Test by Bikram | Mock GATE | Test 1 | Question: 3

Question

Consider the Hasse diagram $D_1$ and $D_2$ as follows:

• $D_1:$ The Hasse diagram for the partial ordering $\left \{ \langle a, b \rangle \mid a \mod \: b=0 \right \}$ on the set of positive divisors of $24 \}$
• $D_2:$ The Hasse diagram for the partial ordering $\left \{\langle a, b\rangle \mid a \mod \: b=0 \right \}$ on the set of positive divisors of $81 \}$

Then which among $D_1$ and $D_2$ is/are a total ordering?

• A. Only $D_1$
• B. Only $D_2$
• C. Both $D_1$ & $D_2$
• D. None $D_1$ & $D_2$

Answer 1

D1: The Hasse diagram for the partial ordering {<a, b> |a mod b=0} on the set of {positive divisor of 24} 

Set of positive divisor of 24 = {1,2,3,4,6,8,12}

D2: The Hasse diagram for the partial ordering {<a, b> |a mod b = 0} on the set of {positive divisor of 81} 
Set of positive divisor of 81={1,3,9,27,81} 

So, D2 is total ordering while D1 is not! Answer: B

Comments

what is the significance of a mod b =0????

Answer 2

in D1: for example 6 and 4 are not comparable.

Comments

"not comparable" for <6,4> because 6 mod 4 ≠ 0 ,right ?

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@amitqy 'not comparable' means in D1 there is no relation betn 6 and 4. In D2 you can see every element is relatable i.e they are comparable.