The diagonal (d) of the unit square is said to have no commensurable value with the sides that equal (s=1). If so explain this.?

Vertical lines drawn from (s) to (d) divide each segment of (d) into a ratio equal to the ratio of (d)/(s). Lines d & s have the same number of divisions but of different lengths. If I multiply the number of segments on (d) by the sq. root of 2, then divide each segment of that quantity by the root of 2, how is it possible that the segments of the diagonal are not equal to the segments of the sides? I say they are the same. Conventional wisdom says this cannot be so. Just to be sure you understand what the problem is, read this example. I have two lines (L1) = 2 and (L2) = 1. I divide them into 10 segments. The segments of L1 are twice the length as those of L2. Now I multiply the segments on L1 two times (or divide L1 by 20 (the same thing). You now see that the segments on L1 are the same length as on L2; however, L1 does have twice as many segments. You see? It is the same with the diagonal of the unit square (d) and the sides (s). Explain to me why I am wrong if you can, Answer-Man. Best Wishes.

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  • 3 weeks ago
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    The issue here is the definition of "commensurable." You seem to be using a different definition from the common one. See

    https://en.wikipedia.org/wiki/Commensurability_(ma...

    • The identity principle as commensurable equality may be trivial but true. If the segments a in (s) are equal in length to the 2 *n number of segments in (d), and those segments each divided by the root value, they must equal each other, congruent segments in  lines s and d.

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