This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: SOLUTIONS FOR HOMEWORK 6 4.13. This is a bonus problem . Here we consider the difference x between a number n = αβγ and its opposite n ( o ) = γβα . When computing this difference, we subtract the smaller number from the larger one. For instance, when x = 239, we have n = x ( o ) − x = 932 − 239. Suppose x = 100 a + 10 b + c , with a negationslash = c . Then x ( o ) = 100 c + 10 b + a . Without loss of generality, we may assume that a > c (otherwise, just relabel x and x ( o ) ). Then n = x − x ( o ) = 99( a − c ) = 100( a − c − 1) + 10 · 9 + (10 − a + c ) . Note that the right hand side above contains the correct decimal expansion of n , as a − c − 1, 9, and 10 − a + c all lie in { , 1 ,... , 9 } . Then n ( o ) = 100(10 − a + c ) + 10 · 9 + ( a − c − 1), and therefore, n + n ( o ) = 100 ( ( a − c − 1) + (10 − a + c ) ) + 2 · 10 · 9 + ( (10 − a + c ) + ( a − c − 1) ) = 100 · 9 + 2 · 10 · 9 + 9 = 1089 . 4.24. NO , h need not be a surjection. As an example, suppose f and g are identity maps – that is, f ( x ) = g ( x ) = x for any x ∈ Z . Clearly, f and g are bijections. However, h ( x ) = x 2 . Thus, h is not a surjection. In fact, it is neither injective (is not a surjection....
View
Full Document
 Fall '09
 ?
 Math, Inverse function, Universal quantification, 2m

Click to edit the document details