6.4 One-to-One Functions
Definition 6.16. A function \(f:X\longrightarrow Y\) is one-to-one. If whenever \(x_1\) and \(x_2\) are distinct elements of \(X\) then \(f(x_1)\neq f(x_2)\).
Equivalently a function is one-to-one if \(f(x_1)=f(x_2)\implies x_1=x_2\).
Proof.
Suppose that \(f(x_1)=f(x_2).\) Then \begin {align*} \frac {6x_1-4}{2x_1-3} & = \frac {6x_2-4}{2x_2-3}\\ (6x_1-4)(2x_2-3) & = (6x_2-4)(2x_1-3)\\ 12x_1x_2-18x_1-8x_2+12 & = 12x_1x_2-18x_2-8x_1+12\\ -18x_1-8x_2 & = -18x_2-8x_1\\ 18x_2-8x_2 & = 18x_1-8x_1\\ 10x_2 & = 10x_1\\ x_2 & = x_1 . \end {align*}
So \(f(x_1)=f(x_2)\) forces \(x_1=x_2\), and \(f(x)=\dfrac {6x-4}{2x-3}\) is one-to-one. □
Proof. Suppose \(f(a)=f(b)\). Then \[\frac {4}{7}a+8=\frac {4}{7}b+8 \qquad \implies \qquad \frac {4}{7}a=\frac {4}{7}b \qquad \implies \qquad a=b,\] so \(f(x)=\dfrac {4}{7}x+8\) is one-to-one. □
Proof. Suppose that \(f(a)=f(b)\). Then \begin {align*} a^2-2a-3 & = b^2-2b-3\\ \implies a^2-2a & = b^2-2b\\ \implies a^2-b^2-2a+2b & =0\\ \implies (a-b)(a+b)-2(a-b) & = 0\\ \implies (a-b)(a+b-2) &=0\\ \end {align*}
Either \(a-b=0\) or \(a+b-2=0\), that is \(a=b\) or \(a=2-b\).
The second possibility is not excluded, so \(f(a)=f(b)\) does not force \(a=b\), and the function is not one-to-one. A single counterexample makes this concrete: taking \(a=0\) and \(b=2\), \[f(0)=-3 \qquad \text {and}\qquad f(2)=4-4-3=-3,\] two distinct inputs with the same output. □
Note 6.20. Contrast this with the two examples before it. There the algebra closed with \(a=b\) as the only possibility, and the functions were one-to-one. Here a second branch survived, and that branch is the counterexample — \(a=2-b\) says the graph is symmetric about \(x=1\), so points equidistant from \(x=1\) share an output.
When an argument of this kind produces an extra factor, do not discard it. Read it as a description of which pairs collide, then pick one such pair to display. A proof that a function is not one-to-one is finished by exhibiting two inputs, not by algebra alone.
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