Below we present three proofs by contradiction that utilize the Well-Ordering Principle. These proofs may be correct or they may have an error. For each proof identify if it is correct, or select the choice which best describes the error if it has one.

Theorem: V13 E ZH'L' 5 11:2. Proof: We proceed by contradiction. Suppose The statement is false, that is assume 3x E 1.x > 1'2. In this case, we'll let 0 be the set of all values of :c which make that statement true, i.e, C : {x '1: > 332 }. Since we assumed that there is at least one value of it such that a: > :62. it must be that C is not empty. By applying the Well-Ordering Principle, it must be that C has a smallest element, we'll call that E. First we observe that n 55 0, since 0 : 02. Next we observe that n must be positive, since otherwise n is negative and n2 is positive, meaning it must be that n n2, so alson 1 > n2 1. Since'n. 2 1 it must also be that n2 1 2 n2 2n + 1, and son 1 > n2 2n + 1. However, since n2 2n + 1 : (n 1)2,we have that n 7 1 > (n 7 IV, and so (n 7 1) E C, contradicting om definition of 'n being the smallest member of 0. Since our assumption that C was non-empty produced a contradiction, it must be that C was empty, and so we can conclude VI 6 Zn: S 1:2. I O This proof is correct. . O This proof incorrectly negates the theorem. . O This proof incorrectly uses the Well-Ordering Principle. I O This proof made an algebra error. WUOWW I O This proof did not arrive at a contradiction