Short Answer:

$a1=2; $b1=10; ($a1 > $b1) ? ($int = ($a1/$b1)) : ($int = ($b1/$a1)); 
+print $int;
[download]

Better Short Answer:

$a1=2; $b1=10; $int = ($a1 > $b1) ? ($a1/$b1) : ($b1/$a1); print $int;
[download]

What it should really be:

use strict   ;
use warnings ;

# Use Better Variable Names. 
my $a1 = 2  ; 
my $b1 = 10 ; 

my $int = ( $a1 > $b1 ) ? ( $a1 / $b1 ) : ( $b1 / $a1 ) ; 

print $int;

[download]

Reasoning:

defined($int = ($a1/$b1)) will always evaluate to true regardless of the values of $a1 and $b1. If you want to check if both $a1 and $b1 are defined you might want to nest it in there like:

my $int = ( defined( $a1 ) and defined( $b1 ) ) ? ( ( $a1 > $b1 ) ? ( 
+$a1 / $b1 ) : ( $b1 / $a1 ) ) : die( '$a1 and $b1 need to be defined'
+ ) ;
[download]

The other thing to note is that the the Ternary Operator, being an Operator, will return something and so you dont need to do the $int = in both the if and else parts.

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