Get ceiling integer from number in linux (BASH)

Why use external script languages? You get floor by default. To get ceil, do

$ divide=8; by=3; (( result=(divide+by-1)/by )); echo $result
3
$ divide=9; by=3; (( result=(divide+by-1)/by )); echo $result
3
$ divide=10; by=3; (( result=(divide+by-1)/by )); echo $result
4
$ divide=11; by=3; (( result=(divide+by-1)/by )); echo $result
4
$ divide=12; by=3; (( result=(divide+by-1)/by )); echo $result
4
$ divide=13; by=3; (( result=(divide+by-1)/by )); echo $result
5
....

To take negative numbers into account you can beef it up a bit. Probably cleaner ways out there but for starters

$ divide=-10; by=10; neg=; if [ $divide -lt 0 ]; then (( divide=-divide )); neg=1; fi; (( result=(divide+by-1)/by )); if [ $neg ]; then (( result=-result )); fi; echo $result
-1

$ divide=10; by=10; neg=; if [ $divide -lt 0 ]; then (( divide=-divide )); neg=1; fi; (( result=(divide+by-1)/by )); if [ $neg ]; then (( result=-result )); fi; echo $result
1

(Edited to switch let ... to (( ... )).)

Here's a solution using bc (which should be installed just about everywhere):

ceiling_divide() {
  ceiling_result=`echo "($1 + $2 - 1)/$2" | bc`
}

Here's another purely in bash:

# Call it with two numbers.
# It has no error checking.
# It places the result in a global since return() will sometimes truncate at 255.

# Short form from comments (thanks: Jonathan Leffler)
ceiling_divide() {
  ceiling_result=$((($1+$2-1)/$2))
}

# Long drawn out form.
ceiling_divide() {
  # Normal integer divide.
  ceiling_result=$(($1/$2))
  # If there is any remainder...
  if [ $(($1%$2)) -gt 0 ]; then
    # rount up to the next integer
    ceiling_result=$((ceiling_result + 1))
  fi
  # debugging
  # echo $ceiling_result
}