math/mpreduce.c: Fix some typos in commentary.

diff --git a/math/mpreduce.c b/math/mpreduce.c
index 2193de6dd5e7201ae651448a1dce0fa0a4be884d..871d3a42826c6928a4b737eac18be79f4e851f0f 100644 (file)
--- a/math/mpreduce.c
+++ b/math/mpreduce.c
@@ -47,7 +47,7 @@ DA_DECL(instr_v, mpreduce_instr);
  *
  * Suppose %$x = x' + z 2^k$%, where %$k \ge n$%; then
  * %$x \equiv x' + d z 2^{k-n} \pmod p$%.  We can use this to trim the
  *
  * Suppose %$x = x' + z 2^k$%, where %$k \ge n$%; then
  * %$x \equiv x' + d z 2^{k-n} \pmod p$%.  We can use this to trim the

- * representation of %$x$%; each time, we reduce %$x$% by a mutliple of
+ * representation of %$x$%; each time, we reduce %$x$% by a multiple of

  * %$2^{k-n} p$%.  We can do this in two passes: firstly by taking whole
  * words off the top, and then (if necessary) by trimming the top word.
  * Finally, if %$p \le x < 2^n$% then %$0 \le x - p < p$% and we're done.
  * %$2^{k-n} p$%.  We can do this in two passes: firstly by taking whole
  * words off the top, and then (if necessary) by trimming the top word.
  * Finally, if %$p \le x < 2^n$% then %$0 \le x - p < p$% and we're done.


@@ -210,7 +210,7 @@ int mpreduce_create(mpreduce *r, mp *p)
    * a perfect power of two, and %$d = 0$%, so again there is nothing to do.
    *
    * In the remaining case, we have decomposed @x@ as %$2^{n-1} + d$%, for
    * a perfect power of two, and %$d = 0$%, so again there is nothing to do.
    *
    * In the remaining case, we have decomposed @x@ as %$2^{n-1} + d$%, for

-   * some positive %$d%, which is unfortuante: if we're asked to reduce
+   * some positive %$d%, which is unfortunate: if we're asked to reduce

    * %$2^n$%, say, we'll end up with %$-d$% (or would do, if we weren't
    * sticking to unsigned arithmetic for good performance).  So instead, we
    * rewrite this as %$2^n - 2^{n-1} + d$% and everything will be good.
    * %$2^n$%, say, we'll end up with %$-d$% (or would do, if we weren't
    * sticking to unsigned arithmetic for good performance).  So instead, we
    * rewrite this as %$2^n - 2^{n-1} + d$% and everything will be good.