109_sp2011_lfsample

109_sp2011_lfsample - are both false. By inspection of the...

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MATH 109 - Sample Long-Form Solution Question. Do there exist propositions p and q such that both “ p implies q ”anditsconverse are true? Do there exist propositions p and q such that both “ p implies q ”anditsconverse are false? Before answering the question, we recall some key deFnitions. The proposition “ p implies q ”, abbreviated as p = q , is deFned by the following truth-table (see p. 11 of Eccles). p q p = q T T T T ± ± ± T T ± ± T The converse of the implication p = q is deFned as q = p (see p. 14 of Eccles). ±irst, we consider whether there can be propositions p and q such that both p = q and q = p are true. If p and q are both true propositions then both p = q and q = p are true by the Frst line of the truth-table. Similarly, if p and q are both false then both p = q and q = p are true by the last line of the truth-table. So, one example of propositions p and q such that both “ p implies q ” and its converse are true is p :‘ 0 < 1’ and q :‘ 2 6 =3’ . Another example is p :‘ π<e ’a n d q :‘ 0=1
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Unformatted text preview: are both false. By inspection of the truth-table, this seems less likely to be possible since there is only one combination of truth values of p and q which makes p = q false. We now prove that it is, in fact, impossible to Fnd such propositions. Proof. By contradiction, suppose that there were propositions p and q such that p = q is false and q = p is also false. Since p = q is false, it must be the case that p is true and q is false (by the third line of the truth-table). Similarly, since q = p is false, q is true and p is false. Thus, p must be both true and false, which is impossible. We arrived at contradiction when we assumed that there were such propositions so we conclude that there are no propositions where the implication and its converse are both false....
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