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  text: '[summary: A unit of a [3gq ring] is an element with a multiplicative inverse.]\n\nAn element $x$ of a non-trivial [3gq ring]%%note:That is, a ring in which $0 \\not = 1$; equivalently, a ring with more than one element.%% is known as a **unit** if it has a multiplicative inverse: that is, if there is $y$ such that $xy = 1$.\n(We specified that the ring be non-trivial. \nIf the ring is trivial then $0=1$ and so the requirement is the same as $xy = 0$; this means $0$ is actually invertible in this ring, since its inverse is $0$: we have $0 \\times 0 = 0 = 1$.)\n\n$0$ is never a unit, because $0 \\times y = 0$ is never equal to $1$ for any $y$ (since we specified that the ring be non-trivial).\n\nIf every nonzero element of a ring is a unit, then we say the ring is a [481 field].\n\nNote that if $x$ is a unit, then it has a *unique* inverse; the proof is an exercise.\n%%hidden(Proof):\nIf $xy = xz = 1$, then $zxy = z$ (by multiplying both sides of $xy=1$ by $z$) and so $y = z$ (by using $zx = 1$).\n%%\n\n# Examples\n\n- In [48l $\\mathbb{Z}$], $1$ and $-1$ are both units, since $1 \\times 1 = 1$ and $-1 \\times -1 = 1$. However, $2$ is not a unit, since there is no integer $x$ such that $2x=1$. In fact, the *only* units are $\\pm 1$.\n- [4zq $\\mathbb{Q}$] is a [481 field], so every rational except $0$ is a unit.',
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