# ordering relation $x\,\,|\, \,y \,⊆\, N\,×\,N\,$ over natural numbers $N$

## Question

Consider the ordering relation $x\,\,|\, \,y \,⊆\, N\,×\,N\,$ over natural numbers $N$ such that $x \, | \,y\,\,$ iff there exists $z$ ∈ $N$ such that $x\,\, .\,\,z = y.$ A set is called lattice if every ﬁnite subset has a least upper bound and greatest lower bound. It is called a complete lattice if every subset has a least upper bound and greatest lower bound.

Then,

1. $|$ is an equivalence relation.
2. Every subset of $N$ has an upper bound under $|$
3. $|$ is a total order.
4. $\left(N, |\right)$ is a complete lattice

5. $(N, |)$ is a lattice but not a complete lattice

### My Attempt

1.$\,\,|\,\,$ cannot be an equivalence relation,because it will not be symmetric,example-: $(2,6)$ is while $(6,2)$ is not.

now

3..$\,\,|\,\,$ cannot be total order.As for total order ,every 2 elelment of the relation must be *comparable** .But it is not.Take example, $(3,9)$ and $(4,16)$ belongs to relation but $3$ and $4$ are not comprable.

i am sure about these 2 options ,I am also sure that it is a Lattice as every pair of element has a Greatest lower bound and a Least upper Bound.

As GLB(Greatest lower bound)=GCD of 2 elements

and LUB(Least upper Bound)=LCM of 2 elements.

both GCD and LCM will belong to $N\,X\,N$

But i am stuck at option $2,4,5$.Please help me out ! Thanks

• Answering questions 2, 4, and 5 will depend on whether $0$ is a natural number or not Commented Apr 25, 2017 at 13:47
• you arre correct ,but for the lower bound !Question is aking about upper Bound(option-2) Commented Apr 25, 2017 at 15:11
• No, $0$, if you consider it to be a natural number, will be the top element: for every other natural $n$, we have $n \cdot 0 = 0$, whence $n | 0$. Commented Apr 25, 2017 at 17:00
• yes , you are right !if it would be the whole number then option 2 would be right which makes option 4 also right !Isn't it? Commented Apr 25, 2017 at 17:14
• Precisely! And then, of course, (5) would be wrong. Commented Apr 25, 2017 at 17:16

All you have to do is solve 2. Then with what you have, you should be done. For that, try to consider $N$ itself. It is a subset of $N$ (although not finite). Does it have an upper bound?