}\) Then, Assume that \(a\) or \(b\) is even - say it is \(a\) (the case where \(b\) is even will be identical). In fact, we can quickly see that \(n = 41\) will give \(41^2\) which is certainly not prime. Since then, we have used some common terminology in mathematics without much explanation. The argument is valid so the conclusion must be true if the premises are true. }\) (Here \(x|y\text{,}\) read â\(x\) divides \(y\)â means that \(y\) is a multiple of \(x\text{,}\) i.e., that \(x\) will divide into \(y\) without remainder). Thus \(a^2\) is even, and as such \(a\) is even. Prove your answers. What if there were? Consider the set of numbers of friends that everyone has. }\) We want this to work for all \(x\text{. Let \(n\) be an arbitrary integer. Prove the statement: For all integers \(n\text{,}\) if \(5n\) is odd, then \(n\) is odd. Proof by contradiction. It will go something like this: Let \(a\text{,}\) \(b\text{,}\) and \(c\) be arbitrary integers. So if the premises are true, then the conclusion must be true. Prove your answer. Suppose you made an even amount of postage. This gives \(-1 = 1\text{. Then \(\sqrt 2\) is equal to a fraction \(\frac{a}{b}\text{. Prove: There are no integers \(x\) and \(y\) such that \(x^2 = 4y + 2\text{. •Proof : Assume that m and n are both squares. \newcommand{\N}{\mathbb N} What can we assume? Both of these are odd, but \(1+3 = 4\) is not odd. Why not? If you draw some number of cards at random you might or might not have a pair (two cards with the same value) or three cards all of the same suit. }\) It must be false as well, which makes \(P\) true! \newcommand{\Imp}{\Rightarrow} A good place to start might be to study a classic. Try something else: write the contrapositive of the statement. This Lecture Now we have learnt the basics in logic. Deduce that \(c^2\) is even, and therefore a multiple of 4 (why? Consider the statement âfor all integers \(a\) and \(b\text{,}\) if \(a + b\) is even, then \(a\) and \(b\) are evenâ. + 1\text{. In fact, we can prove this conjecture is false by proving its negation: âThere is a positive integer \(n\) such that \(n^2 - n + 41\) is not prime.â Since this is an existential statement, it suffices to show that there does indeed exist such a number. Simplify using the rules from the previous sections: As the negation passed by the quantifiers, they changed from \(\forall\) to \(\exists\text{. But the examples do not belong in the proof. }\) Then, Suppose that \(ab\) is even but \(a\) and \(b\) are both odd. \newcommand{\B}{\mathbf B} Suppose you are at a party with 19 of your closest friends (so including you, there are 20 people there). }\) Carefully explain using what we know about logic. Then \(n = 2k\) for some integer \(k\text{. \newcommand{\amp}{&} Additionally, it is equally important to understand the overall structure of the proof. However, if you draw enough cards, you will be guaranteed to have these. Proof: Suppose n is any [particular but arbitrarily chosen] even integer. Prove your answer. \neg \forall a \forall b ((O(a) \vee O(b)) \imp O(a+b))\text{.} + 1 = (p \cdot (p-1) \cdot \cdots 3\cdot 2 \cdot 1) + 1\text{.} \end{equation*}, \begin{equation*} \newcommand{\vr}[1]{\vtx{right}{#1}} }\) In other words, \(b\) is a multiple of \(a\) and \(c\) is a multiple of \(b\text{. }\) There are plenty of examples of statements which are hard to prove directly, but whose contrapositive can easily be proved directly. (and in math, we are nearly always considering statements about infinite sized sets). The idea is to prove that \(P\) is true by proving that \(Q \imp P\) and \(\neg Q \imp P\) for some statement \(Q\text{. By definition of even number, we have. That's easy: 1 and 3. \amp = 2(4k^3 + 3k^2 + 2k)\text{,} Your friend has given you his list of 115 best Doctor Who episodes (in order of greatness). }\) Here is the proof: I claim that \(1 = 3\text{. Let \(n\) be an arbitrary integer. Solution. Let \(ab\) be an even number, say \(ab=2n\text{,}\) and \(a\) be an odd number, say \(a=2k+1\text{.}\). Prove the contrapositive by cases. Then at most there will be \(n\) pigeons. \amp = 8k^3 + 6k^2 + 6k + 1 - 2k - 1\\ In fact, we could generalize this. Consider the following statement: for every prime number \(p\text{,}\) either \(p = 2\) or \(p\) is odd. Math 232 - Discrete Math Notes 2.1 Direct Proofs and Counterexamples Axiom: Proposition that is assumed to be true. A proof in mathematics is a convincing argument that some mathematical statement is true. Proof by contrapositive. At least, you could try to replicate the style of proof used by the pigeonhole principle. For all integers \(n\text{,}\) if \(n\) is a multiple of 3, then \(n\) can be written as the sum of consecutive integers. Case 2: \(n\) is odd. }\)], Therefore there are infinitely many primes. You should always be able to identify how it follows from earlier statements. [by line 6, \(N\) is divisible by a prime larger than \(p\text{. Is the contrapositive of the original statement true or false? \newcommand{\vtx}[2]{node[fill,circle,inner sep=0pt, minimum size=4pt,label=#1:#2]{}} Methods of Proof Lecture 3: Sep 9 2. Then \(a = 2k\) and \(b = 2j\) for some integers \(k\) and \(j\text{. }\) This in turn gives \(2k^2 = (2y + 1)\text{. ], Therefore \(p\) is not the largest prime. }\) But \(2k^2\) is even, and \(2y + 1\) is odd, so these cannot be equal. You do not need to provide details for the proofs (since you do not know what solitary means). As before, if there are variables and quantifiers, we set them to be arbitrary elements of our domain. b^2 = 2k^2\text{.} Suppose you roll all 40 dice. It is hard to know where to start this, because we don't know much of anything about \(n\text{. \end{equation*}, \begin{equation*} \end{align*}, \begin{align*} }\) Dot dot dot. This means that â¦. A proof by contradiction would be reasonable here, because then you get to assume that both \(a\) and \(b\) are odd. True. Here are three examples of proofs by contradiction: Suppose not. \newcommand{\R}{\mathbb R} Your friend's proof a proof, but of what? Is the original statement true or false? Example 1.5.6: a theorem If x2 is odd, then so is x. \newcommand{\card}[1]{\left| #1 \right|}
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