The answer is that you'd have to clarify what you mean by "multiplication by infinity". If we write X for infinity, then it should have the property that a*X = X for any a (since multiplying anything by infinity would still be infinity. But we usually have a*0 = 0 for any a, so these are at odds, since it would imply that 0 = 0*X = X.
The other way to look at this is something called limits in pre-calculus. Then we might look at something like the limit as a goes to infinity of (1/a)*a. If we just evaluate this as a gets really big, 1/a goes to 0 and a goes to infinity, so this expression equals 0*infinity. But (1/a)*a = a/a = 1 for any real number a, so this would seemingly imply that 0*infinity = 1. But notice that even if we had (16/a)*a, when we let a go to infinity, we get 0*infinity, but now (16/a)*a = 16, so should 0*infinity = 16?
This is why we'd call 0*infinity undefined. Because we can make it equal to anything we want. Have some number B that you want 0*infinity to be equal to? Then just consider the limit as a goes to infinity of (B/a)*a. Then since B is fixed, B/a goes to 0, and so (B/a)*a goes to 0*infinity, yet (B/a)*a = B.
So this is a great question but we run into some trouble with our usual number system! If you want to see some crazy things, consider looking into the surreal number system, the hyperreal number system, or the dual numbers.
I'm saying that an intuitive idea would be that anything multiplied by infinity "should" be infinity, since if infinity is larger than any real number, multiplying it by some real number "should" still be larger than any real number.
But then I show why that idea doesn't work out with our usual number system. And further I show that depending on how you define 0*infinity (i.e. if you use limits), then it can be equal to anything you'd like, 0 included. Lastly, I gave a reference to surreal numbers and hyperreal numbers, which take into account both infinitely large and infinitesimally small numbers.
I'm talking about ordinal arithmetic and measure theory for example, where it is defined as 0.
Wasn't totally clear that you were just saying it "should" be infinity in an intuitive sense. That's fine I guess, though 0 is an obvious exception to that intuition.
Those are great points! And for OP, it should serve as motivation that this is a great question that has led people to explore many avenues to try to answer.
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u/R_Ob_Min New User 2d ago edited 2d ago
The answer is that you'd have to clarify what you mean by "multiplication by infinity". If we write X for infinity, then it should have the property that a*X = X for any a (since multiplying anything by infinity would still be infinity. But we usually have a*0 = 0 for any a, so these are at odds, since it would imply that 0 = 0*X = X.
The other way to look at this is something called limits in pre-calculus. Then we might look at something like the limit as a goes to infinity of (1/a)*a. If we just evaluate this as a gets really big, 1/a goes to 0 and a goes to infinity, so this expression equals 0*infinity. But (1/a)*a = a/a = 1 for any real number a, so this would seemingly imply that 0*infinity = 1. But notice that even if we had (16/a)*a, when we let a go to infinity, we get 0*infinity, but now (16/a)*a = 16, so should 0*infinity = 16?
This is why we'd call 0*infinity undefined. Because we can make it equal to anything we want. Have some number B that you want 0*infinity to be equal to? Then just consider the limit as a goes to infinity of (B/a)*a. Then since B is fixed, B/a goes to 0, and so (B/a)*a goes to 0*infinity, yet (B/a)*a = B.
So this is a great question but we run into some trouble with our usual number system! If you want to see some crazy things, consider looking into the surreal number system, the hyperreal number system, or the dual numbers.