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Cоnsider the fоllоwing proposition: If x is а reаl number, then|x−1|>0.{"version":"1.1","mаth":"text{If $x$ is a real number, then} , |x-1| > 0."}Where does the following "proof" of the proposition go wrong?"Proof: Let x be any real number. Case 1: if x>1, then x-1>0, and so |x−1|=x−1>0.{"version":"1.1","math":"|x-1|=x-1 > 0."}Case 2: if x0{"version":"1.1","math":"|x-1| > 0"} in each case, for all real numbers x, |x−1|>0.◼{"version":"1.1","math":"|x-1| > 0. blacksquare"}