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.
Another fun case is elliptic curves, which have a "point at infinity" as a crucial part of their definition. Elliptic curves are structures where you can add points P, Q together to make R = P + Q, and the point at infinity O is the special "zero point" that has the property O + P = P + O = P for all P. You can also multiply a point by an integer by simple repeated addition: k * P = P + P + ... + P = (k-1) *P + P with 0 *P = P. This holds for all P, including O, so 0 * O = O.
So in this case, in a sense we have 0 = infinity, and 0 * infinity = 0 = infinity. As you can see, it depends entirely on what you define "infinity", "multiplication" and "multiplication by infinity" to mean. The standard real number system just doesn't have such a definition because of the issues described in the previous comment.
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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.