{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sam1sol

# sam1sol - k 3 = k 1 2 k 3 the equation is also true...

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

Solution of CSE20 Exercise 1 January 27, 2010 1. (1000110100) 2 2. 53 3. 3 5 × 4 4 , or 0 (3 5 × 4 4 - 1) 4. x y b in b out d 0 0 0 0 0 0 0 1 1 1 0 1 0 1 1 0 1 1 1 0 1 0 0 0 1 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 5. id b2 b1 b0 0 0 0 0 1 0 0 1 2 0 1 1 3 0 1 0 4 1 1 0 5 1 1 1 6 1 0 1 7 1 0 0 6. Signed magnitude & one’s complement: - 127 127 Two’s complement: - 128 127 7. (01000010) 2 1

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

View Full Document
8. x = 12 = 1100 2 y = 14 = 1110 2 = 001110 2 By 6-bit one’s complement, x - y = x +( - y ) = 001100 2 +110001 2 = 111101 2 9. Define eight’s complement so that - x is represented as n-digit 8 n - x . In 5-digit system, - y = 8 5 - 15 8 = 77763 8 , x - y = x + ( - y ) = 00011 8 + 77763 8 = 77774 8 (which equals - 4) 10. (a) Proof without induction: 1 1 × 3 + 1 3 × 5 + · · · + 1 (2 n - 1)(2 n +1) = 1 2 ( 1 1 - 1 3 ) + 1 2 ( 1 3 - 1 5 ) + · · · + 1 2 ( 1 2 n - 1 - 1 2 n +1 ) = 1 2 ( 1 1 ) + 1 2 ( - 1 3 + 1 3 ) + · · · + 1 2 ( - 1 2 n - 1 + 1 2 n - 1 ) + 1 2 ( - 1 2 n +1 ) = 1 2 (1 - 1 2 n +1 ) = n 2 n +1 (b) Proof with induction: When n = 1, 1 1 × 3 = 1 2+1 , the equation is true. Assume for n = k , 1 1 × 3 + 1 3 × 5 + · · · + 1 (2 k - 1)(2 k +1) = k 2 k +1 is true, so when n = k + 1, using our assumption on n = k , 1 1 × 3 + 1 3 × 5 + · · · + 1 (2 k - 1)(2 k +1) + 1 (2 k +1)(2 k +3) = k 2 k +1 + 1 (2 k +1)(2 k +3) = 2 k 2 +3 k +1 (2 k +1)(2 k +3) =
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: k +3) = k +1 2 k +3 the equation is also true. Induction complete. 11. Proof: Let x = q x d + r x , y = q y d + r y , ( x + y )% d = ( q x d + r x + q y d + r y )% d = ( r x + r y )% d = ( x % d + y % d )% d 12. Proof: Let x = q x d + r x , y = q y d + r y , ( x-y )% d = [( q x d + r x )-( q y d + r y )]% d = ( r x-r y )% d = ( x % d-y % d )% d 13. Proof: Let x = q x d + r x , y = q y d + r y , ( x y )% d = [( q x d + r x ) ( q y d + r y )]% d = ( q x q y d 2 + q x r y d + q y r x d + r x r y )% d = ( r x r y )% d = ( x % d y % d )% d 2...
View Full Document

{[ snackBarMessage ]}