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8 October, 16:06

Find the error in the "proof" of the following "theorem." "Theorem": Let R be a relation on a set A that is symmetric and transitive. Then R is reflexive. "Proof ": Let a ∈ A. Take an element b ∈ A such that (a, b) ∈ R. Because R is symmetric, we also have (b, a) ∈R. Now using the transitive property, we can conclude that (a, a) ∈ R because (a, b) ∈ R and (b, a) ∈ R.

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  1. 8 October, 17:19
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    The proof is wrong. The part that says 'take an element b ∈ A such that (a, b) ∈ R' is assuming that such an element b exists, and that may not be the case. We can give the 'theorem' validation by adding the hypothesis ' each element of A is related to any other element of A', otherwise the 'theorem' is false. An example for that is the following relation on the natural numbers:

    'a R b if both numbers are even'

    This relation if crearly symmetric and transitive, but it is not reflexive, an odd number does not relate with itself. In fact, odd numbers dont relate with any number.
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