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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Suppose that $f : A \\to B$ is [[injective|0PVG1CRAJN]], and let $C = f(A)$ be the [[image|0Q3F20XB0A]] of $A$.\n2. Then $f$ is a [[bijection|0SNQEF7Z53]] from $A$ onto $C$: it is injective, and every element of $C$ is a value of $f$.\n3. Composing a numbering of $A$ with $f$ numbers $C$. So $C$ is finite and $|C| = |A|$, as in the [[bijection principle|0GV2573FP4]].\n4. The set $C$ is a subset of $B$. By the [[complement principle|03QC59M99B]], $|B \\setminus C| = |B| - |C| = |B| - |A|$.\n5. The left side is a natural number, but $|B| - |A| \u003c 0$ because $|A| \u003e |B|$. This is a contradiction.\n6. Hence $f$ is not injective. So there are $a \\ne a'$ in $A$ with $f(a) = f(a')$.\n\n## Idea\n\nAn injective map copies $A$ onto a subset of $B$ of the same size.\nA subset of $B$ cannot be larger than $B$, so this is impossible when $A$ has more elements than $B$.\n",
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