r| < 1 and the series converges. The sum of a convergent geometric series is a/(1 − r), which here equals 3/(1 − 0.4) = 3/0.6 = 5.
The real part is negative and the imaginary part is positive, so z lies in the second quadrant. The reference angle is arctan(√3/1) = π/3, and a second-quadrant angle with that reference angle is π − π/3 = 2π/3.
An induction proof has a base case and an inductive step. The inductive step assumes the statement for n = k and uses that assumption to derive the statement for n = k + 1, here by adding the next odd number 2k + 1 and factoring k² + 2k + 1 = (k + 1)².
A set with n elements has exactly 2ⁿ subsets, because each element is independently either included or excluded. Solving 2ⁿ = 32 = 2⁵ gives n = 5, so the set has five elements.
Because 48 = 16 · 3, the product rule for logarithms gives log₂ 48 = log₂ 16 + log₂ 3. Since log₂ 16 = 4 and log₂ 3 = a, the value is 4 + a.
If N = pᵃqᵇrᶜ, every positive divisor has the form pⁱqʲrᵏ with 0 ≤ i ≤ a, 0 ≤ j ≤ b, and 0 ≤ k ≤ c, so the count is (a + 1)(b + 1)(c + 1). Here that product is (4 + 1)(2 + 1)(1 + 1) = 5 · 3 · 2 = 30.
Apply the recursion one step at a time: a₂ = 3(2) − 1 = 5, a₃ = 3(5) − 1 = 14, and a₄ = 3(14) − 1 = 41. Each term must be computed from the immediately preceding one, so no term can be skipped.
The negation of “for all x, P(x)” is “there exists an x such that not P(x)”. So the negation asserts that at least one real number fails to have a multiplicative inverse, which is true of 0 and false of every nonzero real number.
Since 3 and 7 are relatively prime, the congruences n ≡ 2 (mod 3) and n ≡ 2 (mod 7) combine into n ≡ 2 (mod 21). Testing the candidates 2, 23, 44, … modulo 5 shows that 23 ≡ 3 (mod 5), so 23 satisfies all three conditions and is the smallest such positive integer.
Density means that between any two distinct real numbers there is a rational number; one proof picks n with 1/n smaller than the gap and then a suitable multiple of 1/n inside the interval. The irrationals and the reals are both uncountable, and numbers such as √2 are neither rational nor integers.
By De Moivre's theorem, z⁵ = cos 2π + i sin 2π = 1, and z ≠ 1, so z is a primitive fifth root of unity. The sum is geometric with ratio z, so it equals (z⁵ − 1)/(z − 1) = 0/(z − 1) = 0.
If log₂ 3 were rational, say p/q with positive integers p and q, then 2ᵖ = 3ᵠ, contradicting unique prime factorization. Hence log₂ 3 is irrational, while the other three choices equal 2, 4/5, and 4/33 respectively.
If the sequence converges to L, taking limits on both sides of aₙ₊₁ = √(2 + aₙ) gives L = √(2 + L), so L² − L − 2 = 0 and L = 2 or L = −1. Every term is positive, so the limit cannot be negative and must equal 2.
A prime has exactly two positive divisors, 1 and itself. The number 29 is not divisible by 2, 3, or 5, and its only positive divisors are 1 and 29. The other choices factor as 21 = 3·7, 33 = 3·11, and 39 = 3·13.
When multiplying powers with the same base, add the exponents: 2³·2⁴ = 2^(3+4) = 2⁷. This rule applies because both factors have base 2 and are being multiplied, not added or raised to another power.
The converse of a conditional statement interchanges the hypothesis and conclusion. The original statement has hypothesis “n is divisible by 6” and conclusion “n is divisible by 3,” so the converse is “If n is divisible by 3, then n is divisible by 6.”
The intersection A ∩ B contains exactly the elements that belong to both A and B. The numbers 3 and 4 appear in both sets, so A ∩ B = {3, 4}.
In binary, the place values from right to left are 1, 2, 4, 8, and so on. Since 13 = 8 + 4 + 1, the binary numeral is 1101.
The inequality |x − 2| < 3 means the distance from x to 2 is less than 3. This is equivalent to −3 < x − 2 < 3, and adding 2 to all parts gives −1 < x < 5.
A proof by contradiction begins by assuming the negation of the desired conclusion and shows that this assumption leads to an impossibility. Once the contradiction is reached, the original statement must be true.
The contrapositive of P → Q is ¬Q → ¬P. Here P is “f is differentiable” and Q is “f is continuous,” so the contrapositive is “If f is not continuous, then f is not differentiable.”
A counterexample must be an odd integer that is not prime. The number 9 is odd, but 9 = 3·3, so it is composite; thus it disproves the claim.
The floor of −2.3 is the greatest integer less than or equal to −2.3, which is −3. The ceiling of 2.3 is the least integer greater than or equal to 2.3, which is 3. Their sum is −3 + 3 = 0.
Let x = 0.474747… Multiplying by 100 gives 100x = 47.474747…, so 100x − x = 47, or 99x = 47. Therefore x = 47/99.
The units digits of powers of 3 repeat in the cycle 3, 9, 7, 1 with period 4. Since 2024 is divisible by 4, 3²⁰²⁴ has the same units digit as 3⁴, which is 1.
If a ≡ 3 and b ≡ 5 modulo 7, then ab ≡ 3·5 = 15 modulo 7. Since 15 = 2·7 + 1, the remainder is 1.
The first 50 positive even integers are 2, 4, …, 100, an arithmetic series with 50 terms. Their sum is 50(2 + 100)/2 = 50·51 = 2550.
By the binomial theorem, the term containing x⁵ in (x + 2)⁷ is C(7,5)x⁵·2². Since C(7,5) = 21 and 2² = 4, the coefficient is 21·4 = 84.
Use inclusion–exclusion. There are ⌊1000/3⌋ = 333 multiples of 3, ⌊1000/5⌋ = 200 multiples of 5, and ⌊1000/15⌋ = 66 multiples of both; the count is 333 + 200 − 66 = 467.
Euler’s totient function for 1000 = 2³·5³ is φ(1000) = 1000(1 − 1/2)(1 − 1/5) = 1000·1/2·4/5 = 400.
R contains (1,1), (2,2), and (3,3), so it is reflexive; whenever (a,b) is in R, (b,a) is also in R, so it is symmetric; and the required transitive combinations hold. It is therefore an equivalence relation.
The next simultaneous flash occurs after the least common multiple of the two flashing intervals. Since 18 = 2 · 3² and 30 = 2 · 3 · 5, the LCM is 2 · 3² · 5 = 90. Thus the lights next flash together 90 seconds after t = 0.
Substitute −2 for every x and follow the order of operations: 3(−2)² = 3·4 = 12, then −2(−2) = +4, and finally add the constant 1. This gives 12 + 4 + 1 = 17. The most frequent error is mishandling the sign when the middle term is multiplied by a negative input, or squaring only the numeral 2 and discarding the negative sign.
In an arithmetic sequence each term is obtained by adding a fixed constant to the previous term, so subtract any term from the one that follows it. Here 11 − 7 = 4, and the next difference 15 − 11 = 4 confirms the constant is 4. Consistency across several consecutive differences is what verifies an arithmetic pattern rather than a coincidental one.
Slope is the vertical change divided by the horizontal change between two points: (11 − 3) divided by (6 − 2), which is 8/4 = 2. The subtractions must be performed in the same order in the numerator and the denominator; reversing one but not the other changes the sign of the result and produces a different line.
Write y = 2x + 5, interchange x and y to obtain x = 2y + 5, then solve for y by subtracting 5 and dividing by 2, giving (x − 5)/2. Because f and its inverse undo each other, the operations must be reversed and performed in reverse order. Checking a value such as f(1) = 7 with f⁻¹(7) = 1 confirms the rule.
The expression log₂ 32 asks for the exponent to which 2 must be raised to obtain 32. Repeated doubling gives 2, 4, 8, 16, 32, which requires five multiplications by 2, so 2⁵ = 32 and the logarithm equals 5. Reading a logarithm as an exponent rather than as a quotient prevents most errors on this type of item.
The vertex form y = a(x − h)² + k has vertex (h, k). Here the squared expression is zero when x = 3, and at that input the output equals the constant 4, so the vertex is (3, 4). Note that the horizontal coordinate takes the opposite sign of the number appearing inside the parentheses, while the vertical coordinate keeps the sign of the outside constant.