00:01
So we're given a relation on two bracelets, where each bracelet has 3b, red, white, or blue.
00:18
So the relation says v1 and b2 are related.
00:23
If you can transform b1 to b2 by rotating b1 or by rotating b1 and then reflecting it.
00:33
So we have to show it's an equivalent solution.
00:37
So reflexivity.
00:38
So let's see i have a bracelet b inside the set of all the bracelets.
00:53
Let's call it a.
00:59
Well, if you have a bracelet and you rotate it to zero degrees, well then you still have the same bracelet.
01:09
So under zero degrees rotation, bracelet b becomes itself and transforms into itself.
01:20
Well, you can also do any multiple of 360 degrees.
01:25
Degrees in the store work, 720 degrees would work, 360 would work.
01:32
So under 0 degrees rotation, b becomes itself, so that implies b is related to itself and r is reflexive.
01:51
Now for symmetry, let's say i have b1 related to b2 to that meaning by definition.
02:05
I can transform b1 into b2 either by rotation or rotation plus reflection.
02:24
So if you rotate b1, if we rotate b1, we'll say, let's see.
03:02
So you want to show that b2 can also be rotated into b1.
03:15
So this will show that b2 is related to b1 and that we have similar to b1 and that we have symmetry so how can we do well if you rotate by some degree then you can get b1 from b2 by going back by rotating in the opposite direction and if you rotate and then reflect then you can get b1 from b2 by doing the opposite by reflecting and then rotating so if i in the first case so we have two cases right first case is just rotating only so if i rotate by some x degree, right? then i can get b2 to become b1 just by rotating minus x degrees.
04:41
I can just go in the opposite direction.
04:44
And for the second case of rotating plus reflecting, i can get b2 from b1.
05:04
So if i rotate, so if i rotate by again some x degree, and then i make some reflection across any line.
05:27
So across some line.
05:40
Then if i just do the opposite, i can get from b2 to b1.
05:48
If i do, how do you undo your rotation and reflection? well, i can rotate minus x degrees.
06:07
And then i can reflect across the inverse of whatever line that i rotated on.
06:30
So that implies i can get b2 to b1 using either rotation or rotation plus reflection.
06:40
So b2 is related to b1, so r is symmetric.
06:53
Now for the transitive case, for transitivity, let's say i had 3 beats this time.
07:06
We have 3 bead.
07:07
And let's say first bead is related to the second bead and the second bead is related to the third bracelet of the same bead bracelet.
07:22
So by definition, i can, by definition of r, i can get b1 into b2 to transform into b2 by either strategy one.
07:46
I'll call s1 or s2, strategy 2, which is rotating or rotating reflection.
07:56
And it also means that i can transform b2 into b3 by strategy 1 or strategy 2.
08:09
Well, then if i use on the first case with, if i, on the first case, which is using strategy 1, if i rotate by some degree and i get b2 to b1, then we can rotate by another degree and rotate again, let's say by another degree, let's say n, to get b2 to b3.
09:17
Well, this should be b1.
09:18
To get b1 to b2.
09:20
Yeah, so we need this transitivity here.
09:23
And then i can rotate again to get b2 to b3.
09:26
That means if i rotate in minus n degrees, i can get from b3 to b2, and then i can rotate minus x degrees to get v2 to b1.
10:01
So this means we've shown transitivity with the first case of strategy 1, because this means i can go from b1 to b3 just by doing a series of two rotations of rotating by x and then rotating by n.
10:40
Now let's show it with the second case, with the case two using strategy two.
10:51
So here we can get b1 to b3 by rotating x and then rotating end.
11:06
And if the second strategy, well, if we can rotate by some degree x to get b1 to b2, and let's say we do a reflection on the second...