# Download Discrete math exercise Indonesian, Kenneth Rosen

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```Doanda Dresta Rahma
5025201049
Matematika Diskrit C
Exercise 2.3: 27
a) Prove that a strictly decreasing function from R to itself is one-to-one.
b) Give an example of a decreasing function from R to itself that is not one-to-one.
a) Proof:
ð strictly decreasing mengimplikasikan : jika ðĨ < ðĶ, maka ð(ðĨ) > ð(ðĶ).
Definisi one-to-one, untuk tiap pasangan x dan y, jika ð(ðĨ) = ð(ðĶ) maka ðĨ = ðĶ
Asumsikan ð(ð) = ð(ð).
Jika ð < ð, maka dari definisi strictly decreasing didapatkan ð(ð) > ð(ð). Jadi tidak
mungkin ð < ð apabila ð(ð) = ð(ð).
Jika ð < ð, maka dari definisi strictly decreasing didapatkan ð(ð) > ð(ð). Jadi tidak
mungkin ð < ð apabila ð(ð) = ð(ð).
Karena ð < ð tidak benar dan ð < ð tidak benar, maka ð = ð.
Dari definisi one-to-one, telah dibuktikan ð one-to-one.
b) ð(ðĨ) = ââðĨâ adalah decreasing function, tetapi tidak one-to-one karena
memungkinkan beberapa nilai ðĨ yang terpetakan ke ð(ðĨ) yang sama. (âðĨâ = ceiling
function)
ð(1) = ââ1â = â1
ð(0.5) = ââ0.5â = â1
Exercise 8.5: 7
There are 2504 computer science students at a school. Of these, 1876 have taken a
course in Java, 999 have taken a course in Linux, and 345 have taken a course in C.
Further, 876 have taken courses in both Java and Linux, 231 have taken courses in both
Linux and C, and 290 have taken courses in both Java and C. If 189 of these students
have taken courses in Linux, Java, and C, how many of these 2504 students have not
taken a course in any of these three programming languages?
Misalkan:
J : Murid yang mengambil kursus Java.
L : Murid yang mengambil kursus Linux.
C : Murid yang mengambil kursus C.
Kita tahu bahwa |J| = 1876, |L| = 999, |C| = 345, |J âĐ L| = 876, |L âĐ C| = 231, |J âĐ C| =
290, |J âĐ L âĐ C| = 189. Dengan rumus inklusi-eksklusi:.
Doanda Dresta Rahma
5025201049
Matematika Diskrit C
|J âŠ L âŠ C| = |J| + |L| + |C| â |J âĐ L| â |L âĐ C| â |J âĐ C| + |J âĐ L âĐ C|
|J âŠ L âŠ C| = 1876 + 999 + 345 â 876 â 231 â 290 + 189
|J âŠ L âŠ C| = 2012
Murid yang tidak mengambil kursus dari ketiga programming language adalah:
2504 â |J âŠ L âŠ C| = 2504 â 2012 = 492
Exercise 8.5: 7
How many students are enrolled in a course either in calculus, discrete mathematics,
data structures, or programming languages at a school if there are 507, 292, 312, and
344 students in these courses, respectively; 14 in both calculus and data structures; 213
in both calculus and programming languages; 211 in both discrete mathematics and data
structures; 43 in both discrete mathematics and programming languages; and no student
may take calculus and discrete mathematics, or data structures and programming
languages, concurrently?
Misalkan:
A1 : Murid yang mendaftar calculus
A2 : Murid yang mendaftar discrete mathematics
A3 : Murid yang mendaftar data structures
A4 : Murid yang mendaftar programming languages
|A1| = 507
|A2| = 292
|A3| = 312
|A4| = 344
| A1âĐA2 | = 0 âĶâĶ (1)
| A1âĐA3 | = 14
| A1âĐA4 | = 213
| A2âĐA3 | = 211
| A2âĐA4 | = 43
| A3âĐA4 | = 0 âĶâĶ (2)
Doanda Dresta Rahma
5025201049
Matematika Diskrit C
Dari (1) dan (2), didapatkan:
| A1âĐA2âĐA3 | = 0
| A1âĐA2âĐA4 | = 0
| A1âĐA3âĐA4 | = 0
| A2âĐA3âĐA4 | = 0
| A1âĐA2âĐA3 âĐA4 | = 0
Dengan prinsip inklusi-eksklusi:
|A1âŠA2âŠA3 âŠA4|
= |A1| + |A2| + |A3| + |A4| â |A1âĐA2| â |A1âĐA3| â |A1âĐA4|
â |A2âĐA3| â |A2âĐA4| â |A3âĐA4| + |A1âĐA2âĐA3|
+ |A1âĐA2âĐA4| + |A1âĐA3âĐA4| + |A2âĐA3âĐA4|
â |A1âĐA2âĐA3 âĐA4|
= 507 + 292 + 312 + 344 â 0 â 14 â 213 â 211 â 43 â 0 + 0 + 0
+0+0â0
= 974
Jadi, terdapat 974 murid mengambil calculus, discrete mathematics, data structures atau
programming languages.
Exercise 2.3: 36
Find f âĶ g and g âĶ f , where f (x) = x2 + 1 and g(x) = x + 2, are functions from R to R.
(ð â ð)(ðĨ) = ð(ð(ðĨ)) = (ð(ðĨ))² + 1 = (ðĨ + 2)² + 1 = ðĨ² + 4ðĨ + 4 + 1 = ðĨ² +
4ðĨ + 5
(ð â ð)(ðĨ) = ð(ð(ðĨ)) = ð(ðĨ) + 2 = ðĨ 2 + 1 + 2 = ðĨ 2 + 3
Exercise 2.3: 37
Find f + g and fg for the functions f and g given in Exercise 36.
(ð + ð)(ðĨ) = ð(ðĨ) + ð(ðĨ) = ðĨ 2 + 1 + ðĨ + 2 = ðĨ 2 + ðĨ + 3
(ðð)(ðĨ) = ð(ðĨ) â ð(ðĨ) = (ðĨ 2 + 1)(ðĨ + 2) = ðĨ 3 + 2ðĨ 2 + ðĨ + 2
Doanda Dresta Rahma
5025201049
Matematika Diskrit C
Exercise 2.3: 38
Let f (x) = ax + b and g(x) = cx + d, where a, b, c, and d are constants. Determine
necessary and sufficient conditions on the constants a, b, c, and d so that f âĶ g = g âĶ f .
(ð â ð)(ðĨ) = ð(ððĨ + ð) + ð = ðððĨ + ðð + ð
(ð â ð)(ðĨ) = ð(ððĨ + ð) + ð = ðððĨ + ðð + ð
(ð â ð)(ðĨ) = (ð â ð)(ðĨ)
ðððĨ + ðð + ð = ðððĨ + ðð + ð
ðð + ð = ðð + ð
ðð â ð = ðð â ð
ð(ð â 1) = ð(ð â 1)
ð
ð
=
ðâ1 ðâ1
ð âĶ ð = ð âĶ ð jika dan hanya jika konstanta a, b, c, d memenuhi persamaan
ð
ð
=
ðâ1 ðâ1
ð â  1, ð â  0, ð â  1, ð â  0
```
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