What is the negation of "At least $K$ numbers are larger than $W$"?

Suppose if we set $j=$ number of numbers larger than $W$. Then, $$j \ge K$$ which negates as, $$j<K$$ which translates to $$\text{at most }K\text{ numbers are larger than }W$$

Am I right?


  • 1
    $\begingroup$ You are not quite negating the way one should when doing predicate calculus, but that aside: the negation of "there are at least $K$ integers greater than $W$" over the domain of the integers is "there are at most $K-1$ integers greater than $W$". Your inequalities, while not very well defined, give this; just note "at most" is inclusive while your inequality is not inclusive, so you must use $K-1$. You could also say "at most $K$ integers greater than $W$ exclusive", though this is sure to lead to confusion. $\endgroup$ May 30, 2020 at 9:50
  • $\begingroup$ Thanks, it cleared me well. $\endgroup$
    – Dinesh
    May 30, 2020 at 9:54

1 Answer 1


The negation may be given by

Less than $K$ numbers are larger than $W$


Not up to $K$ numbers are larger than $W$


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