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Infinity doesn't need to be in some "algebraic field" for it to be patently true that an infinite amount of nothing is still nothing, and that adding zero to it

Infinity doesn't need to be in some "algebraic field" for it to be patently true that an infinite amount of nothing is still nothing, and that adding zero to itself over and over again for an infinitely long time will never give you a result other than zero. It's only impossible to define if you overthink it, and/or maintain a needlessly narrow definition of what a "number" is. Or, if you really insist on speaking in mathematician-ese, an infinite series of zero is zero, and a zero-bounded summation is zero regardless of the summand: x · y ≡ Σ(y, i = 1) x = y times { x + x + … + x } ≡ Σ(x,i=1) y = x times { y + y + … + y } ∞ · 0 ≡ Σ(0, i = 1) ∞ = 0 times { ∞ + ∞ + … + ∞ } = 0 0 · ∞ ≡ Σ(∞, i = 1) 0 = ∞ times { 0 + 0 + … + 0 } = 0 Or, if you prefer programmer-ese: julia> i = 0 0 julia> while true println(i) global i += 0 end 0 0 0 0 0 0 0 0 0 ...and on and on until the heat death of the universe or you hit Ctrl-C. Either way, seems pretty straightforward to define if you have a clear definition of what multiplication is in the first place (and what either zero or infinite iterations of that definition will produce).

ryandv 2y ago
Ok, let's assume you are correct and that ∞ · 0 = 0. Consider then the two sequences sₙ = n, tₙ = 1/n. By Definition 3.15 as provided in my last post, sₙ ⟶ +∞, and you will have to take it for granted that tₙ ⟶ 0 [0]. Intuitively we can see that the terms of {sₙ} are 1, 2, 3, ... tending to +∞; for {tₙ} we have 1, 1/2, 1/3, ... tending to zero, for progressively larger values of n. Now I ask what happens if we multiply the "infinite'th" terms of both sequences together. The first few terms of this product would be 1 · 1, 2 · 1/2, 3 · 1/3, and so on; I ask what the value x is in the limit sₙ · tₙ ⟶ x as we evaluate further and further "nth" terms of both sequences. You may have observed from the first three terms evaluated that sₙ · tₙ = n(1/n) = 1. Thus, as we continue to increase the value of n, it's always the case that sₙ · tₙ ⟶ 1 and the product tends to 1, because the product is constant and irrespective of n; we've "cancelled it out." The limit of the product is the product of the limits [1]; that is, sₙ · tₙ ⟶ +∞ · 0, as we first established that sₙ ⟶ +∞, tₙ ⟶ 0. If we thus take your supposition that +∞ · 0 = 0 for granted, we obtain sₙ · tₙ ⟶ 0, which contradicts our previous result that sₙ · tₙ ⟶ 1. Thus we can either dispense with the cited established theorems of analysis used to deduce that sₙ · tₙ ⟶ 1, or conclude that the supposition +∞ · 0 = 0 must be false. It might be the case that Σ(∞, i = 1) 0 = 0, but you can't extend this to conclude +∞ · 0 = 0 in general. Lots of intuitions from informal mathematics and even calculus start to break down once you examine the lower-level "machine code" of proof and analysis, especially once you start talking about concepts like infinity. [0] See Theorem 3.20 (a) in Rudin: https://archive.org/details/principles-of-mathematical-analysis-walter-rudin/page/56/mode/2up https://archive.org/details/principles-of-mathematical-analy... [1] See Theorem 3.3 (c) in Rudin: https://archive.org/details/principles-of-mathematical-analysis-walter-rudin/page/48/mode/2up https://archive.org/details/principles-of-mathematical-analy...
- yellowapple 2y ago
  > Now I ask what happens if we multiply the "infinite'th" terms of both sequences together. In that case, you would've reached their respective limits, and you're back to adding one of those limits into itself an other-limit number of times. If sₙ · tₙ ⟶ 1, then that only holds true if tₙ hasn't actually reached 0. > Thus we can either dispense with the cited established theorems of analysis used to deduce that sₙ · tₙ ⟶ 1 You don't need to do that. You just need to accept that zero is just as much of a mathematical special case as infinity - unsurprisingly, since it's the inverse of infinity and vice versa. > It might be the case that Σ(∞, i = 1) 0 = 0, but you can't extend this to conclude +∞ · 0 = 0 in general. Sure you can, unless you've got some other definition of multiplication that's impossible to express as self-summation. Even if you go with the alternative definition of multiplication as a scaling operation (wherein you're computing m × n by taking the slope from (x=0,y=0) to (x=1,y=m) and then looking up y where x=n), if m is zero then the line being drawn never stops being vertical, and if n is zero then you never leave (0,0) in the first place. Doesn't matter if the other factor is infinitely far in either the x or y axis; you're still ending up with zero no matter how hard you try and fight it. > Lots of intuitions from informal mathematics and even calculus start to break down once you examine the lower-level "machine code" of proof and analysis, especially once you start talking about concepts like infinity. Sure, but in this case, it's the intuition that multiplying something by its inverse (a.k.a. dividing something by itself) is always 1 that breaks down, not the above-verifiable and inescapable fact that multiplying something by zero is always zero. 0 ÷ 0 = n looks like it should correct for any value of n (incl. n = 1), since multiplying both sides by zero to eliminate that divide-by-zero will always produce a correct equation, but since m ÷ n ≡ m × (1/n), if m is zero then anything on the RHS must be zero, because of that inescapable nature of nothingness - thus, 0 ÷ 0 = 0 × (1/0) = 0, with all other possible alternatives having been rendered impossible.
  - ryandv 2y ago
    > you're back to adding one of those limits into itself an other-limit number of times. > some other definition of multiplication that's impossible to express as self-summation. Ok. What happens if I multiply a number by pi? What does it mean to add something to itself, pi times? > If sₙ · tₙ ⟶ 1, then that only holds true if tₙ hasn't actually reached 0. I mean... it is in fact the case that tₙ never actually reaches zero; otherwise, if 1/n = 0 for some n, then by multiplying both sides by n we obtain 1 = 0. What's meant by tₙ ⟶ 0 is that any neighborhood centered about 0 of any radius (call the radius "epsilon") always contains at least one point from the sequence {tₙ}. To hammer the point that sₙ · tₙ ⟶ 1 home, and since you are fond of using a computer to perform arithmetic (note: not prove mathematical statements), here's what computers have to say about the limit of n · (1/n): https://www.wolframalpha.com/input?i=limit+as+n-%3Einfinity+of+n(1%2Fn) https://www.wolframalpha.com/input?i=limit+as+n-%3Einfinity+... > You just need to accept that zero is just as much of a mathematical special case as infinity - unsurprisingly, since it's the inverse of infinity and vice versa. > the inverse of infinity You again throw around words like "inverse" whose meaning you don't understand. Do you mean a multiplicative inverse, where a number and its multiplicative inverse yield the multiplicative identity, in which case +∞ · 0 = 1? Or an additive inverse that yields the additive identity, in which case +∞ + 0 = 0? Or some other pseudomathematical definition of "inverse" pulled out of a hat, like your definitions of +∞ · 0? > if m is zero then the line being drawn never stops being vertical Drawing pictures is different from putting together a formal, airtight proof in first-order logic that can be (in principle) machine-verified. Maybe I'll make an exception for compass-and-straightedge proofs, but that's not what you're presenting here. Rudin was published in 1953, there are probably very good reasons for why this text has withstood refutation for over 70 years. Maybe you can rise to the task; publish a paper with your novel number system in which +∞ · 0 = 0 and 0 ÷ 0 = 0 and wait for your Fields Medal in the mail. Maybe you can collaborate with Terrence Howard and get a spot on Joe Rogan.

ryandv 2y ago

Ok, let's assume you are correct and that ∞ · 0 = 0. Consider then the two sequences sₙ = n, tₙ = 1/n. By Definition 3.15 as provided in my last post, sₙ ⟶ +∞, and you will have to take it for granted that tₙ ⟶ 0 [0]. Intuitively we can see that the terms of {sₙ} are 1, 2, 3, ... tending to +∞; for {tₙ} we have 1, 1/2, 1/3, ... tending to zero, for progressively larger values of n. Now I ask what happens if we multiply the "infinite'th" terms of both sequences together. The first few terms of this product would be 1 · 1, 2 · 1/2, 3 · 1/3, and so on; I ask what the value x is in the limit sₙ · tₙ ⟶ x as we evaluate further and further "nth" terms of both sequences. You may have observed from the first three terms evaluated that sₙ · tₙ = n(1/n) = 1. Thus, as we continue to increase the value of n, it's always the case that sₙ · tₙ ⟶ 1 and the product tends to 1, because the product is constant and irrespective of n; we've "cancelled it out." The limit of the product is the product of the limits [1]; that is, sₙ · tₙ ⟶ +∞ · 0, as we first established that sₙ ⟶ +∞, tₙ ⟶ 0. If we thus take your supposition that +∞ · 0 = 0 for granted, we obtain sₙ · tₙ ⟶ 0, which contradicts our previous result that sₙ · tₙ ⟶ 1. Thus we can either dispense with the cited established theorems of analysis used to deduce that sₙ · tₙ ⟶ 1, or conclude that the supposition +∞ · 0 = 0 must be false. It might be the case that Σ(∞, i = 1) 0 = 0, but you can't extend this to conclude +∞ · 0 = 0 in general. Lots of intuitions from informal mathematics and even calculus start to break down once you examine the lower-level "machine code" of proof and analysis, especially once you start talking about concepts like infinity. [0] See Theorem 3.20 (a) in Rudin: https://archive.org/details/principles-of-mathematical-analysis-walter-rudin/page/56/mode/2up https://archive.org/details/principles-of-mathematical-analy... [1] See Theorem 3.3 (c) in Rudin: https://archive.org/details/principles-of-mathematical-analysis-walter-rudin/page/48/mode/2up https://archive.org/details/principles-of-mathematical-analy...

yellowapple 2y ago
> Now I ask what happens if we multiply the "infinite'th" terms of both sequences together. In that case, you would've reached their respective limits, and you're back to adding one of those limits into itself an other-limit number of times. If sₙ · tₙ ⟶ 1, then that only holds true if tₙ hasn't actually reached 0. > Thus we can either dispense with the cited established theorems of analysis used to deduce that sₙ · tₙ ⟶ 1 You don't need to do that. You just need to accept that zero is just as much of a mathematical special case as infinity - unsurprisingly, since it's the inverse of infinity and vice versa. > It might be the case that Σ(∞, i = 1) 0 = 0, but you can't extend this to conclude +∞ · 0 = 0 in general. Sure you can, unless you've got some other definition of multiplication that's impossible to express as self-summation. Even if you go with the alternative definition of multiplication as a scaling operation (wherein you're computing m × n by taking the slope from (x=0,y=0) to (x=1,y=m) and then looking up y where x=n), if m is zero then the line being drawn never stops being vertical, and if n is zero then you never leave (0,0) in the first place. Doesn't matter if the other factor is infinitely far in either the x or y axis; you're still ending up with zero no matter how hard you try and fight it. > Lots of intuitions from informal mathematics and even calculus start to break down once you examine the lower-level "machine code" of proof and analysis, especially once you start talking about concepts like infinity. Sure, but in this case, it's the intuition that multiplying something by its inverse (a.k.a. dividing something by itself) is always 1 that breaks down, not the above-verifiable and inescapable fact that multiplying something by zero is always zero. 0 ÷ 0 = n looks like it should correct for any value of n (incl. n = 1), since multiplying both sides by zero to eliminate that divide-by-zero will always produce a correct equation, but since m ÷ n ≡ m × (1/n), if m is zero then anything on the RHS must be zero, because of that inescapable nature of nothingness - thus, 0 ÷ 0 = 0 × (1/0) = 0, with all other possible alternatives having been rendered impossible.
- ryandv 2y ago
  > you're back to adding one of those limits into itself an other-limit number of times. > some other definition of multiplication that's impossible to express as self-summation. Ok. What happens if I multiply a number by pi? What does it mean to add something to itself, pi times? > If sₙ · tₙ ⟶ 1, then that only holds true if tₙ hasn't actually reached 0. I mean... it is in fact the case that tₙ never actually reaches zero; otherwise, if 1/n = 0 for some n, then by multiplying both sides by n we obtain 1 = 0. What's meant by tₙ ⟶ 0 is that any neighborhood centered about 0 of any radius (call the radius "epsilon") always contains at least one point from the sequence {tₙ}. To hammer the point that sₙ · tₙ ⟶ 1 home, and since you are fond of using a computer to perform arithmetic (note: not prove mathematical statements), here's what computers have to say about the limit of n · (1/n): https://www.wolframalpha.com/input?i=limit+as+n-%3Einfinity+of+n(1%2Fn) https://www.wolframalpha.com/input?i=limit+as+n-%3Einfinity+... > You just need to accept that zero is just as much of a mathematical special case as infinity - unsurprisingly, since it's the inverse of infinity and vice versa. > the inverse of infinity You again throw around words like "inverse" whose meaning you don't understand. Do you mean a multiplicative inverse, where a number and its multiplicative inverse yield the multiplicative identity, in which case +∞ · 0 = 1? Or an additive inverse that yields the additive identity, in which case +∞ + 0 = 0? Or some other pseudomathematical definition of "inverse" pulled out of a hat, like your definitions of +∞ · 0? > if m is zero then the line being drawn never stops being vertical Drawing pictures is different from putting together a formal, airtight proof in first-order logic that can be (in principle) machine-verified. Maybe I'll make an exception for compass-and-straightedge proofs, but that's not what you're presenting here. Rudin was published in 1953, there are probably very good reasons for why this text has withstood refutation for over 70 years. Maybe you can rise to the task; publish a paper with your novel number system in which +∞ · 0 = 0 and 0 ÷ 0 = 0 and wait for your Fields Medal in the mail. Maybe you can collaborate with Terrence Howard and get a spot on Joe Rogan.