• (disco) in reply to David_C
    David_C:
    the two cats would work together.

    It's almost as if they were.... organized!

    Filed Under: I won the Buckeye Newshawk Award!

  • (disco) in reply to David_C
    David_C:
    two cats would work together

    That alone proves evolution.

  • (disco) in reply to xaade

    Or homes where the litter box is in the bathroom.

  • (disco) in reply to xaade
    xaade:
    .9999.... cannot be properly expresses in finite space. Therefore we chose to represent it as 1.

    0.999... can be properly expressed in finite space. You just did, twice. 0.999... and 1 are two finite representations of the exact same number.

  • (disco) in reply to YellowOnline
    YellowOnline:
    water on your toothbrush before or after putting the tooth paste?",

    Um. Neither.

  • (disco) in reply to David_C
    David_C:
    But if you have pet cats, that can be an invitation for them to unwind the entire roll onto the floor. When it hangs on the wall-side of the roll, a cat pulling on the front of the roll won't end up unwinding anything.

    My dog (when he was still a puppy) discovered the joys of toilet paper... Let's just say the roll was destroyed, not just unrolled. And his poop was a little more textured for a while.

    The roll now sits on the window sill.

  • (disco) in reply to RaceProUK
    RaceProUK:
    Assuming an infinite recurrence of 9s, you can prove that using logarithms

    Or simple algebra. But I'm happy to have hooked 2 people so far. :fishing_pole_and_fish:

  • (disco) in reply to anotherusername

    Well, I mean, representing all the digits distinctly.

    Anything can be represented in finite space.

    (/) = 2304897203874019472097610289470194279174.1249147202049871042981048928738701 + pi + 99.9999... + 2/39.

  • (disco) in reply to boomzilla
    boomzilla:
    Or simple algebra. But I'm happy to have hooked 2 people so far. :fishing_pole_and_fish:

    Including my beloved. But then she did walk straight into it ;)

    Since this thread is now about maths, I found this: [image] Can anyone spot the issue with that proof? Other than the fact it's bollocks?

  • (disco) in reply to RaceProUK

    4(x) = 3(x) x = 0

  • (disco) in reply to Magus

    No. Bloody close though ;)

  • (disco) in reply to RaceProUK

    That's definitely a problem with it. There's no way three of something is the same as four of that thing unless that thing is zero, or you've overridden the + operator.

  • (disco) in reply to RaceProUK

    I'm having trouble with step 1. How do you figure 4*a - 3*a = a? To my knowledge the only value of a that satisfies this condition is 0.

    Oh derp, Hanzo'd again, courtesy of @Magus.

    Why do I insist on writing replies before reading the rest of the thread?

  • (disco) in reply to Tsaukpaetra
    Tsaukpaetra:
    I'm having trouble with step 1. How do you figure `4*a - 3*a = a`?
    Basic algebra.

    Also, I got the 'proof' from here.

  • (disco) in reply to RaceProUK

    Yeah, the second and third step aren't even important. If a + b = c, a + b - c = 0.

  • (disco) in reply to RaceProUK

    Hmm.... Yeah, I can't determine a level 1 algebraic rule from the primary algebraic interpreter dataset that allows splitting an arbitrary variable in this manner. The best rules would be:

    • 5G$2-2.4, which stipulates that a unary multiplicative group with no static value specifier (i.e. a) can be represented with the static value of ?Unit (i.e. 1) like so: 1*a.
    • 5G$119-1.1, which stipulates that for a unary multiplicative group with a static value, the static value can be split into a sub-expression so long as said sub-expression does not contain any dynamic values and returns the same static value it's replacing (i.e. 2 can be replaced by the sub-expression (1+1)).
    • 3B$12-19.3, which stipulates that a sub-expression polynomial can expand out of a sub-expression so long as the remaining portions of the unary multiplicative group it is applied to are replicated along with it (i.e. (1+1)*a can become 1a+1a).

    At best, this allows us the following:

    a+b=c
(4-3)(a+b)=(4-3)(c)
4a-3a-3b+4b=4c-3c

    Which, (If you decide to combine terms under rule 4A$1-10.33) simply results in a+b=c If not, well you can't eliminate a group of dynamic values in a sub-expression out in one step (Forbidden under 4A$2-15.3, sub-expressions in a unary multiplicative group cannot be eliminated from an equation by means of division or multiplication unless all terms in the subexpression are represented by static values), but if you could:

    4a+4b-4c=3a+3b-3c
4(a+b-c)=3(a+b-c)
4/3=(a+b-c)/(a+b-c)

    Assuming a+b-c is non-zero, this means that 4/3 = 1 :interrobang: New proof? Nope, because we broke a rule. :stuck_out_tongue:

  • (disco) in reply to Tsaukpaetra
    Tsaukpaetra:
    a+b=c (4-3)(a+b)=(4-3)(c) 4a-3a-3b+4b=4c-3c

    Which, (If you decide to combine terms under rule 4A$1-10.33) simply results in a+b=c

    You've actually hit on the semantic trick here, at least in regards to the first transition. As given in the first transition of the original, it is violating the rule of equalities. If you have a + b = c, and wish to arbitrarily distribute a pair of values across it, you would have the transitions

    a + b = c 1(a +b) = 1c, x - y = 1 (x - y)(a + b) = (x - y)c

    which is, as you pointed out, a tautology. but the original was based on the assumption of

    1(a +b) = 1c, x - y = 1 (x + y)(a + b) = (x + y)c

    which is invalid. (The specific sign, plus or minus, is irrelevant, you could have written it as

    1(a +b) = 1c, x + y = 1 (x - y)(a + b) = (x - y)c

    and the incorrect transition would still be there. It's the change of sign from the underlying equality that is at fault).

    Since it did not show that explicitly, but skipped to the transition

    xa + ya + xb + yb = xc + yc

    with the incorrect transition cascading through it, it gives a transition that looks correct but is actually invalid from the original rule of a + b = c.

  • (disco) in reply to ScholRLEA

    Wow, the error happened earlier than I thought then? Guys, this is why I don't do math for my job; the company would go down in my meager hands...

    Instead, I'm just doing database and programming. No math to be done here!

  • (disco) in reply to ScholRLEA

    Where did you get the change of sign from?

    (x - y)(a + b) = xa - ya + xb - yb

    does it not? And if you say x = 4 and y = 3, you get

    (4 - 3)(a + b) = 4a - 3a + 4b - 3b

    which is in the original.

    So far, @Magus is the closest; he got the whole a + b - c = 0 thing. Just one more step, and he'd have the answer…

  • (disco)

    /schol-r-lea kicks self Damn, I see what I did wrong, yeah. I misread the original, somehow. Sorry about that.

  • (disco) in reply to RaceProUK
    RaceProUK:
    Can anyone spot the issue with that proof? Other than the fact it's bollocks?

    As established earlier, a + b - c = 0.

    So, 40 = 30 is true.

    But then you divide both sides by (a + b - c), which is dividing by zero. :facepalm:

  • (disco) in reply to xaade

    1 represents all the digits of 0.999... distinctly.

  • (disco) in reply to RaceProUK

    a + b - c is zero.

    4(a + b - c) = 3(a + b - c) 4 * 0 = 3 * 0 0 = 0

    You cannot reduce it to 4 = 3 because that would require dividing by zero. Actually, as soon as you get to 4(something) = 3(something) you have proven that something has to be zero.

    ScholRLEA:
    You've actually hit on the semantic trick here, at least in regards to the first transition. As given in the first transition of the original, it is violating the rule of equalities. If you have a + b = c, and wish to arbitrarily distribute a pair of values across it, you would have the transitions

    a + b = c1(a +b) = 1c, x - y = 1(x - y)(a + b) = (x - y)c

    which is, as you pointed out, a tautology. but the original was based on the assumption of

    1(a +b) = 1c, x - y = 1(x + y)(a + b) = (x + y)c

    which is invalid.

    No. You don't even have to go that far in.

    4 - 3 = 1 (4 - 3)a = 1a 4a - 3a = a

    You can then substitute 4a - 3a directly for a in the equation, which is what was done. Similar to replace b and c.

    a + b = c 4a - 3a + 4b - 3b = 4c - 3c 4a + 4b - 4c = 3a + 3b - 3c 4(a + b - c) = 3(a + b - c) 4 = 3 (true only if a + b - c is not equal to zero)

    And there lies the rub, because a + b - c does equal zero, so you can't take that step.

  • (disco) in reply to anotherusername

    Actually it's a good proof of why 0/0 != 1, and why dividing by 0 is undefined.

    If dividing by 0 was allowed, then 0/0 is either infinity or 1 or 0

    If it were infinity, then 4 * infinity = 3, then infinity = 3/4, and consequently, infinity equals any number. If it were 1, then 4 = 3. If it were 0, then 4 = 0, 3 = 0. And consequently, everything = 0.

  • (disco) in reply to xaade
    xaade:
    If dividing by 0 was allowed, then 0/0 is either infinity or 1 or 0

    That does not necessarily follow. You could permit dividing a nonzero number by zero (resulting in infinity) but still let 0/0 be undefined.

  • (disco) in reply to anotherusername

    https://what.thedailywtf.com/uploads/default/original/3X/d/a/da4016d432917d89f70b06659f68790a31694c21.png

    4(a + b - c) = 3(a + b - c) 4(0) = 3(0) 0(4/0) = 3 0 * infinity = 3 0 = 3.

  • (disco) in reply to xaade

    0 * infinity is not defined. Any number times infinity is not defined in standard algebra. What's 0.5 * infinity? Less than infinity but still infinite? How about 2 * infinity? Is that larger than infinity?

    Actually infinity is not a number. You can't multiply by it. It's just sometimes used to represent an infinite result when a number was expected.

  • (disco) in reply to anotherusername

    But that means that it makes no sense to allow infinity to be introduced into an algebraic expression.

    So defining x/0 = infinity, just creates more problems.

    1 + infinity? 1 - infinity? etc.

    You just made the expression meaningless.

    One more reason to add to the "math isn't real" pile

  • (disco) in reply to xaade

    You're right, it can't be introduced into an algebraic expression.

    x/0 isn't literally equal to infinity, it's just convenient shorthand to say that it is because the result of the division would be infinite (assuming x isn't 0; if x was 0, then the result would be undefined). You can't use it in any algebraic expression though. None of the standard arithmetic operators have any meaning when used with infinity; like I said, it's actually not a number at all.

  • (disco) in reply to anotherusername

    You can introduce it, if you define things properly. You do inevitably lose some of the universality of algebraic laws, but it can be done and is useful in some contexts.

    If you're dealing with non-negative numbers, you can introduce an ∞ such that x/0 = ∞ (for x > 0). To make things work in the most sensible and useful way, you need to have x + ∞ = ∞ for any x (including x = ∞) and x * ∞ = ∞ for any x>0. You still can't sensibly define 0*∞.

    Another example is the extended real line, which is the set of all real numbers together with {-∞, +∞}. Operations are defined as sensibly as possible, to whit: x + ∞ = ∞ (for any x > -∞) x - ∞ = -∞ (for any x < ∞) ∞ - ∞ is, of course, undefined. x * ∞ is ∞ if x > 0, -∞ if x < 0, undefined if x = 0. x / 0 is undefined

    Remember, what is or is not a number is entirely a consequence of what set of numbers you want to work with. In some contexts ∞ is a number, in some contexts (albeit the more common ones) it is not.

  • (disco) in reply to xaade

    That does not necessarily follow. Consider exponentiation (0^b=0, a^0=1, 0^0=?)

  • (disco) in reply to redwizard
    redwizard:
    dividing by zero
    [image]

    That is the answer!

  • (disco) in reply to RaceProUK

    Dividing by zero is almost always the answer in "0=1" style proofs. (I'd estimate at least 95% of the time, probably more like 98%.) Occasionally you see one that uses x² = y² => x = y when really x = -y. I can't remember any other tricks I've seen for it.

  • (disco) in reply to Scarlet_Manuka
    Scarlet_Manuka:
    Dividing by zero is almost always the answer in "0=1" style proofs.
    True, but if you don't know that, it's surprisingly hard to spot it
  • (disco) in reply to anotherusername
    anotherusername:
    You could permit dividing a nonzero number by zero (resulting in infinity) but still let 0/0 be undefined.

    That's exactly the way it is done in function theory.

    I put in the enthymeme in xaade's post:

    xaade:
    If dividing by 0 was allowed for all numbers, then 0/0 is either infinity or 1 or 0

    It follows that we can't define 0 / 0, 0 * ∞, ∞ / ∞, ∞ + ∞, and similar.

  • (disco) in reply to xaade
    xaade:
    If you have cats that can open doors.... you should look around for some way to profit off of that.

    Both of my cats can, I'm eager to hear of a way to get rich quick I'm apparently missing.

    xaade:
    residents don't know how to close doors

    Not sure how cats shitting on the floor outside the toilet would help.

  • (disco)

    There is another approach to give a value for 1/0 that is a bit easier to spot, provided you know the difference between variables and functions:

    We start with the Euler number e.

    Any number x divided by 1 is x, therefore

         e
e = ---
     1

    Any (real) number x > 0 fulfils the identity

    x = exp ln x

    Therefore

         exp ln e
e = ----------
     exp ln 1

    (here should follow some hocus-pocus stuff to distract the reader)

    Now we know that ln e = 1, and ln 1 = 0.

    Therefore

         exp 1
e = -------
     exp 0

    (better a more complicated expression, that will be reduced to this one.)

    As we know from function theory, exp never equals zero. (Or draw its graph etc.)

    Thus, we can cancel exp:   exp 1 e = -------   exp 0

    That means,

         1
e = ---
     0

    The problem with this is, anyone capable of understanding exp and ln, should be capable of understanding the concept of functions. That doesn't keep some students from trying to cancel function names like sin, exp, etc.

  • (disco) in reply to xaade
    xaade:
    So defining x/0 = infinity, just creates more problems.

    that's why it's not defined. there is no way to properly define it.

    you can define "the number that this formula approaches as you reduce X arbitrarily close to 0" and that can be infinite for some formulae, it will be -infinity for others. still others it will be perfectly mundane numbers like 0 and 12 and -7 and 42

  • (disco) in reply to anotherusername
    anotherusername:
    You just did, twice. 0.999... and 1 are two finite representations of the exact same number.
    [Or][1]:

    \begin{aligned} 0.\overline9 \end{aligned}

    Or:

    \begin{aligned} 0.\dot9 \end{aligned} Or:

    \begin{aligned} 0.(9) \end{aligned}

    :imp:

  • (disco) in reply to tarunik
    tarunik:
    That would be an interesting optimization question for our resident compiler experts. @dkf?

    You could, but you probably wouldn't bother for most languages; the effort to implement it would usually not be justified. And if it is just so that you can do comparisons against literals at compile time in the first place, you don't need an RB-tree; a simple list of comparisons will do.

    Assuming I've not totally misunderstood what you're asking about. :smiley:

  • (disco) in reply to accalia
    accalia:
    that's why it's not defined. there is no way to properly define it.

    Most modern mathematics defines x/0 for x not equal to zero. You can determine this by looking at the family of graphs for K/x, where for K non-zero, the behaviour as x tends to zero or infinity is an asymptotic approach to one of the axes.

    On the other hand, 0/0 is completely meaningless. It is any value at all and none. If someone is using that, they're writing something meaningless. If you have it in your program, it's one of the things in IEEE arithmetic that properly produces a NaN (or an exception; the spec is very fuzzy about what happens in this case).

  • (disco)

    https://www.youtube.com/watch?v=F0ov4sv9wPI

    MATH SUCKS!

  • (disco) in reply to dcon

    Continuing the discussion from Listicle:

    dcon:
    David_C:
    But if you have pet cats, that can be an invitation for them to unwind the entire roll onto the floor. When it hangs on the wall-side of the roll, a cat pulling on the front of the roll won't end up unwinding anything.

    My dog (when he was still a puppy) discovered the joys of toilet paper... Let's just say the roll was destroyed, not just unrolled. And his poop was a little more textured for a while.

    The roll now sits on the window sill.

    I knew a great dane who ate a pair of tights (= panty-hose) once. That was fun when it emerged at the other end. I laughed till I puked.

  • (disco) in reply to Matt_Westwood
    Matt_Westwood:
    Continuing the discussion from Listicle:

    No doubt.

  • (disco) in reply to xaade
    xaade:
    Actually it's a good proof of why 0/0 != 1, and why dividing by 0 is undefined. If dividing by 0 was allowed, then 0/0 is either infinity or 1 or 0
    0/0, using basic math is undefined.

    Using calculus, however, 0/0 can have a specific value, given the equations that produced the two zeros.

    For instance, if you have the equation y = 2x (plotting a straight line), then y/x will represent that line's slope (1/2). When x=0, y/2x is 0/0, but the slope of that line at x=0 is still 1/2 - it's not undefined for that specific point. Basic calculus resolves the dillema. When you have a 0/0 situation, you take the derivative of the terms that generated the numerator and denominator and divide them to get the value. In this case the derivative of y/2x is 1/2 - a constant, and the slope of the line, as we intuitively know it must be.

    For more complicated equations, there may be multiple values of x and y which produce 0/0. The derivatives will produce equations that may produce non-zero values for those values of x and y that produce 0/0 in the original equation, and that would be your answer. Or the first derivative doesn't then the second derivative might, etc.

    Of course, there are some systems where you can never get rid of the 0/0, no matter how many derivatives you take. These are typically curves where there is a sharp angle or a discontinuity, and therefore a place where the slope is undefined. For instance y=|x| (the absolute value of x (producing a V shape plot with the point at 0,0.) At the point x=0, y=0, the slope really is undefined, because it's a point. All values to the left of 0 have a slope of -1, and all values to the right of 0 have a slope of 1, but the slope is undefined at 0. If you take the derivative of y/|x|, you end up with 1/(x/|x|) - with another 0/0 where x=0. If we try to take the derivative of that, we end up with dy/(1/(x/|x|)) - with yet another 0/0, ad infinitum.

  • (disco) in reply to David_C

    The joys of L'Hopital's rule...

  • (disco) in reply to Scarlet_Manuka
    Scarlet_Manuka:
    Occasionally you see one that uses x² = y² => x = y when really x = -y.
    More high school math. When there are powers involved, then there are multiple answers. If you have the system x² = y², and you want to solve for x, there are two answers: y and -y. Any answer that mentions only one is wrong.
  • (disco)

    Everyone knows that:

    [image]

    Divided by:

    [image]

    Is Zero and two Halves of Zero.

  • (disco) in reply to Magus

    Makes me want to make a comic where X trips on himself and gets cut in half by a saw blade, and then complains about his feet being massive.

    "Who made me this way" "Who makes something that has to walk with feet this big" "Why!?!?!!?!?"

  • (disco) in reply to David_C
    David_C:
    Using calculus, however, 0/0 can have a specific value, given the equations that produced the two zeros.

    That's not an accurate way to say it. 0/0 still does not have a specific value. It's undefined.

    Using calculus, you can sometimes find a limit at a point whose actual value is undefined. That's because a limit describes what an expression does as you get close to the point you're examining, not necessarily what value occurs at the point you're examining.

    In fact, the limit at the point doesn't even have to be the same as the value if the value is defined at that point. For instance:

    http://i.imgur.com/R6eU4eK.png

    This results in all 3 of the following being simultaneously true:

    http://i.imgur.com/01BxpN8.png

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