Question 1107353
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Proof By Contradiction:


Let p and q be two integers where q is nonzero. Any rational number is of the form {{{p/q}}} (ratio of two integers)


If {{{2+sqrt(3)}}} was rational, then {{{2+sqrt(3)=p/q}}} for a certain (p,q) value. 


Multiplying both sides by q leads us to {{{q(2+sqrt(3)) = p}}} which becomes {{{2q+q*sqrt(3) = p}}}


The right side of {{{2q+q*sqrt(3) = p}}} is an integer (p is an integer)


The left side of {{{2q+q*sqrt(3) = p}}} should be an integer as well; however, it is not an integer. The expression {{{2q}}} is surely an integer because 2 times any integer is an integer, but the expression {{{q*sqrt(3)}}} is NOT an integer. It is only an integer if q was of the form {{{k*sqrt(3)}}}, but it is not. Again, q is an integer with no irrational parts. 


In short, {{{2q+q*sqrt(3)}}} on the left side is NOT an integer while {{{p}}} on the right side is an integer.


So this is where the contradiction lies. This contradiction makes the claim {{{2+sqrt(3)=p/q}}} to be false, therefore {{{2+sqrt(3)}}} is not rational. The only thing left is that {{{2+sqrt(3)}}} must be irrational. 
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