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Set notation question.
Legionnaired
Registered User regular
Registered User regular
So I know that |A| Would be 'the number of elements in A', so |{^, a, aabb}| = 3
I have a question on my homework where I have to discuss the statement |AB| = |A||B|
How am I to understand this question? Is this valid in English:
"The number of strings formed by concatenating a string from set A with a string from set B is equal to (???)"
I have a question on my homework where I have to discuss the statement |AB| = |A||B|
How am I to understand this question? Is this valid in English:
"The number of strings formed by concatenating a string from set A with a string from set B is equal to (???)"
Legionnaired on
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Seems reasonable and correct. Given that on the left side we're only allowed to take one string from A and concatenate one string from B, this is correct because the number of possible combinations would be the number of choices for A times the number of choices for B.
is the correct interpretation. It's not true though, because if A={a,aa} and B={a,aa}, then AB={aa,aaa,aaaa}. Also, what if one of the sets is the null set?
Yeah, there is that because you end up with four results (a, aaa, aaa, aaaa) but obviously two of them are identical. It depends on whether or not you cull the results from your AB concatenation before counting the elements. It's been a number of years since I've dealt with any set theory stuff, so I'm not sure what you'd do.
Though if a set is null you get the correct result of 0, because you can't concatenate any elements of B onto A because there are no elements in B. If however, set B contains a null string, then it has one element and again the result is correct.
Truth.
A = {1, 2, 3} and B = {3, 4, 5}
AB = {1, 2, 3, 4, 5}
|AB| = 5, but |A||B| = (3)(3) = 9.
Well, no. First off, in this case AB is talking about concatenation, not union. So with your example, the concatenation AB = {13,14,15,23,24,25,33,34,35} which is 9 elements. This is dealing with string/language sets. Though as seen in the example already given it's not always the case that |AB| = |A||B|