Print (echo) variable in a single line

Hi,
I have written this code
------------------------------------------------

# !/bin/ksh
i=0
while [ $i -ne 10 ]
do
 j=$i
 while [ $j -ge 0 ]
 do
    echo  -e  $j
   #printf "%d",$j
    j=`expr $j - 1`
 done
  echo
 i=`expr $i + 1`
done

----------------------------------------------------
The ouput which i get is

-n 0
-n 1
-n 0
-n 2
-n 1
-n 0
-n 3
-n 2
-n 1
-n 0
-n 4
-n 3
-n 2
-n 1
-n 0
-n 5
-n 4
-n 3
-n 2
-n 1
-n 0
-n 6
-n 5
-n 4
-n 3
-n 2
-n 1
-n 0
-n 7
-n 6
-n 5
-n 4
-n 3
-n 2
-n 1
-n 0
-n 8
-n 7
-n 6
-n 5
-n 4
-n 3
-n 2
-n 1
-n 0
-n 9
-n 8
-n 7
-n 6
-n 5
-n 4
-n 3
-n 2
-n 1
-n 0

----------------------------------------
------------------------------------------

I want to print the ouput as

0
1 0
2 1 0
3 2 1 0
4 3 2 1 0
and so on till
9 8 7 6 5 4 3 2 1 0

Please help !!

# !/bin/sh
i=0
while [ $i -ne 10 ]
do
    j=$i
    while [ $j -ge 0 ]
    do
        echo -e "$j \c"
        j=`expr $j - 1`
    done
    echo
    i=`expr $i + 1`
done

Instead of invoking 'expr', try using ((j--)) and ((i++))

[root@dist unix]# cat a.sh 
#!/bin/bash
n=0
for i in {1..9};do
    n=${i}" "${n}
    echo ${n}
done 
[root@dist unix]# bash a.sh 
1 0
2 1 0
3 2 1 0
4 3 2 1 0
5 4 3 2 1 0
6 5 4 3 2 1 0
7 6 5 4 3 2 1 0
8 7 6 5 4 3 2 1 0
9 8 7 6 5 4 3 2 1 0