ma8_wk1_friday_notes_2010

# ma8_wk1_friday_notes_2010 - MATH 8 SECTION 1 WEEK 1...

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Unformatted text preview: MATH 8, SECTION 1, WEEK 1 - RECITATION NOTES TA: PADRAIC BARTLETT Abstract. These are the notes from Friday, Oct. 1st’s lecture. In this talk, we study one last example of induction, discuss proofs by contrapositive, and examine the field axioms of R (specifically, we look for other objects that satisfy these rules, and ask when such things can exist.) 1. Random Question Question 1.1. Can you cover the plane R 2 with closed disks 1 of positive radius, so that no two disks intersect at more than one point? 2. Proofs by Induction: One More Example In our last lecture, we discussed proofs by induction. Just to make sure that the concept of an inductive proof is solid, we offer one more example here: Claim 2.1. (Bernoulli) (1 + a ) n ≥ 1 + na , for any a ∈ R ,n ∈ N . Proof. We proceed by induction. Base case: n = 1. In this case, our claim is trivial, as (1+ a ) 1 = 1+ a = 1+1 · a , for any a . Inductive step: Assuming that (1+ a ) k ≥ 1+ ka , for k between 1 and n , we want to prove that (1 + a ) n +1 ≥ 1 + ( n + 1) a. How do we do this? Well: if we factor out a copy of (1 + a ) from the left hand side, we have that (1 + a ) n +1 = (1 + a ) · (1 + a ) n . Why would we do this? Well, the nice thing about our factoring is that it breaks our left hand side into two parts, one of which is really simple [i.e. (1+ a )] and one of which we can apply our inductive hypothesis to [i.e. (1+ a ) n ]. Having done this, we can then apply our inductive hypothesis, which gives us (1 + a ) n +1 = (1 + a ) · (1 + a ) n ≥ (1 + a )(1 + na ) = 1 + a + na + na 2 = 1 + ( n + 1) a + na 2 . 1 A closed disk is just a filled-in circle in R 2 ; explicitly, it’s a collection of points { ( x,y ) : ( x- a ) 2 + ( y- b ) 2 ≤ r } for some a,b,r ∈ R . 1 2 TA: PADRAIC BARTLETT Because na 2 is positive, we can discard this term (as it is only making the right-hand side larger,) to get (1 + a ) n +1 ≥ 1 + ( n + 1) a as claimed. 3. Proofs by Contrapositive The last proof method we have to introduce is, conveniently, one of the easiest – the concept of “proof by contrapositive.” The structure of a proof by contrapositive is remarkably simple: (1) Suppose we want to prove some statement of the form P ⇒ Q ....
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ma8_wk1_friday_notes_2010 - MATH 8 SECTION 1 WEEK 1...

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