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## Here's the Solution to this Question

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