Visionem Ex Illuminatio

Wow, terrible misconception. I thought you meant 12 inches vs 1 inch. My fault; should've read closer. Well, Greg H. made a good explanation. Where did you get 12 x 12 from? I said the relationship between the hypotenuse length ( c ) and the leg length (a) is [math]a\times\sqrt{2}=c[/math]. In this case' date=' it's 12 times sqrt(2), which as others have explained is just as irrational as sqrt(2). Both are, in the way you stated, "non-computable", and require approximations. [math']12\times\sqrt{2}[/math] in inches is the same as [math]\sqrt{2}[/math] in feet... as you see, the contrasting coefficients are 12 and 1. Makes sense

That's what Wolfram Alpha will give you if you type in 1/0. The thing is, infinity's not always considered an actual number, let alone an imaginary/complex one. Most of the time it really isn't a number, depending on which area of math you're using it and the number fields used therein. But merely a mathematical concept, meaning "limitlessness", e.g. in limits, as the graph approaches a certain point, it may be going to infinity. It can't actually go to infinity, since you can't calculate infinity in normal equations like [math]f(x)=\dfrac{x^2}{x+1}-x[/math]. Since infinity is the concept of "without bounds", the graph will go up to infinity, without bounds, not to an actual "infinite" number. It's all okay as long as you don't take it to the realm of elementary algebra/arithmetic. Here, it will cause all sorts of anarchy. So numerical infinities are considered illegal there.

Hey ZeroZero Let's take a look at what you said... (if you find the math below intimidating, don't worry; it always looks scary when it actually represents a simple geometric situation) The first right triangle has legs of 12 units and 12 units. The second right triangle has legs of 1 unit and 1 unit. Now, these triangles are not the same, as you said. They do not have the exact same dimensions. But, they are similar, meaning their dimensions are proportional. You're implying that they're congruent, which in this case would result in [math]12=1[/math] D: There's a theorem in geometry that says that if two triangles have two similar legs and one congruent angle, then the triangles are similar. In other words, all you have to know are the 2 legs and the 90 degree angle and if those are proportional, then the triangles are also proportional (you don't need to know the other two angles or the length of the hypotenuse; that's it!) [math]a^2+b^2=c^2[/math] is the Pythagorean Theorem, as you probably know. However, since you've picked the interesting property that the legs are equal to each other, we can make this: [math]a^2+a^2=c^2[/math] since [math]a[/math] equals [math]b[/math] in this case. Then we further simplify to [math]2a^2=c^2[/math]. Now here's the important part... Look at [math]2a^2=c^2[/math]. We can take the square root of both sides and make it [math]\sqrt{2a^2}=\sqrt{c^2}[/math]. On the right side, it would just simplify to [math]c[/math], since the square root of a number is squared is just that number. On the left, with a little algebra rule, we know that [math]\sqrt{2a^2}[/math] is the same as [math]\sqrt{2}\times\sqrt{a^2}[/math]. (again, the [math]\sqrt{a^2}[/math] just becomes [math]a[/math]) So now! We're left with [math]a\sqrt{2}=c[/math]. Remember, [math]c[/math] is the length of the hypotenuse. So basically, [math]c[/math] is just equal to the length of one of the legs times the square root of 2. [math]\sqrt{2}[/math] is irrational, and any multiple of it is also irrational. So no matter what integers you have for the length of the legs, the answer will be irrational. The first triangle you showed has an irrational hypotenuse, just like the second one. It's not that the first was fine and computable while the second one wasn't, both of them are irrational and "not computable" in that way. The 12-12-? triangle has a hypotenuse of [math]12\times\sqrt{2}[/math] (as seen from the simpler equation we made). Yes, it's approximately 16.97056...etc. However, [math]\sqrt{2}[/math] is no problem either; it's approximately 1.414. Don't worry, there's no Pythagorean conundrum. Sorry if this was much too long, but I felt that too many people get confused with algebra on the square root part, so... hoped this help!

John. Thanks for the constructive statements. "Infinity is not the multiplicative inverse of zero". Yes. You're correct; sorry for the misconception. It is, however, in the way I defined infinity (you could accept it or reject it, it's just my personal speculation outside the respect of established mathematics). Your statement's true, because in the RNS, no number times zero equals one. [math]f(x)=x^{-1}[/math] shows that [math]f(0)=?[/math] undefined, as there is no limit as the function increases/decreases indefinitely to [math]\pm\infty[/math]. That's why I thought the inverse of zero is infinity (+- infinity actually), which is why I looked into this, and made the proof for 0 times ∞ equals the real numbers. And looking upon even weirder properties of infinity I haven't discussed here; I concluded that it all makes sense in its own right, intuitively and with the help of new definitions. Lol. Thanks, I guess. And there's nothing wrong with reality TV shows

What? Not sure what you're saying. Whatever the statement, why not? You have to understand that in this case we're defining a new value with its own rules. You can't negate the proof with rules which only apply to the real numbers. How does it break the definition? Infinity is still [math]\infty>n, n\in\mathbb{R}[/math]. Again, we're not including [math]\pm\infty[/math] in the RNS, because 1) for convenience I won't change the definition of the RNS (thus changing the cardinality of the set and completely overturning Cantor's work on its head), and 2) it has special properties not universally granted to the real numbers. However, I do want you to think it's "real" in the sense that it's not an imaginary, complex, hyperreal, etc. number. It still has the same essence as 1, 2, pi, e, whatever. I never defined this value as less than infinity. It is still infinite, and it is still greater than all real numbers. However, it has the interesting indeterminate property that no matter what real number you add or subtract to it, it remains infinity (it does not change, greater or lesser). This is solely based on the inductive idea that any finite addition to infinity would still keep infinity, infinity. [math]1+\infty=\infty[/math] is what I defined. Of course, as I said, this is an indeterminate form. You cannot reduce it by subtracting infinity from both sides. Again [math]1+\infty\not>\infty[/math] but equal to infinity. BTW, though infinity doesn't work well in elementary arithmetic of the real numbers, there are standard, accepted indeterminate forms of expressions which involve it. (also zero, which is strange, since the multiplicative inverse of zero is infinity). [math]\dfrac{0}{0}, 0\times\infty, \infty-\infty, 0^0, 1^\infty[/math]

Well this is of course speculative territory. And I'm also not sure why "reality" plays any part into this, since mathematics is abstract. Infinity does actually work in some fields of standard math, the weirdest involving real projective geometry. [math]\sqrt{-1}[/math] also used to have no valid numerical value, but now it is standard mathematics. Not that this is necessarily going to become universally standard, but the point is when we're heading into forbidden land, we have to make new laws so we don't break any rules and run into problems.

John (edit: sorry for the broken tex quotes below, not sure why it does that!) As I said, let's assume that there exists a valid, numerical value for infinity (but not call it necessarily a real number, more like a special value in an extension of the RNS). What happens when we add 1 to infinity? Or 2? Or 3? Well, of course this would make no sense in the, should I say, standard view of mathematics, but let's say it still remains infinite. Because infinity plus 1 should still be infinite. And we can apply this to all real numbers (again + and - infinity are excluded and treated as valid values exceptional properties). I defined [math]n\pm\infty=\pm\infty, -\infty<n<\infty[/math] where the positive/negative are independently respective. For the finite real numbers, any number times 0 is zero, but infinity's different in this case, since remember that [math]0\times n=n-n[/math] making [math]0\times\infty=\infty-\infty[/math], which would be satisfied by any real number (again, excluding the special values of +/- infinity). Yes, sorry for that. I was referencing to the property I defined in the beginning... I meant [math]n\in\mathbb{R}[/math]. And I don't find this to in any way demote the credibility of my (quite unorthodox) proof; I hope you don't see it like that. 0 is a satisfying value, but as I showed above, it could also be 1, -1, googol, e, pi ... in an indeterminate form, like 0/0. When I made that statement, I was referring to the idea that "infinity is merely a concept, and makes no sense in arithmetic, like 5+justice". I just meant that we should accept it (for the purpose of this proof) with a valid mathematical meaning. I didn't necessarily mean to include it with the real numbers in the sense that it would have the same universal properties of real numbers. I do hope that it would be seen as a real number in this proof (so it would seem comparable to the finite reals and not abstract), but of course with different rules. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Demonstrated as speculated. I really don't remember where I got that from But it seemed to fit. Sorry for your italics problem. I hope you get better. Lol. Thank you Thanks I'm pretty sure it's not about "lack of interest". I think any mathematician would be exceedingly interested to expand infinity to areas of math where it is forbidden (in a figurative sense). And yes, it's the contradictions. [math]0\times1=0\times2[/math], by "simplifying out" the zero you would get 1 = 2. Which, since 0 times any real number is 0, would make every number in the real number system equal to each other, causing a collapse of mathematics so disastrous as to cause an annoying internet meme of nuke-like explosions, bottomless pits in the ground, and black holes. Really, I don't think that argument is very convincing, since pretty much anything involving zero and an arbitrary value also involves indeterminate forms. It's a given that you will get different values equal to each other if you try reducing the equation. It's as valid as this... [math]x^2=9[/math] If I square root both sides, I'll get roots of 3 and -3. Does that mean I could say "Well, that would make 3=-3. Meaning every real number is equal to its negative. That doesn't make sense, so square rooting is undefined.[/math] In fact, this argument can be applied to any equation where two different values share a common function. That does not make them equal though, most especially in the case of indeterminate forms. As for usefulness, I'm certain we'll never use infinity in our everyday lives. Infinity itself now is useful enough for what we have in limits, sums, fractals, etc. However, if we allow it to be a tangible value it might help in describing certain things. E.g. An infinite amount of nothing is something. An infinite amount of 1-dimensional lines (no area) makes up a 2-dimensional plane (an actual finite area).

[math]0\times n=n-n[/math] by the iterative definition of multiplication. True for all numbers. Assume the existence of a valid, numerical infinity, as well as obvious properties like [math]n\pm\infty=\pm\infty[/math] positive/negative respectively. True for all real numbers. Now we want to find the value of [math]0 \times \infty[/math]. Let's call it [math]x[/math] [math]0\times \infty=\infty - \infty[/math] [math]x=\infty-\infty[/math] [math]x+\infty=\infty[/math] [math]x\in\mathbb{R}[/math], since we inductively established the postulate [math]n+\infty=\infty, x\in\mathbb{R}[/math]. In conclusion, [math]0\times\infty=\mathbb{R}[/math] indeterminate. Demonstratur ut putantur. Of course, this is only valid by establishing new grounds. Infinity itself isn't even a number in most cases, so why not go this far? Still, it's all meaningless without formally accepting infinity and solving the consequential contradictions we run into. And that would be pointless, since it wouldn't really help us in any way.

Zero is not a number? One could speculate such off the fact that it represents an empty set. However, this does not violate its status as a number. By the Peano Axioms, zero is a number (specifically stated in the axioms as a natural number). And upon what is mathematics analytically founded on? Axioms. Which is also the reason why one cannot truly define [math]\dfrac{1}{0}[/math]. I have to admit, it does seem as though the value is infinity. But by assuming division by zero (for the dividend is not zero itself) at all is a legitimate operation, we run into serious problems. (My first real post by the way. Hi everyone ) EDIT: Oops. Fixed LaTex tags. Used to other forum.

Hello everyone! New here. About me... Hmm, I love mathematics (I'm no expert though) and I'm 15 years old (I'm not an awkward nerd though, I act different than how I think. Haha) ... and I'm in high school. I've been on one other science forum in the past. I hope to learn from and contribute to this forum in the time I'm here. Yeah, this username is probably too long, and in the other forum I had an incredibly (stupidly) simple username. But it fits, I guess. Hope to gain progress with you all. Bye