Q33: $(a, b)R(c, d) \Leftrightarrow ad = bc$ on $\mathbb{N} \times \mathbb{N}$⁠, or $f(x) = \frac{x}{1 + |x|}$ is one-one and onto

Reflexive means every element is related to itself. To test $(a, b)R(a, b)$⁠, put $c = a$ and $d = b$ in the rule $ad = bc$⁠. It becomes $ab = ba$⁠, which is always true.

Symmetric means: if $(a, b)R(c, d)$⁠, then $(c, d)R(a, b)$⁠.

Read the rule carefully. $(p, q)R(r, s)$ means $ps = qr$⁠: first of the first pair times second of the second pair equals second of the first times first of the second. For $(c, d)R(a, b)$ this is $cb = da$⁠.

We get it from $ad = bc$ by swapping the two sides and the order in each product.

Transitive means: if $(a, b)R(c, d)$ and $(c, d)R(e, f)$⁠, then $(a, b)R(e, f)$⁠. Write the rule for each given pair. The target is $af = be$⁠.

Multiply the two equations side by side. Both sides now contain the factor $cd$⁠:

$c$ and $d$ are natural numbers, so $cd \ne 0$⁠, and we may divide by it.

Watch out: cancelling is only allowed for a factor that cannot be 0. Say why, as the bracket does.

The equivalence class of $(2, 6)$ is the set of all pairs related to $(2, 6)$⁠.

By the rule, $(x, y)R(2, 6)$ means $x \cdot 6 = y \cdot 2$⁠. Dividing by 2 gives $y = 3x$⁠. Every natural number $x$ gives such a pair, so the class is infinite.

For example $x = 1$ gives $y = 3$⁠, and $x = 3$ gives $y = 9$⁠.

Remove the modulus first: $|x| = x$ when $x \ge 0$⁠, and $|x| = -x$ when $x < 0$⁠. So $1 + |x|$ is $1 + x$ or $1 - x$⁠.

The plan: prove one-one and onto on each piece, then check that the two pieces never share a value.

One-one means equal outputs come only from equal inputs: $f(x) = f(y) \Rightarrow x = y$⁠.

Cross-multiply (both denominators are positive): $x(1 + y) = y(1 + x)$⁠. Expand, and the $xy$ terms cancel.

These two lines show that the values on this piece lie in $[0, 1)$⁠. The rewrite is

Split it into two fractions:

Since $\frac{1}{1 + x} > 0$⁠, the value is less than 1.

It is $\ge 0$ because the top $x \ge 0$ and the bottom $1 + x > 0$⁠.

Onto means every value in the target is reached. To find the input, solve $y = \frac{x}{1 + x}$ for $x$⁠:

As $0 \le y < 1$⁠, the top is $\ge 0$ and the bottom is positive, so $x \ge 0$⁠, as Case 1 needs.

In the check, the bottom is $1 + \frac{y}{1 - y} = \frac{1}{1 - y}$⁠, so the fraction is $y$⁠.

Cross-multiply (both denominators are positive, since $x, y < 0$⁠). Expand, and the $xy$ terms cancel:

Same steps as Case 1. For $x < 0$⁠, $1 - x > 0$⁠, so the cross-multiplying is safe, and

(The top is $-1 + x + 1 = x$⁠.) Split it into two fractions:

Since $\frac{1}{1 - x} > 0$⁠, the value is more than $-1$⁠, and it is negative because the top is negative.

Solve $y = \frac{x}{1 - x}$ for $x$⁠: $y - xy = x$⁠, so $x(1 + y) = y$ and $x = \frac{y}{1 + y}$⁠.

In the check, the bottom is

As $-1 < y < 0$⁠, the top is negative and the bottom is positive, so $x < 0$⁠, as Case 2 needs.

Cases 1 and 2 only compare inputs from the same piece. A non-negative input and a negative input give outputs of opposite signs, so they can never give the same output. Together the three cases cover every pair of different inputs.