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  "body": "\n## Proof\n\n1. Let $x$ and $y$ be real numbers with $x \u003c y$.\n2. By the [[existence of square roots|068BZBXE0C]], there is a real $s \\ge 0$ with $s^2 = 2$. Since $s^2 \\ne 0$, we have $s \\ne 0$, so $s \u003e 0$.\n3. The number $s$ is irrational, by the [[irrationality of the square root of 2|0DZGSDHD92]].\n4. Dividing by $s \u003e 0$ keeps the order in the [[ordered field|0QSPNASE50]] $\\R$. So $x/s \u003c y/s$.\n5. By the [[density of the rationals|0STX0WRDNG]], there is a rational $q_1$ with $x/s \u003c q_1 \u003c y/s$. Again by density, there is a rational $q_2$ with $q_1 \u003c q_2 \u003c y/s$.\n6. Since $q_1 \\ne q_2$, at least one of them is nonzero. Call it $q$. Then $q$ is rational, $q \\ne 0$ and $x/s \u003c q \u003c y/s$.\n7. Multiply by $s \u003e 0$: $x \u003c q s \u003c y$. Let $z = q s$.\n8. Suppose $z$ is a [[rational number|0NCJ5WXN6Z]] $r$. Then $s = r/q$ is a quotient of rationals with $q \\ne 0$, so $s$ is rational. This contradicts step 3.\n9. So $z$ is an [[irrational number|080T2TXVQV]] with $x \u003c z \u003c y$.\n\n## Idea\n\nA nonzero rational multiple of $\\sqrt{2}$ is irrational. Scale the interval by $1/\\sqrt{2}$, find a nonzero rational inside it, and scale back.\n",
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