Question

Math 2300 - Homework 5 Due: Thursday, 10/10/2019 at 12:30 pm Notes: This homework consists of 6 problems. Write your answers in full sentences. Before you start a problem set up the goal and explain how will you proceed and why. Write all details thinking that the grader knows nothing about Mathematics. Pictures and graphs are great ideas to start a proof but will not be sufficient without written explanation. If you are referring to some result or theorem, please cite. Use a discussion and a formal proof of one of mathematical proofs we have learned in Chapter 3 to show each of the followings. These are: Trivial Proof, Vacuous Proof, Di- rect Proof, Proof by Contrapositive, and Proof by Cases. 1. Let r ER. Prove that if r-5r - 1 > 0, then (x - 1)(x-3) > -2. 2. Let r e Z. Prove that if r is even if and only if is even. 3. Prove that if n e Z, then n? - 3n +9 is odd 4. Let r e Z. Prove that 3.r + 1 is even if and only if 5x – 2 is odd. 5. Let a, b e Z. Prove that if ab and a + b are even, then both a and b are even. 6. Give a direct proof of the following: Let n e Z. If lin -5 is odd, then n is even.

Answer 1

Even integer can be written as 2k , kt Z. integer = 2k + [arak 1] APCO- - KE Z Page geren x-5x-120 Now, we will prove (x-1) (x-3) >-2 for - every XER and so for XEIR satisfying ( New, (x-1) (-3) = 1² - 4x +3 =(x-27 - > o-1 Luare of an (x-1) (x-3) -1 real numbers (x-1) (x-3) ? 2. for every KER [: 17-2] Hence, the desired result holds. 2 To prove x is even if and only if a even. Let us take x x = 2m is even. some MEZ : x = (am) = 8 m F = 2(4 m²) = ak [k= 4 m² E2 Thus, x is even whenever x is even Conversely, let & is even, we have to show We will prove this by contradiction-

Page So, let us suppose x is odd. So, K= 2kt for some k EZ 23 - (2k + 2 z 8 k3 + 1+ 3 (sk) [ak + 1) 28 k + 1 + 12 ke +6k 22k [4 k + 6k+ 3] + L = 2m+1 where m= k (46² +61 +3) EZ (lez) Thus, N3 is odd , which is a contradiction Thus, our assumption that x is odd, was wrong So, a is even. Hence, the result NEZ To show = n -3nta és odd Case ( ú n is even integer nz 2k for some k EZ i n2-3n+9 = (ak) - 3 labe) +9 2 46²-6k+ 8 + 1 = 2 26 - 3k + 4 + 1 za mti where me 2k - 3k + 4 € Z like Z) Thus, n - 30th is odd.

Date Page Case (ii) n is odd integer - n = set KEZ n-3n+ 9 = (2k + 1)²-3(2k + 1) + 4 = 4k + 1 + 4k - k-3+q = 4k - 2k + 7 = 4k?- 2k + 6 + 22 [ak - k + 3] + 1 - where pe 2k²-k+ 3 EZ 1 = k c 2] Thus, n2-3nt is odd. Hence for every nez, n-3nth is odd. To prove: 3xt is even if and only if 1 [ By contrapositive ] Assume So, that 5X-2 54-2= 2k es even. , ha ez 50 = 2 k + 2. 32 = 5n - 2x = (ak+2) -an. = 2(k+ 1-x ] [From (1) ] t 3x+1 = 2 [k+1-x] + 1 ie, 3x+1 is odd. (=k+1- 8E7) which is a contradiction. Hence, the result.

Date Page ab to As and atb are prove = af b ab is even even. both are even. ع ه ه ه ا ن م a be even het Now, if b is odd Then , atb will be odd. [ even & odd = odd] Thes, But atb is given even. So, b is also evene Thus, both a and 6 are even. NEZ To prove: If he lin-5 is odd, then n is even. 1 By Contrapositive / Let n be odd. Then, n= akti, kEZ 11n-5 = 1182k+1)-5 = 22k + 11-5 = 22k + 6 22 [uk+31 = 2m me lik+3 € 2 ie, il n is odd, then 110-5 is even Which is a contradiction, as lin-5 is odd. Thus, our assumption that n is add was wrong.