The following binary operations are defined as a set F, of frogs living in a Pond:?

Question

x+y = (the heavier frog of x and y)
xºy = (the older frog of x and y)
So if Henry is a 398 day old frog weighing 23 grams, and Rebecca is a 453 day old frog weighing 19 grams,

Henry + Rebecca = Henry
Henry º Rebecca = Rebecca

No two frogs have exactly the same age or weight.
State whether or not each of the following is true. In each case give a brief explanation if it is true, or an example in which it does not work if it is false.

a) + is commutative
b) º is commutative
c) + is associative
d) º is associative
e) x º (y+z) = x º y + x º z
f) x + (y º z) = (x+y) º (x+z)

Answer 1

a-d are true: these operations are commutative and associative. The oldest/heaviest of two is the oldest/heaviest no matter what order they are in (commutativity). The oldest/heaviest of three will "rise to the top" which ever order they show up in associativity.

Suppose x=2 y/o and 3#, y=3 y/o and 2#, z=1 y/o and 1#. Then y+z = heaviest y&z =y and then x*(y+z) = x*y = oldest x&y = y. BUT x*y = oldest x&y = y and x*z = oldest x&z =x and x*y+x*z = y+x = heaviest y&x =x. So x*(y+z) <> x*y+x*z. (e) is false.

Answer 2

Both operations amount to taking the max of a pair of numbers. So they're both commutative -- the order doesn't change which one is bigger. And they're both associative -- you wind up taking the max of a set of three numbers no matter what order you do the operations.

Note also that e and f will either both be true or both be false for the same reason, whatever that reason is. This is again because both operations do such similar things. I think they're both false, but I'm too lazy right now to construct and check an example. I recommend you try taking three frogs, who are heaviest to lightest in one order and youngest to oldest in the opposite order, and seeing what happens. If not, play around with the two orders for other examples.