Relations and Functions

Equal ordered pairs have equal corresponding coordinates: x/2=5/3 gives x=10/3; y−1/3=1/3 gives y=2/3.

n(A×B)=n(A)n(B)=3×3=9.

Choose the first coordinate from the first set and the second from the other; reverse the roles for H×G. Order matters.

A cartesian product contains every ordered pair, not just pairs with different entries.

Each of the three coordinates has two choices, giving 2³=8 ordered triples.

Collect the first coordinates to find A and the second coordinates to find B.

B∩C={5,6}; the products intersect in precisely those pairs. Since A⊂B and C⊂D, every pair in A×C is in B×D.

There are 4 ordered pairs. A four-element set has 2⁴=16 subsets, including ∅ and the entire set.

The three distinct first coordinates exhaust A; the two second coordinates exhaust B, matching the specified sizes.

n(A)²=9 gives n(A)=3. The given pairs require −1,0,1 as the three elements. List the 3×3 product and remove the given two.

The rule is y=3x. Keep pairs whose two entries are in A; 3x≤14 allows x=1,2,3,4.

Assume N={1,2,3,…}. Apply y=x+5 to each allowed natural input x<4.

An odd difference requires one even and one odd number. All A-values are odd; 4 and 6 are even, while 9 is odd.

Read each arrow as an ordered pair, keeping the starting value first: (5,3), (6,4), (7,5).

In every pair, the first value is 2 more than the second, so x−y=2 describes the relation.

The domain is the set of first coordinates; the range is the set of second coordinates.

For each a, list members of A that are multiples of a. Since every a divides itself, every member appears as a first and second coordinate.

Substitute each allowed input into x+5.

The primes below 10 are 2,3,5,7; cube each for the second coordinate.

A×B has six ordered pairs. Each subset is a relation, giving 2⁶=64 relations.

Every integer pair satisfies the condition because the difference of two integers is an integer.

A repeated output is allowed. A repeated input with different outputs violates the definition of a function.

A real square root needs 9−x²≥0; the negative modulus is never positive.

Substitute: 2(0)−5=−5; 14−5=9; −6−5=−11.

Substitute the first three values. For 212=9C/5+32, subtract 32 and multiply by 5/9: C=100.

Positive x makes 2−3x less than 2 with no lower bound. Since x²≥0, x²+2≥2. The identity function returns every real input.

The pieces defining f do not overlap and cover the domain, so each input has one value. In g, x=2 lies in both pieces and gives two different outputs.

((1+x)²−1)/x=(x²+2x)/x=x+2. The original denominator requires x≠0.

The denominator factors as (x−4)(x+3); exclude x=4 and x=−3.

Every real input is allowed and each maps to 5.

Exclude x=1 because it makes the denominator zero. The output cannot be zero; conversely, each nonzero y is achieved by x=1+2/y.

Let t=x²≥0. Then f=t/(t+1)=1−1/(t+1), which is 0 when t=0 and approaches but never reaches 1.

Combine values at the same input. For division, exclude the input where g(x)=0.

f(0)=b=−1. Then f(1)=a+b=1, so a=2.

For each claim, use the definition a=b²; a single counterexample disproves a universal claim.

All first coordinates are in A and second coordinates in B, so f⊂A×B. Input 2 has two outputs (9 and 11), so f is not a function.

A function must give one output per input. The same product ab can result from different pairs with different sums.

9=3² gives 3; 10=2×5 gives 5; 11 and 13 are prime; 12=2²×3 gives 3. Collect distinct highest prime factors.