- #1
Perion
This should be so easy but I'm having trouble understanding why {{x},{x,y}} defines an ordered pair (x,y). I'm trying to work through the following problem from the Zakon Series pdf textbook:
I'm just learning to do formal proofs so not sure about where to start. Since it's an iff equivalence I assume that I need to show that
1. if (x,y)=(u,v) then x=u and y=v
AND then also show that
2. if x=u and y=v then (x,y)=(u,v).
Is this the correct overall approach for proving the iff biconditional?
I'm OK with #2 since it's trivial to show that {{x},{x,y}}={{u},{u,v}} for x=u and y=v.
#1 also seems pretty simple when x=y but I'm not sure how to go about it for the case when x is not equal to y. Can you help me with this please.
Thank you,
Perion
[This isn't homework - I'm studying free math e-books on my own.]
Code:
If (x,y) denotes the set {{x},{x,y}} [B]prove that, for any x, y, u, v,
we have (x,y) = (u,v) iff x=u and y=v[/B]. Treat this as a definition of an
ordered pair.
[Hint: Consider separately the two cases (x equals y) and
(x not equal to y), noting that {x,x}={x}]
I'm just learning to do formal proofs so not sure about where to start. Since it's an iff equivalence I assume that I need to show that
1. if (x,y)=(u,v) then x=u and y=v
AND then also show that
2. if x=u and y=v then (x,y)=(u,v).
Is this the correct overall approach for proving the iff biconditional?
I'm OK with #2 since it's trivial to show that {{x},{x,y}}={{u},{u,v}} for x=u and y=v.
#1 also seems pretty simple when x=y but I'm not sure how to go about it for the case when x is not equal to y. Can you help me with this please.
Thank you,
Perion
[This isn't homework - I'm studying free math e-books on my own.]