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To those who insist on the wrong answer

Anonymous_IIII

@KhrivenilSSP said ^

Given two disjoint sets {a, b} and {c, d}. Take their union—how many elements are in the resulting set? If there are 5, what is the fifth element?

There is no fifth element!

Of course there are only 4.

Then wtf u meant??

I meant that 2+2=4, not 5.

Okaaay 2 + 2 is 1

@KhrivenilSSP said [^](/forum/redirect/post/GO4n0Eio) > > > > > Given two disjoint sets {a, b} and {c, d}. Take their union—how many elements are in the resulting set? If there are 5, what is the fifth element? > > > > > > > > There is no fifth element! > > > > > > Of course there are only 4. > > > > Then wtf u meant?? > > I meant that 2+2=4, not 5. Okaaay 2 + 2 is 1
O_olo_O

Y'all took math rightz? 2+2=4 and 2^2=4 and 2*2=4 but 2! is also =2 and 2-2=0

Y'all took math rightz? 2+2=4 and 2^2=4 and 2*2=4 but 2! is also =2 and 2-2=0
CSKA_Moscou

@KhrivenilSSP said ^

what is the fifth element?

Aether.

@KhrivenilSSP said [^](/forum/redirect/post/XyBjtnNE) > what is the fifth element? Aether.
KhrivenilSSP

@CSKA_Moscou said ^

what is the fifth element?

Aether.

But we didn't define anything with a lame with more than one letter.

@CSKA_Moscou said [^](/forum/redirect/post/oQ7Lroko) > > what is the fifth element? > > Aether. But we didn't define anything with a lame with more than one letter.
KhrivenilSSP

@O_olo_O said ^

Y'all took math rightz? 2+2=4 and 2^2=4 and 2*2=4 but 2! is also =2 and 2-2=0

Yes, 2! = 1×...×2 = 1×2 = 2, and 2-2 indeed is 0. Oh, and also, 2 ÷ 2 = 1.

@O_olo_O said [^](/forum/redirect/post/NxbqVctR) > Y'all took math rightz? 2+2=4 and 2^2=4 and 2*2=4 but 2! is also =2 and 2-2=0 Yes, 2! = 1×...×2 = 1×2 = 2, and 2-2 indeed is 0. Oh, and also, 2 ÷ 2 = 1.
O_olo_O

@KhrivenilSSP said ^

Y'all took math rightz? 2+2=4 and 2^2=4 and 2*2=4 but 2! is also =2 and 2-2=0

Yes, 2! = 1×...×2 = 1×2 = 2, and 2-2 indeed is 0. Oh, and also, 2 ÷ 2 = 1.

Yep, and sin(2)=0.909...

@KhrivenilSSP said [^](/forum/redirect/post/qLUWoAIh) > > Y'all took math rightz? 2+2=4 and 2^2=4 and 2*2=4 but 2! is also =2 and 2-2=0 > > Yes, 2! = 1×...×2 = 1×2 = 2, and 2-2 indeed is 0. Oh, and also, 2 ÷ 2 = 1. Yep, and sin(2)=0.909...
Noflapsedited

The null set is not an "element" of the union of two disjoint sets, each with a cardinality of two, provided, as in the illustrated case, that neither a, b, c or d was intended to represent the null set.

The null set is a "subset" of that union, since that union is, itself, a set.

And even if a, b, c, or d (any one of them) WERE silently intended to represent the null set, there would still just be four "elements" and the null set would still be a subset, as well.

It's important not to confuse "elements" and "subsets."

It's also important not to confuse "union" with "intersection."

The union of the two sets would have four elements.

The intersection of the two sets, since they are said to be "disjoint," has zero elements. It is (you guessed it) the null set.

This is why we must not read math books like a novel, unless we are Ramanujan. We must take a long time with each page. And study example problems slowly and carefully. And do our homework with care, as well.

Oh, what's the use. Soon kids will be having AI do everything for them and not carrying around much independent knowledge. It would make me want to shout "get off my lawn," if only I had a lawn.

I know, I know, what good did the knowledge do me if I don't even have a lawn. I ponder that myself. I DO have a spaniel, however. And that's not nothing!

And I play the hippopotamus opening. And that, even by itself, is enough to make life worth living.

Oh, and by the way, the null set is not "nothing." It is a SET containing nothing. And a set is something!

The null set is not an "element" of the union of two disjoint sets, each with a cardinality of two, provided, as in the illustrated case, that neither a, b, c or d was intended to represent the null set. The null set is a "subset" of that union, since that union is, itself, a set. And even if a, b, c, or d (any one of them) WERE silently intended to represent the null set, there would still just be four "elements" and the null set would still be a subset, as well. It's important not to confuse "elements" and "subsets." It's also important not to confuse "union" with "intersection." The union of the two sets would have four elements. The intersection of the two sets, since they are said to be "disjoint," has zero elements. It is (you guessed it) the null set. This is why we must not read math books like a novel, unless we are Ramanujan. We must take a long time with each page. And study example problems slowly and carefully. And do our homework with care, as well. Oh, what's the use. Soon kids will be having AI do everything for them and not carrying around much independent knowledge. It would make me want to shout "get off my lawn," if only I had a lawn. I know, I know, what good did the knowledge do me if I don't even have a lawn. I ponder that myself. I DO have a spaniel, however. And that's not nothing! And I play the hippopotamus opening. And that, even by itself, is enough to make life worth living. Oh, and by the way, the null set is not "nothing." It is a SET containing nothing. And a set is something!
HuaTianXiangQi

@O_olo_O said ^

Y'all took math rightz? 2+2=4 and 2^2=4 and 2*2=4 but 2! is also =2 and 2-2=0

Yes, 2! = 1×...×2 = 1×2 = 2, and 2-2 indeed is 0. Oh, and also, 2 ÷ 2 = 1.

Yep, and sin(2)=0.909...

You're using radians. In degree system, sin(2) ≈ 0.0348994967.

@O_olo_O said [^](/forum/redirect/post/orrMEbAl) > > > Y'all took math rightz? 2+2=4 and 2^2=4 and 2*2=4 but 2! is also =2 and 2-2=0 > > > > Yes, 2! = 1×...×2 = 1×2 = 2, and 2-2 indeed is 0. Oh, and also, 2 ÷ 2 = 1. > > Yep, and sin(2)=0.909... You're using radians. In degree system, sin(2) ≈ 0.0348994967.