Answer to Question #231513 in Discrete Mathematics for paa

Question

1.  Show if  satisfy the algebraic axioms:

            identity and inverse under addition.

2.  Show if satisfy all (communicative, associative, identity, and inverse) the algebraic axioms under both multiplication and addition.

3.    Which number set can you find the inverse of integers under multiplication?

4.    Which number set can you find the inverse of natural numbers under multiplication?

 

5.    Find the prime factors of the following numbers:

(a) 2003

(b) 1560

(c)  5680

(d) 3050

 

6.    Calculate gcd of the following using Euclidean Algorithm:

(a) (572, 279)

(b) (138, 114)

(c)  (578, 255)

(d) (688, 212)

 

 

 

Answer 1

Part 1

$\N = \{1,2,3,4,5,...\}$

If x is an additive identity , then

$x+n =n$ for all $n \epsilon \N$

$\implies x=n-n=0\\ \implies x=0$

Then 0 is an additive identity that does not belong to $\N$. Hence $\N$ does not condition of having identity under addition

Similarly additive inverse of 1 is -1 because 1+(-1)=0. Hence it doesn't satisfy the algebraic axiom of inverse under addition.

Part 2

$For \space a,b \epsilon \Z \\ a+b=b+a ; a*b=b*a$

So $\Z$ is communicative under multiplication and addition.

$For \space a,b, c \epsilon \Z \\ (a+b)+c=a+(b+c) ; (a*b)c=a*(b*c)$

So $\Z$ is associative under multiplication and addition.

$For \space a \epsilon \Z \\ a+0=a ; a*1=a$

So $\Z$ has an identity '0' under addition and has an identity '1' multiplication.

$For \space a \epsilon \Z \\$ - a also belong to $\Z$

and $a+(-a)=0$

So $\Z$ has an inverse under addition

$For \space a \epsilon \Z \\$

$ab=1\\ \implies b = \frac{1}{a}$

So $\Z$ does not have an inverse under multiplication.

Part 3

We can't have a set containing the inverse of all elements of it under multiplication.(as 0 € Z does not have inverse under multiplication).

If we take Z^* then we can have a set containing the inverse of all its elements. i.e. Q (set of rationals)

Part 4

For N, the set containing its inverse elements is Q⁺ .(set of positive rationals)

Part 5

a. 2003 is a prime number. So factors are 1 and 2003.

b. All Factors of 1560:

1, 2, 3, 4, 5, 6, 8, 10, 12, 13, 15, 20, 24, 26, 30, 39, 40, 52, 60, 65, 78, 104, 120, 130, 156, 195, 260, 312, 390, 520, 780, 1560

c. All Factors of 5680:

1, 2, 4, 5, 8, 10, 16, 20, 40, 71, 80, 142, 284, 355, 568, 710, 1136, 1420, 2840, 5680

d. All Factors of 3050:

1, 2, 5, 10, 25, 50, 61, 122, 305, 610, 1525, 3050

Part 6

Part a

Set up a division problem where a is larger than b.

a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.

$572 ÷ 279 = 2 R 14 (572 = 2 × 279 + 14)\\ 279 ÷ 14 = 19 R 13 (279 = 19 × 14 + 13)\\ 14 ÷ 13 = 1 R 1 (14 = 1 × 13 + 1)\\ 13 ÷ 1 = 13 R 0 (13 = 13 × 1 + 0)\\$

When remainder R = 0, the GCF is the divisor, b, in the last equation. GCF = 1

Part b

Set up a division problem where a is larger than b.

a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.

$138 ÷ 114 = 1 R 24 (138 = 1 × 114 + 24)\\ 114 ÷ 24 = 4 R 18 (114 = 4 × 24 + 18)\\ 24 ÷ 18 = 1 R 6 (24 = 1 × 18 + 6)\\ 18 ÷ 6 = 3 R 0 (18 = 3 × 6 + 0)$

When remainder R = 0, the GCF is the divisor, b, in the last equation. GCF = 6

Part c

Set up a division problem where a is larger than b.

a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.

$578 ÷ 255 = 2 R 68 (578 = 2 × 255 + 68)\\ 255 ÷ 68 = 3 R 51 (255 = 3 × 68 + 51)\\ 68 ÷ 51 = 1 R 17 (68 = 1 × 51 + 17)\\ 51 ÷ 17 = 3 R 0 (51 = 3 × 17 + 0)$

When remainder R = 0, the GCF is the divisor, b, in the last equation. GCF = 17

Part d

Set up a division problem where a is larger than b.

a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.

$688 ÷ 212 = 3 R 52 (688 = 3 × 212 + 52)\\ 212 ÷ 52 = 4 R 4 (212 = 4 × 52 + 4)\\ 52 ÷ 4 = 13 R 0 (52 = 13 × 4 + 0)$

When remainder R = 0, the GCF is the divisor, b, in the last equation. GCF = 4