Find real number x such that x+2 = x

Method 0 For any real number  x,  we know  x+2 > x. Therefore it  has no solutions. Method 1  （反证法）  Suppose  that it has a  solution, i.e. there exists an x,  such that       x = x+2 =>  0=2 Contradiction. so It has no solution. Meyjod 2  （反证法）

Suppose it has a solution, i.e.,  there exists an x,  such that     x  = x+2         （1）  =>  x^2 = (x+2)^2  =>  x=-1            代入（1） => -1=1 Contradiction. So  it has no solution.

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