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Tristram Shandy

  1. Nov 19, 2015 #1
    A minor write stated the following argument about this famous puzzle:

    "Tristram Shandy, who writes his autobiography so slowly that he covers only one day of his life in a year of writing. the set of days written about cannot in fact always be a subset of the set of days past. Consider any day n. Suppose Tristram Shandy writes about day n and he finishes writing about day n on day n + m. Then for any day n + i, he finishes writing about n + i on day n + m + 365i. To find the day d on which Tristram Shandy writes about day d, we must solve for i in n + i = n + m + 365i; if the solution is I, then d = n + I = n + m + 365I. The solution is I = -m/364. So d = n -m/364 = n + m -365m/364- or their integral parts, [n -m/364] = [n + m -365m/364.]. On days I later than d, Tristram Shandy writes about his past (about days between days d and I); on day d, he writes about day d; and on days e earlier than d, he writes about his future (about days between days e and d). For any i, day n + i is covered by the end of day n + m + 365i-or, equivalently, any day x, past, present or future, is covered by the end of day f(x) = n + m + 365(x -n), a monotonically increasing function of x."

    I am lost as to his argument, but I don't see how his position is correct starting from what he is trying to prove. Can someone help me with his argument here?
  2. jcsd
  3. Nov 19, 2015 #2


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    The argument seems to be for the proposition that every day, past, present and future is eventually covered in the autobiography. It accepts without comment the notion that in the year prior to his birth, Tristram was writing about the day prior to his birth.

    That understanding may be faulty. After all, it hardly seems necessary to introduce variables i, d, n, m and l in order to phrase an argument that f(x) = x/365 is bijective.
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