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)
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Here's the Solution to this Question
Part 1
If x is an additive identity , then
for all
Then 0 is an additive identity that does not belong to . Hence 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
So is communicative under multiplication and addition.
So is associative under multiplication and addition.
So has an identity '0' under addition and has an identity '1' multiplication.
- a also belong to
and
So has an inverse under addition
So 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.
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.
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.
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.
When remainder R = 0, the GCF is the divisor, b, in the last equation. GCF = 4