2. Number Theory. Please provide complete and correct mathematical solution done on computer or by hand. Please, emphasis on correct solution. This is all the information provided. Thank you very much.
show that if a and b are positive integers then a!bb!/(ab)!.(H)
Prove or disprove: If a, b belong to the set of positive integers, and if a divides b and b divides a, then a=b. Does this hold if if a,b are not necessarily positive? Why or Why not?
Here is what I have:
If a and b are integers, we say a divides b if there exists an integer k so that b=ak. Thus:
Since a divides b and b divides a, we know that k,j must both be integer, which means k and j must both equal 1 or -1. This means:
This proves that a=b when we assume that a,b are both positive integers.
For the second part of the question, I am asked if I can still say a=b if the assumption about a, b is relaxed so that a,b belong to the integers (not the positive integers.) I think The answer is no, I cannot say that a=b for every case when we allow a,b to also be negative integers.
Here is my reasoning (I use the same idea that a=bj and b=ak here)
So |a|=|b||a|=|b| but not a=b
Prove or disprove: If a, b belong to the set of positive integers, and if a divides b and b divides a, then a=b. Does this hold if if a,b are not necessarily positive? Why or Why not? Here is what I have: If a and b are integers, we say a divides b if there exists an integer k so that b=ak. Thus: b=akb=ak a=bja=bj a=akja=akj kj=1kj=1 Since a divides b and b divides a, we know that k,j must both be integer, which means k and j must both equal 1 or -1. This means: a=ba=b This proves that a=b when we assume that a,b are both positive integers. For the second part of the question, I am asked if I can still say a=b if the assumption about a, b is relaxed so that a,b belong to the integers (not the positive integers.) I think The answer is no, I cannot say that a=b for every case when we allow a,b to also be negative integers. Here is my reasoning (I use the same idea that a=bj and b=ak here) +/a=akj+/a=akj +/1=kj+/1=kj So |a|=|b||a|=|b| but not a=b |