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An irrational number cannot be written as a ratio of two integers and has a non-terminating, non-repeating decimal. The square root of 7 is not a perfect square, so it is irrational, while the others are rational.
The integers are the whole numbers together with their negatives, including zero. They do not include fractions, decimals, or irrational values.
Absolute value measures distance from zero, so |−9| = 9 and |4| = 4. Adding them gives 13, which is positive.
The associative property states that regrouping addends does not change the sum. Only the grouping parentheses move here; the order of the numbers is unchanged.
The additive inverse of a number is the value that sums with it to zero. Since −12 + 12 = 0, the additive inverse is 12.
A number lies between −2 and −1 when it is greater than −2 and less than −1. The value −1.5 satisfies both conditions.
Natural numbers are the positive counting numbers 1, 2, 3, and so on. Among the choices, only 7 belongs to that set.
Multiplication is performed before addition, so 4 × 2 = 8 first. Then 3 + 8 = 11.
A repeating decimal can always be written as a fraction, so 0.333… equals 1/3 and is rational. Non-repeating, non-terminating decimals are irrational.
Subtracting a negative is the same as adding its opposite, so −5 + 8 + 3. Combining gives 3 + 3 = 6.
The equation |x| = 5 has two solutions, 5 and −5. The additional condition x < 0 eliminates 5, leaving −5.
The distributive property multiplies a factor across each term inside parentheses: 4·x + 4·3 = 4x + 12.
Since −3/2 = −1.5, it is the smallest value. Then comes −0.5, followed by 0.25 and 1 in increasing order.
The multiplicative inverse is the reciprocal, found by flipping the fraction while keeping the sign. The reciprocal of −4/5 is −5/4, and their product is 1.
Rising means adding the change to the starting value: −7 + 12 = 5. The result is positive because the rise exceeds the magnitude of the starting value.
Any integer n can be written as n/1, a ratio of integers, so every integer is rational. The reverse containment does not hold.
In −2³ the exponent applies only to 2, giving −8. Then (−2)² = 4, so −8 + 4 = −4.
Whole numbers are 0, 1, 2, … while natural numbers are usually 1, 2, 3, …. Zero is a whole number and an integer but not a natural (counting) number.
First evaluate the parentheses: 3 − 5 = −2. Squaring gives 4, then 2 × 4 = 8, and 8 ÷ 4 = 2.
The product of two negative numbers is positive, so a × b is positive regardless of the values. Sums and differences of negatives may be negative.
A square root of a non-perfect-square positive integer is irrational. Since 11 is not a perfect square, √11 has a non-terminating, non-repeating decimal.
If the product were rational, dividing by the nonzero rational factor would make the irrational factor rational, a contradiction. Hence the product must be irrational.
Convert −3/4 to −0.75. On the number line, −0.75 is to the left of −0.7, so −0.75 < −0.7.
First simplify inside each absolute value: −3 − 7 = −10, so |−10| = 10. Then −2 + 5 = 3, so |3| = 3, and 10 − 3 = 7.
The rationals are dense: given two distinct rationals, their average is a rational number strictly between them. This can be repeated indefinitely.
Every element of A also belongs to B, so A is a subset of B; because B contains 4 and 5 which are not in A, the subset relation is proper. Equal sets must contain exactly the same elements, which is not the case here. Disjoint sets share no elements, but A and B share three.
The union collects every element that appears in either set, and repeated elements are listed only once. Combining P and Q gives a, b, c, d, and e with the shared c counted a single time. The intersection, by contrast, would be just {c}.
The complement of A consists of all elements of the universal set that are not in A. Removing 2, 4, and 6 from U leaves 1, 3, and 5. The empty set would be the complement only if A already equaled U.
A set with n elements has 2^n subsets, counting the empty set and the set itself. Here n = 3, so there are 2^3 = 8 subsets. Listing them confirms this: the empty set, three singletons, three pairs, and R itself.
A statement in propositional logic must be a declarative sentence that is either true or false. The conditional form 'If P, then Q' is a standard compound statement built from two simpler propositions. Questions and scrambled word strings are not truth-valued statements.
A conjunction is true only when both of its parts are true. Since q is false, the conjunction p AND q is false regardless of p. Conjunction is commutative, so the order of the parts cannot change the result.
An inclusive disjunction is true whenever at least one part is true, so it fails only when both parts are false. All other combinations, including one true and one false, make the disjunction true. This is the standard truth table for OR.
Set-builder notation names a set by stating a property that its elements satisfy. The numbers 2, 4, 6, 8, and 10 are exactly the even integers strictly between 1 and 11. Each other description yields numbers that do not match E.
The intersection contains only the elements that belong to both sets. Scanning M and N, the shared values are 3 and 4. The other elements appear in just one of the sets, so they are excluded.
By the inclusion-exclusion principle, the union equals the sum of the two groups minus the overlap: 25 + 18 - 10 = 33. Subtracting the overlap is necessary because the 10 bilingual students were counted in both groups. So 33 students study at least one language.
Disjoint sets have no element in common, so their intersection is empty. The odd numbers 1, 3, 5 and the even numbers 2, 4, 6 share nothing, making them disjoint. Each other pair shares at least one element.
The contrapositive of 'If P, then Q' is 'If not Q, then not P,' and it is logically equivalent to the original. Negating and swapping gives 'If a number is not even, then it is not divisible by 4.' The converse and inverse are different statements and are not equivalent to the original.
The negation of a universal statement 'All S are P' is the particular statement 'Some S are not P.' Reversing the negation returns 'All cats are friendly.' The other choices concern different quantifier patterns and are not logical negations of the given statement.
A conditional is false in exactly one case: when the antecedent is true and the consequent is false. Here p is true and q is false, which is that case, so the conditional is false. All other combinations of truth values make the conditional true.
The power set is the collection of all subsets of T, and a set with n elements has 2^n subsets. With n = 5, the power set has 2^5 = 32 elements. The value 64 would correspond to a set with 6 elements.
Inclusion-exclusion gives the overlap as 20 + 15 - (number owning at least one). To minimize the overlap, maximize the union, but the union cannot exceed the 30 people available. So the overlap is at least 20 + 15 - 30 = 5.
De Morgan's law states that p OR q is equivalent to not (not p AND not q), because negating both parts and switching the connective preserves the truth value. Both expressions are false only when p and q are both false. The other options produce different truth tables.
A partition divides a set into nonempty blocks that are pairwise disjoint and whose union is the whole set. That means each element lies in exactly one block, as in {a, b}, {c}, and {d, e}. Blocks need not be equal in size, and the empty set is never a block.
The symmetric difference contains the elements that belong to exactly one of the two sets. The shared elements 2 and 3 are removed, leaving 1 from A and 4 from B. Thus A △ B = {1, 4}.
Negating a universal statement 'All S are P' produces the existential statement 'Some S are not P,' meaning at least one counterexample exists. So the negation is that at least one mammal does not live on land. The other options either restate the claim or make a different, stronger assertion.
The number in A or B is 30 + 20 - 8 = 42 by inclusion-exclusion. Elements outside both sets are the remainder of the universal set: 50 - 42 = 8. So 8 elements lie in the region outside both circles.
A conditional 'If P, then Q' permits valid inference only in the direction P to Q, or equivalently from not Q to not P. Knowing Q (that the figure is a rectangle) does not determine P. Affirming the consequent is a logical fallacy, so no conclusion about being a square follows.
If the union of A and B equals A, then every element of B must already lie in A, which means B ⊆ A. If B contained an element outside A, the union would be strictly larger than A. The condition does not require A to be empty or the sets to be disjoint.
All 12 calculus students also take physics, so they are counted inside the 20 physics students. The number taking at least one subject is therefore 20, and adding the 5 who take neither gives 20 + 5 = 25. Wait: the union is 20, so the class has 25 students; the correct total is 25.
The symmetric difference keeps elements in exactly one set, which is the union with the shared elements removed. So A △ B = (A ∪ B) - (A ∩ B). Removing the intersection from the union eliminates precisely the elements counted in both sets.