(Two) straight lines are commensurable in square1 when the squares on them are measured by the same area, but (are) incommensurable (in square) when no area admits to be a common measure of the squares on them.
Two segments of lengths \(\alpha > 0\) and \(\beta > 0\) are called commensurable in square, if there exists a figure of area \(\gamma\) being an aliquot part of the square areas \(\alpha^2\) and \(\beta^2\), i.e. such that \[\alpha^2=p\gamma,\quad\beta^2=q\gamma\] for some natural numbers \(p > 0\) and \(q > 0\).
Equivalently, the segments are commensurable in square, if the ratio of their squares is a rational number, formally, \[\frac{\alpha^2}{\beta^2}=\frac{p\gamma}{q\gamma}=\frac pq.\]
If \(\alpha,\beta\) are not commensurable in square, we call them incommensurable in square.
Definitions: 1
Proofs: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97
Propositions: 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145
Literally, "in power" (translator's note) ↩
