Proofs Involving Quantifiers. Proof Let B be an arbitrary member Proof Somehow show that there is a value

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1 Proofs Involving Quantifiers For a given universe Y : Theorem ÐaBÑ T ÐBÑ Theorem ÐbBÑ T ÐBÑ Proof Let B be an arbitrary member Proof Somehow show that there is a value of Y. Call it B œ +, say Þ ÐYou can continue of Bß say B œ +, from Y that makes TÐ+Ñ to call it B if you like; switching to a name true. ( You can announce + and verify like + just makes the psychological that it works, construct an + that must point that its no longer a variable but work, or argue that you theoretical a particular object chosen from Y. That reasons there must be such a value B œ + + is arbitrary means just that you know even though the proof doesn't explicitly nothing special about + beyond the fact say how to find the value B œ +Þ) that it's a member of YÞ) Prove that TÐ+Ñis true. Sometimes these theorems can be proved by contradiction Theorem ÐaBÑ T ÐBÑ Theorem ÐbBÑ T ÐBÑ Proof Assume µ ÐaBÑ T ÐBÑ, or Proof Assume µ ÐbBÑ T ÐBÑ equivalently, that ÐbBÑ µ T ÐBÑÞ or, equivalently, that ÐaBÑ µ T ÐBÑÞ Choose such an BÀ call it Bœ+Þ From this, argue somehow to a Then we know that TÐ+Ñis false for this contradiction. value +. From this, argue somehow to a contradiction.

2 Example Theorem Ða Ñ Ð is even Ñ Ê Ð is evenñ Proof Assume is a natural number. For this, prove that Ð is even Ñ Ê Ð is even Ñ.... We have done proofs like this before. Earlier, the theorem was worded in such a way as to avoid using the quantifier. For example, the following wording gives an equivalent theorem, and the proof starts out the same way. Theorem Assume is a natural number. Prove that if is even, then is even. Proof Assume is a natural number. For this, prove that Ð is even Ñ Ê Ð is even Ñ.... Example Theorem ÐaB ÑÐÐ! B Ñ ÊÐtan B cot B ÑÑ Proof ( by contradiction) Assume µ ÐaB ÑÐÐ! B Ñ ÊÐtan B cot B ÑÑ, which is equivalent to ÐbBÑ µ Ð Ð! B Ñ Ê Ð tan B cot B ÑÑ which is equivalent ( why? ) to ÐbBÑ Ð! B Ñ Ð tan B cot B Ÿ ÑÑ Pick such a real number B and ( optional) call it + ). tan + Since! +, tan +! and cot +œ!þ Therefore we have! tan + cot +Ÿ., Squaring both sides gives! tan + cot +Ÿ, so that tan + cot +Ÿ. But this is a contradiction ( a sum of squares can't be negative). So assuming the theorem is false leads to a contradiction, and therefore the theorem is true. ñ

3 Example Theorem Ðb Ñ Ð can be written as a sum of cubes in two different waysñ (We could write this more formally as: for the universe À $ $ $ $ ÐbÑ Ð Ðb:ÑÐb;ÑÐb<ÑÐb=ÑÐ Ð œ : ; œ < = : Á < : Á =ÑÑ Proof Let $ $ $ $ œ (*Þ This proves the theorem because (* œ! * œ Þ ñ This theorem comes from an anecdote about the English mathematician G.H. Hardy and the Indian mathematician Srinivasa Ramanujan set around the time of World War I. Ramanujan was an untaught natural mathematical genius who had sent some of his work, unsolicited, to Hardy at Cambridge. Hardy was sufficiently impressed to bring him to England where, after a while, he became ill. Visiting him in the hospital, Hardy remarked that the number of the cab in which he had arrive was 729, not a very interesting number. To which Ramanujan immediately responded, No, no Hardy. It is a very interesting number. It is the smallest number that can be expressed as a sum of cubes in two different ways. You can read a bit about Hardy and Ramanujan in Hardy's little book, A Mathematician's Apology. Also, there was recently published an historical novel about Ramanujan called The Indian Clerk by David Leavitt. Also see the course web syllabus for a little more about Ramanujan.

4 The Intermediate Value Theorem In Calculus I, you should have seen the Intermediate Value Theorem. It was probably motivated and presented as intuitively true (which it is). It's usually not proven until a more advanced course (Math 4, for example) because the proof depends on a careful, proof-oriented definition for continuity and on a fairly subtle property of the real number system. Intermediate Value Theorem Suppose that 0 is a continuous function whose domain is the closed interval Ò+ß,ÓÞ If 0Ð+Ñ and 0Ð,Ñ have opposite signs, then there exists a number < in the open interval Ð+ß,Ñ where 0Ð<Ñ œ!þ More formally: if Y is the universe of all continuous functions defined on a given closed interval Ò+ß,Óß then the Intermediate Value Theorem says: Ða0Ñ Ò Ð0Ð+Ñ 0Ð,Ñ!Ñ Ê Ðb<Ñ ÐÐ+ <,Ñ Ð0Ð<Ñ œ!ññ Ó Note that the theorem does not use the quantifier Ðb x <Ñ because there might be more than one < that works.

5 Theorem (b! BÑT ÐBÑ Proof Such a proof usually has two parts: i) Existence Prove (b! BÑT ÐBÑ ii) Uniqueness Prove ÐaBÑÐaÐCÑ Ð ÐT ÐBÑ T ÐCÑÑ Ê ÐB œ CÑ Ñ & Example Prove that the equation B B œ! has a unique root between and!. More formally, with Yœthe set of real numbers, À & Prove that Ðb! <ÑÒÐ < Ñ Ð< < œ!ñ The figure on the next page is not used in the proof; but having it there helps the reader to understand the argument. The proof assumes that the reader knows certain facts from Calculus I. & Proof i) Existence Let 0ÐBÑ œ B B Þ The function 0 is continuous on the interval Ò ß Ó (because 0 is a polynomial) and 0Ð Ñ œ ( and 0Ð Ñ œ Þ Since 0Ð Ñ and 0Ð Ñ have opposite signs, the Intermediate Value Theorem says that there is a number < is the interval Ð ß Ñfor which 0Ð<Ñœ!, that is, for which < is a root of the equation & B B œ!. w % ii) Uniqueness 0 ÐBÑœ&B Þ Since B, we know that &B &ßso w 0 ÐBÑ! on the interval Ð ß ÑÞ Therefore 0 is an increasing function on theis interval so its graph can cross the B -axis at most once. Therefore there is at most one < in the interval for which 0Ð<Ñ œ!þ By i) and ii) B B œ! has a unique root in the interval Ð ß ÑÞ ñ Note that part ii) does not, by itself, show that there exists an < that works, only that there is at most one such <. The job of part i) is to show that there is at least one such <. Taken together, i) and ii) show that there is exactly one < that works. The preceding proof is an example of what mathematicians sometimes call a pure existence proof À this refers to a proof of ÐbBÑT ÐBÑ or Ðb! BÑT ÐBÑ where it is argued that some value Bœ<must exist that makes TÐ<Ñ true, but the proof does not actually give the value of < or show an exact method for finding it.

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7 Example (Multiple quantifiers) Let the universe Y œ Theorem Ða% ÑÐÐ%!ÑÊÐÐb5 ÑÐa Ñ Ð 5ÑÊÐ % Ñ Ñ Proof Let % be an arbitrary real number and assume %!. We now need to show that Ð%!Ñ Ê Ð Ðb5 Ñ Ða Ñ Ð 5Ñ Ê Ð % Ñ is true We now need to show that Ðb5 Ñ Ða Ñ Ð 5Ñ Ê Ð % Ñ is true We do some scratchwork: We want to force %. This inequality will be true if or Þ We can't use for 5 because 5 needs to be an % % % integer. But any 5 % will work: if 5 %, then % so % Þ Just to be as efficient as possible, we'll pick the smallest integer k %, which can be denoted using the ceiling function À 5 œ Ü % Ý Let 5 œü % ÝÞ For this 5, we need to show that Ða Ñ Ð 5ÑÊÐ % Ñ is true Pick an arbitrary natural number. For this n (and the k already chosen ) we need to show that Ð 5Ñ Ê Ð Ñ is true % % % % Assume 5. Then, so n and therefore Þ ñ (OVER Ä Ä )

8 Notes: ) All the italicized material is commentary. The whole proof, without commentary, is very short: Proof Consider an arbitrary %!, and let 5œÜ % Ý. If 5, then % and therefore % Þ ñ 2) In the original statement of the theorem: % starts as an arbitrary real number, but then the following conditional if %!, then... immediately forces you to assume, in addition, that % is positive. Because of this observation, a mathematician would probably build-in that %! using notation like this: Instead of Ða% ÑÐÐ%!ÑÊÐÐb5 ÑÐa Ñ Ð 5ÑÊÐ % Ñ Ñ write Ða%!Ñ Ð Ðb5 Ñ Ða Ñ Ð 5Ñ Ê Ð % Ñ Ñ In the second version, the informal quantifier notation Ða%!Ñ *** is just a shorthand for Ða% Ñ Ð%!Ñ Ê *** 3) Suppose we have a sequence + œ ß + œ ß ÞÞÞ ß + œ ß ÞÞÞ In the more informal language of Calculus II Ä Ða%!Ñ Ð Ðb5 Ñ Ða Ñ Ð 5Ñ Ê Ð % Ñ Ñ says that lim œ!þ ÐAgreed? Ñ