Relations and functions

R={(1,3),(2,6),(3,9),(4,12)}.

(1,1)∉R; (1,3)∈R but (3,1)∉R; (1,3),(3,9)∈R but (1,9)∉R.

The relation consists of (1,6),(2,7),(3,8).

It is not reflexive or symmetric; no second coordinate can start another pair, so transitivity holds vacuously.

Every number divides itself.

1|2 but 2∤1; and x|y,y|z implies x|z.

The difference of any two integers is an integer, so every ordered pair is related.

The relation is A×A, which has all three properties.

Everyone works at the same place as themself.

“Same workplace” is mutual and chains across people.

Each person lives in the same locality as themself.

Sharing a locality works in either order and passes through a third person.

Nobody is 7 cm taller than themself.

If x is 7 cm taller than y and y is 7 cm taller than z, x is 14 cm taller than z.

No person is their own wife, and a wife relation does not reverse.

A chain x is wife of y and y is wife of z cannot occur under the usual meaning, so the transitivity condition has no counterexample.

No one is their own father; fatherhood does not reverse.

If x is y’s father and y is z’s father, x is z’s grandfather, not father.

Reflexivity fails at a=1/2 because 1/2>1/4.

(0,1)∈R but (1,0)∉R; (2.2,1.5),(1.5,1.3)∈R but (2.2,1.3)∉R.

Pairs are (1,2),(2,3),(3,4),(4,5),(5,6).

No pair is diagonal; (1,2) has no reverse; (1,2),(2,3) lack (1,3).

a≤a for every real a.

a≤b and b≤c imply a≤c; 1≤2 but 2≰1.

At a=1/2, 1/2≤1/8 is false.

(0,1) is in R but (1,0) is not; 8≤2³ and 2≤1.5³, but 8≰1.5³.

The two pairs reverse one another, so R is symmetric.

It omits every diagonal pair; (1,2),(2,1) would require (1,1), which is absent.

A book has the same page count as itself.

Equality of page counts is symmetric and transitive.

|a−a|=0; reversing the difference changes only its sign.

A sum of two even differences is even, so parity classes are {1,3,5} and {2,4}.

The condition means a and b have the same remainder upon division by 4.

The elements of A congruent to 1 mod 4 are 1,5,9.

a=a, equality reverses, and a=b,b=c imply a=c.

Only 1 equals 1, so no other element is related to 1.

Each pair’s reverse is included.

No diagonal pairs are included, and chaining the two pairs would require (1,1).

There is no two-pair chain that could violate transitivity.

It omits diagonal pairs and omits reverse pair (2,1).

Including all diagonal pairs makes it reflexive; including both directions makes it symmetric.

(1,2),(2,3) are present but (1,3) is absent.

Both diagonal pairs are present.

The only nontrivial order step is 1→2; its reverse (2,1) is absent.

No pair violates symmetry or transitivity, so both implications hold.

Reflexivity requires (1,1), which is absent.

Every point has its own distance; equality of distances is symmetric and transitive.

Q is related to P exactly when OQ=OP, the circle equation with radius OP.

Every triangle is similar to itself; similarity reverses and chains.

(6,8,10)=2(3,4,5), while the side ratios of (5,12,13) differ.

A polygon has the same side count as itself; equality is symmetric and transitive.

The given polygon has 3 sides, so all and only 3-sided polygons are related to it.

Every line is parallel to itself; parallelism is symmetric and transitive.

A line parallel to y=2x+4 has the same slope 2, so it is y=2x+c.

All four diagonal pairs are present, so it is reflexive.

The only two-step off-diagonal chain is (1,3),(3,2), and its result (1,2) is present; (1,2) has no reverse.

Check b>6 and a=b−2 for each pair.

For (6,8), 8>6 and 6=8−2; the other pairs fail a condition.

If 1/x₁=1/x₂, nonzero denominators allow x₁=x₂.

For any y≠0, x=1/y maps to y; positive natural inputs cannot produce negative outputs.

For positive naturals, x₁²=x₂² implies x₁=x₂.

A nonsquare natural such as 2 is not an output.

f(1)=f(−1)=1, so it is not injective.

Negative integers, such as −1, are not squares.

f(1)=f(−1), so it is not injective.

No real square equals −1, so it is not onto R.

The cube function is strictly increasing on positive numbers.

2 is not a cube of a natural number.

Equal cubes have equal real cube roots, so f is injective.

2 is not the cube of an integer, so f is not onto.

⌊1.2⌋=⌊1.8⌋=1, so it is not one-one.

Its outputs are integers, so a noninteger real such as 1/2 is not reached.

|1|=|−1|, so it is not one-one.

Absolute values are nonnegative, so negative reals are missing from the range.

f(1)=f(2)=1, so it is not one-one.

Its only outputs are −1,0,1, so values such as 2 are missed.

The images of 1,2,3 are 4,5,6.

These images are all distinct, so equal outputs can only come from equal inputs.

If 3−4x₁=3−4x₂, then x₁=x₂.

For any y∈R, x=(3−y)/4 is real and maps to y.

f(1)=f(−1)=2.

Every output is at least 1, so 0 is not attained.

Swapping twice returns the original ordered pair.

Every (b,a)∈B×A is reached from (a,b)∈A×B.

Odd n=2k−1 maps to k; even n=2k maps to k.

Actually both 2k−1 and 2k map to k, so it is not one-one; every k is attained, so it is onto.

If y=(x−2)/(x−3), solving gives x=(3y−2)/(y−1).

For every y≠1 this x is real and x≠3; the formula gives a unique preimage.

f(1)=f(−1)=1, so it is not one-one.

Its outputs are nonnegative, so it cannot reach −1.

3x₁=3x₂ implies x₁=x₂.

For any y∈R, x=y/3 satisfies f(x)=y.

For x≥0, f(x)=x/(1+x); for x<0, f(x)=x/(1−x).

Each branch is strictly increasing and maps onto [0,1) or (−1,0); together they cover (−1,1).

Assume x₁³=x₂³.

The real cube function is strictly increasing, so x₁=x₂.

A⊆A and A⊆B,B⊆C imply A⊆C.

Choose distinct x∈X: ∅⊆{x}, but {x}⊄∅.

An onto function between finite sets of the same size must also be one-one.

The possible bijections are the n! permutations of the codomain elements.

Evaluate each function at all four domain values.

The resulting output lists match, so the functions are equal.

Reflexivity forces all three diagonal pairs; symmetry forces (2,1),(3,1).

Including (2,3) and (3,2) makes the relation transitive, so omit both for the required nontransitive relation.

An equivalence relation corresponds to a partition into classes.

1 and 2 must share a class: either {1,2}|{3} or {1,2,3}.