{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Hout1 - Answer key No.1 to selected homework problems Math...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Answer key No.1 to selected homework problems: Math. 61 1.7 1. Suppose the formula for n and try to prove the formula for n + 1: 1(1!) + 2(2!) + Β·Β·Β· + n ( n !) + ( n + 1)(( n + 1)!) = ( n + 1)!- 1 + ( n + 1)(( n + 1)!) = ( n + 1)! + ( n + 1)(( n + 1)!)- 1 = ( n + 1)!(1 + n + 1)- 1 = ( n + 2)!- 1 . 9. First check the equality: 1 a 2- 1 = 1 2( a- 1)- 1 2( a + 1) . And use this formula for a = n + 2 in the induction process. 13. Write S n for the fraction of the LHS (the left hand side). Then S n +1 = S n Γ— (2 n + 1) 2( n + 1) ≀ ( n + 1)- 1 / 2 Γ— (2 n + 1) 2( n + 1) ≀ ( n + 2)- 1 / 2 . The last inequality: ( n + 1)- 1 / 2 Γ— (2 n +1) 2( n +1) ≀ ( n + 2)- 1 / 2 is equivalent to (2 n + 1) 2( n + 1) ≀ ( n + 1) 1 / 2 ( n + 2)- 1 / 2 , which in turn equivalent to: (2 n + 1) 2 4( n + 1) 2 = ( n + 1 2 ) 2 ( n + 1) 2 ≀ ( n + 1) ( n + 2) , which can be checked by computation. Indeed ( n + 1 2 ) 2 ( n + 1) 2 ≀ ( n + 1) ( n + 2) is equivalent to n 3 + 3 n 2 + 9 4 n + 1 2 = ( n + 1 2 ) 2 ( n + 2) ≀ ( n + 1) 3 = n 3 + 3 n 2 + 3 n + 1 and RHS-LHS = 3 4 n + 1 2 β‰₯ 0 if n β‰₯ 0. 14. Check the starting point of the induction by your self. Now suppose: 2 n + 1 ≀ 2 n and prove: 2( n + 1) ≀ 2 n +1 as follows: 2( n + 1) + 1 = 2 + 2 n + 1 ≀ 2 + 2 n ≀ 2 n + 2 n = 2 Γ— 2 n = 2 n +1 . 2.1 56. Let P ( X ) denote the set of all subsets in a set X . Then by Theorem 2.1.6, we know that |P ( X ) | = 2 n if | X | = n . The only non-proper subset in P ( X ) is X itself. Thus the answer is 2 n- 1 . 78. Suppose X, A, B, C to be subsets of a universe U . We prove the statement A C = B C β‡’ A = B . By definition, we have X C = ( X βˆͺ C )- ( X ∩ C ) = ( X- C ) βˆͺ ( C- X ) . Pick an arbitrary element a in X . Then there are two possibilities: (i) a ∈ C or (ii) a ∈ C. 1 2 If a ∈ C , then a ∈ X C . If a ∈ C , then a ∈ X C . Since (i) and (ii) never happen at the same time, we see that for a ∈ X (a) a ∈ C ⇔ a ∈ X C and a ∈ C ⇔ a ∈ X C. Pick an arbitrary element x outside X . Then (b) x ∈ X C ⇔ x ∈ C and x ∈ X C ⇔ x ∈ C. Suppose a ∈ A ∩ C . Then by (a) applied to X = A , a ∈ B C = A C . Then by (b) applied to X = B , a ∈ B ⇔ a ∈ C , a contradiction because a ∈ A ∩ C . Thus a ∈ B . This shows that A ∩ C βŠ† B ∩ C . Since the roles of A and B are interchangeable, we conclude A ∩ C βŠ‡ B ∩ C . This shows that A ∩ C = B ∩ C . Now suppose that a ∈ A- C . Then by (a) applied to X = A , we see a ∈ A C = B C . Since a ∈ C , we know a ∈ B- C because B C = ( B- C ) βˆͺ ( C- B ). Thus A- C βŠ† B- C . Interchanging the role of A and B , we get A- C βŠ‡ B- C . This shows A- C = B- C ....
View Full Document

{[ snackBarMessage ]}

Page1 / 6

Hout1 - Answer key No.1 to selected homework problems Math...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online