{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol7 - MATH 135 SOLUTIONS TO HOMEWORK SET#3 1(page 133 2...

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

MATH 135: SOLUTIONS TO HOMEWORK SET #3 1. (page 133, # 2) Show that in Fig. 32 we have J ( m, n ) = [1 + 2 + · · · + ( m + n )] + m = 1 2 [( m + n ) 2 + 3 m + n ] Proof. We demonstrate that 0 = J (0 , 0) and the J ( m, n ) + 1 = J ( m + 1 , n - 1) if n > 0 and that J ( m, n ) + 1 = J (0 , m + 1) if n = 0. Certainly, J (0 , 0) = 1 2 [(0 + 0) 2 + 3(0) + 0] = 0. We compute, J ( m, n ) + 1 = 1 2 [( m + n ) 2 + 3 m + n ] + 1 = 1 2 [( m + n ) 2 + 3 m + n + 2] = 1 2 [([ m + 1] + [ n - 1]) 2 + 3( m + 1) + ( n - 1)] = J ( m + 1 , n - 1) So, if n > 0, this gives the desired result. We compute, J ( m, 0) + 1 = 1 2 [( m + 0) 2 + 3 m + 0] + 1 = 1 2 [ m 2 + 3 m + 2] = 1 2 [( m + 1) 2 + ( m + 1)] = 1 2 [(0 + [ m + 1]) 2 + 3(0) + ( m + 1)] = J (0 , m + 1) 2. (page 133, # 3) Find a one-to-one correspondence between the open unit interval (0 , 1) and R that takes rationals to rationals and irrationals to irrationals. f ( x ) = - 1 4 x if 0 < x 1 4 1 - 4 x if 1 4 < x < 1 2 0 if x = 1 2 4 x - 2 if 1 2 < x 3 4 1 4 x - 3 if 3 4 < x < 1 3. (page 138, # 6) Let κ be a nonzero cardinal. Show that there is no set K for which | X | = κ X K . 1

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

View Full Document
2 MATH 135: SOLUTIONS TO HOMEWORK SET #3 Proof. Suppose that such a K exists. Let A be some set with | A | = κ . Let a A be any element (such exists as κ = 0). For each set x we see that κ = | A | = | ( A \{ a } ) ∪{ x }| via the bijection f : A ( A \{ a } ) ∪{ x } given by f ( y ) = y if y = a
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

sol7 - MATH 135 SOLUTIONS TO HOMEWORK SET#3 1(page 133 2...

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

View Full Document
Ask a homework question - tutors are online