Answer
$For\,\,all\,\,set\! s\,\,A,A\times \varnothing =\varnothing \\
To\,\,prove\,\,that\,\,a\,\,set\,\,A\times \varnothing\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,A\times \varnothing\,\,has\,\,no\,\,elements. \\ To\,\,
do\,\,this, suppose\,\,A\times \varnothing \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\
suppose\,\,\,(x,y)\in A\times \varnothing \\
by\,\,def.\,\,of\,\,cartesian\,\,product\,\,x\in A \,\,,y\in \varnothing \\
but\,\,y\in\varnothing (this\,\,is\,\,a\,\,contradiction)\\
so\,\,no\,\,elements\,\,(x,y)\,\,such\,that\,\,x\in A\,\,,y\in\varnothing \\
so\,\,A\times \varnothing =\varnothing
$
Work Step by Step
$For\,\,all\,\,set\! s\,\,A,A\times \varnothing =\varnothing \\
To\,\,prove\,\,that\,\,a\,\,set\,\,A\times \varnothing\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,A\times \varnothing\,\,has\,\,no\,\,elements. \\ To\,\,
do\,\,this, suppose\,\,A\times \varnothing \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\
suppose\,\,\,(x,y)\in A\times \varnothing \\
by\,\,def.\,\,of\,\,cartesian\,\,product\,\,x\in A \,\,,y\in \varnothing \\
but\,\,y\in\varnothing (this\,\,is\,\,a\,\,contradiction)\\
so\,\,no\,\,elements\,\,(x,y)\,\,such\,that\,\,x\in A\,\,,y\in\varnothing \\
so\,\,A\times \varnothing =\varnothing
$
