{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

This preview shows pages 1–2. Sign up to view the full content.

4.12 (a) False. One counterexample is f ( x ) = e - x . (b) False. Take f ( x ) = 0. (Any constant function will work.) (c) False. Consider f ( x ) = x if 0 6 = x 6 = 1 , 1 if x = 0 , 0 if x = 1 . (d) True. If | f ( x ) | ≤ M for every x R , then M + 1 is not in the image of f . (e) False. Take f ( x ) = x 2 . 4.13 We assume without loss of generality that n > rev( n ) so that a > c . Then x = n - rev( n ) = (100 a + 10 b + c ) - (100 c + 10 b + a ) = (100 a + 10( b - 1) + (10 + c )) - (100 c + 10 b + a ) = (100( a - 1) + 10(10 + b - 1) + (10 + c )) - (100 c + 10 b + a ) = 100(( a - 1) - c ) + 10((9 + b ) - b ) + ((10 + c ) - a ) = 100(( a - c ) - 1) + 10 · 9 + (10 - ( a - c )) . (Note that these algebraic manipulations are motivated by the notion of “borrowing” as in the standard algorithm for subtraction from grade school.) Since 0 c < a 9, we have 1 a - c 9, which implies that 0 ( a - c ) - 1 8 and 1 10 - ( a - c ) 9. Hence the digits of x are ( a - c ) - 1 , 9, and 10 - ( a - c ). This yields x + rev( x ) = [100(( a - c ) - 1) + 10 · 9 + (10 - ( a - c ))] + [100(10 - ( a - c )) + 10 · 9 + (( a - c ) - 1)] = 100 · 9 + 2 · 10 · 9 + 9 = 1089 . 4.22 We first show f is surjective. We have f (1 / 2) = 0, so it remains to show that if z R , z 6 = 0, then there is some y (0 , 1) such that f ( x ) = z . We claim that y = ( z - 1 + z 2 + 1) / 2 z will work.

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

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

{[ snackBarMessage ]}

### Page1 / 3

HW9solns - 4.12(a False One counterexample is f(x = e-x(b...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online