course Mth 151
april 1 around 2:30
Mary Sholesassignment 21-QA
q001:
we would start with the 2@5. we would have 2*5 to get 10. Double it to get 20. and then divide by 3 to get 6 with the remainder of 2. so 2@5 is 2 because it is the remainder. 3@8 is 24. double that to get 48. and divide by 3 to get 12. so 3@8 would be 0 because it is the remainder. now we have 7@13. we would take 7*13 to get 91. Double that to get 182. and divide that from 3 to get 60 with a remainder of 2. so 7@13 would be 2 because it is the remainder.
*3
q002:
I would use the same process as the previous problem and the table would show: @234
2201
3000
4102
*2
q003:
it would be from [0,1,2] because those are the only remainders you could get from dividing by 3.
*2
q004:
they do not all come from [2,3,4] the only one that would could be 2.
*2
q005:
the operation is closed on set R due to to S and T resulting with it.
*1
q006:
if all the numbers are under its set, just like the ones in previous problems, then they are closed because the numbers all go in the set.
*2
q007:
putting in order or multiplying x and y will not make any difference. so therefore x@y = y@x
*2
q008:
subtraction would not have forms of commutative property because most times they will not equal
*3
q009:
this operation is not closed because sometimes when we subtract whole numbers we will get negative numbers.
*2
q010:
this operation is closed because we will always get a whole number when we add whole numbers.
*2
q011:
the exact number that we divided one by would be the answer.
*3
q012:
no number would be changed due to the identity property.
*1
q013:
no other number but 0 would have this property for addition.
*2
&#Please let me know if you have questions. &#