![We\ are\ given\ that,\\A\ is\ the\ set\ of\ squares.\\x\ belongs\ to\ the\ Set\ A\\y\ belongs\ to\ the\ Set\ A,\ too.\\Hence,\\Lets\ consider\ that\ \sqrt{x} =a, \sqrt{y}=b\\Hence,\\a^2=x, b^2=y\\Before\ proceeding\ we\ may\ use\ the\ following\ Laws\ Of\ Exponents:\\* For\ any\ two\ real\ numbers,\ m\ and\ n, m^a*n^a=(mn)^a\\* For\ any\ two\ real\ numbers,\ m\ and\ n, \frac{m^a}{n^a}=[\frac{m}{n}]^a\\](/tpl/images/1005/4376/48bd8.png)
![Now,\\When\ we\ intend\ to\ multiply\ x\ and\ y,\\x*y=a^2*b^2\\xy=a^2*b^2\\xy=(ab)^2\\As\ the\ new\ term\ xy\ is\ the\ square\ of\ product\ of\ a\ and\ b\,\ it\\ is\ a\ square\ too.\ Hence,\ it\ belongs\ in\ the\ set\ of\ squares(A).\\Similarly,\\When\ we\ intend\ to\ divide\ x\ by\ y,\\\frac{x}{y} =\frac{a^2}{b^2}\\Hence,\\\frac{x}{y} =[\frac{a}{b}]^2\\As\ the\ new\ term\ \frac{x}{y}\ is\ the\ square\ of\ the\ quotient\ of\ a\ and\ b\ i.e\ \frac{a}{b}, the\ term\\ \frac{x}{y}\ too\ belongs\ toA](/tpl/images/1005/4376/eca95.png)
![Why\ not\ 1)\ and\ 2)?\\Because\ by\ just\ adding/subtracting\ x\ and\ y:\\x+y=a^2+b^2\\x-y=a^2-b^2\\It\ doesnt\ really\ prove\ that\ their\ sum\ say\ c\ is\ a\ perfect\ square.(except\\ in\ the\ case\ of\ Pythagorean\ Triplets [Ex. 3,4,5\ or\ 6,8,10]. Hence, As\\ not\ all\ numbers\ form\ integral\ pythagorean\ triplets,\\ (x+y)\ and\ (x-y)\ are\ not\ always\ perfect\ squares\ and\ hence,\\ dont\ belong\ to\ Set\ A.](/tpl/images/1005/4376/202c2.png)
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