[HN Gopher] A visual proof that a^2 - b^2 = (a + b)(a - b)
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A visual proof that a^2 - b^2 = (a + b)(a - b)
Author : beardyw
Score : 378 points
Date : 2024-12-15 14:01 UTC (8 hours ago)
(HTM) web link (www.futilitycloset.com)
(TXT) w3m dump (www.futilitycloset.com)
| ujikoluk wrote:
| That seems to show that there exist a and b such that the
| equality holds. But not that it holds for all a and b.
| wrsh07 wrote:
| Which constraints on a,b (besides positive) does this proof
| require?
| davrosthedalek wrote:
| b<a Edit: and b,a element of R, but ok...
| phoe-krk wrote:
| _> (besides positive)_
|
| You can chart a and b on a 2D coordinate system, where
| they're allowed to be negative. Even positivity is not
| strictly required here.
| wrsh07 wrote:
| I'm not sure I quite understand what the visual proof looks
| like if they're negative
| phoe-krk wrote:
| Check the quadrant which the result ends up in.
| xigoi wrote:
| a and b have to be real numbers, whereas the identity works
| for any commutative ring.
| martin293 wrote:
| And how exactly did you come to the conclusion that this is
| relevant here?
| davrosthedalek wrote:
| By the fact that the geometric proof in the link wants to
| proof the formula, but only does so for a small subset of
| all a,b for which the formula is correct. This makes it a
| partial proof, at best.
| martin293 wrote:
| Ok nvm I can't resist wasting my time and typing stuff on
| the internet again, probably gonna regret it later.
|
| How is it not obvious to the dullest of the dull that
| this visual proof is not supposed to work for goddamn
| commutative rings lmao
|
| It's probably not even supposed to work for negative
| reals, 0 or the case b>a. It's supposed to demonstrate
| the central _idea_ of the visual proof. Also yes, by
| choosing suitable ways to interpret the lengths shown in
| the diagrams it 's absolutely possible to extend the
| proof to all reals but I'm not convinced it's meant to be
| interpreted like that.
|
| But bringing commutative rings into this... man you're
| funny
| notorandit wrote:
| Really? It even holds true for either a=0 or b=0.
| Scarblac wrote:
| But not for b > a.
| wcrossbow wrote:
| Just rename a to b and b to a.
| kleiba wrote:
| Why the downvote? That's a correct argument.
| supernewton wrote:
| It is not. a and b are not symmetric in this equation,
| you can't just swap them.
| olddustytrail wrote:
| Of course you can. What do you mean?
| davrosthedalek wrote:
| 3^2-2^2 =!= 2^2-3^2.
|
| (You can exchange a and b in, say a^2+b^2, because
| 2^2+3^2=3^2+2^2)
| wcrossbow wrote:
| This is not what I meant. What is being proved is:
| a^2-b^2 - (a+b)(a-b) = 0. If you swap a and b you end up
| with a sign switch on the lhs which is inconsequential.
| davrosthedalek wrote:
| That is not what the proof proves. The proof proves the
| equivalence how it was originally stated, and assumes for
| that b<a.
|
| Your rewriting is of course true for all a,b and might be
| used in an algebraic proof. But this transformation is
| not at all shown in the geometric proof.
| olddustytrail wrote:
| Did you think that I meant you can switch them on one
| side of the equation but not the other?
|
| That's not what anyone is saying.
| davrosthedalek wrote:
| No, of course not.
| henry2023 wrote:
| You can swap them without loss of generality (WLOG).
| davrosthedalek wrote:
| No, this is not correct. WLOG means: I assume one of the
| possible cases, but the proof works the same way for
| other cases. But that's not true here. The proof, as
| shown, only works for a>b>0, it does not work (without
| extra work or explanation) for a<b. The proof for a<b is
| similar, but not the same. [And it certainly does not
| show it for a,b element of C]
| edflsafoiewq wrote:
| WLOG just means the other cases follow from the one case.
| There is no implication about how hard it is to get to
| the other cases, although generally it is easy and you
| don't bother spelling it out exactly.
| nuancebydefault wrote:
| Is that allowed? You will prove another equation. You
| cannot swap pi and e either.
| Ma8ee wrote:
| You are not swapping the values, you are swapping the
| names.
| kelnos wrote:
| pi and e aren't names; they're values. Of course you
| can't swap values.
| Ma8ee wrote:
| (a^2 - b^2) = -(b^2 - a^2)
|
| use the same visual proof but with a and b switched to get
|
| -(b + a)(b - a) = (a + b)(a - b)
| notorandit wrote:
| We need to introduce these simple demonstrations in math courses.
| Someone wrote:
| Aren't they already used in early high-school math books?
|
| They aren't used later on because, for more complicated
| expressions, manipulating the algebraic formulas is easier than
| the geometric method, or completely breaks down (e.g. when the
| powers aren't integers or are variables)
|
| For example, I think anybody who can visualize what (a+2b)3 or
| (a+b)4 looks like geometrically can also, and easier, do the
| expansion algebraically.
| pwg wrote:
| > Aren't they already used in early high-school math books?
|
| No, or at least they were not in the high-school math books I
| was assigned. There were no explicit referrals in any way to
| visual representations of the algebra involved.
| konschubert wrote:
| The expression immediately follows from
|
| a ( b+c) = ab + ac.
|
| I agree it's important that students intuitively feel that
|
| a ( b+c) = ab + ac
|
| is true.
|
| But once we are at that, why would we even bother to teach or
| "proof" all variations of that expression?
| GuB-42 wrote:
| Visual proofs can be deceptive. For example, you have the
| famous missing square puzzle, where rearranging pieces of a
| triangle causes a square to vanish, something that should be
| impossible.
|
| It doesn't mean it is not possible to have rigorous visual
| proofs, the one in the article is, but doing so can be
| deceptively hard. It is also common for visual proofs to be
| less complete, for instance, here, it only covers the cases
| where a>0, b>0 and a>b, there are no such limitations when
| doing so with algebra. I guess you can tweak the visual proof
| to account for these cases, but it will become far less
| elegant.
|
| So I understand why teachers avoid visual proofs in math
| classes, they actually want you to stay away from them. They
| have entertainment value, that's why you see them so much in
| pop-science, and I think it is important, and also some
| historical value, as they were much more present in ancient
| times, but to actually learn maths, not so much. That is unless
| you are studying visual proofs specifically, but I think it is
| an advanced topic you would tackle well after you can master
| simple equations like a^2 - b^2 = (a + b)(a - b).
| FartyMcFarter wrote:
| This reminds me of this video about why we need to be very
| careful when inspecting visual proofs:
| https://www.youtube.com/watch?v=VYQVlVoWoPY (it includes a
| "proof" that pi equals exactly 4).
|
| In this case, as someone else pointed out below, this proof has
| unjustified assumptions in it (at least it assumes that b < a).
| orangepanda wrote:
| It still holds if you're not bothered by negative-area
| rectangles
| FartyMcFarter wrote:
| But I definitely am bothered.
|
| A visual proof is supposed to appeal to our visual intuition
| - I don't know about you, but negative areas are not
| something that is visually intuitive to me.
| Out_of_Characte wrote:
| Negative area's can be the only practical solution. Imagine
| the worst case: a 10*10 room with a pillar in the middle.
| How would you calculate that?
| FartyMcFarter wrote:
| I don't see that as negative area. I see it as
| subtracting two areas. Both the room and the pillar have
| a positive area, none of them has negative lengths or
| areas.
|
| Kids learn subtraction before they learn negative numbers
| - once you learn negative numbers, you know that addition
| and subtraction are almost interchangeable, but this is
| not necessarily intuitive to begin with.
| ses1984 wrote:
| Subtracting two areas is like adding two areas where one
| is negative...right?
| FartyMcFarter wrote:
| Yes , but see my second paragraph. Just because something
| is true it doesn't mean it's intuitive.
| trimethylpurine wrote:
| That's interesting. To me it seems intuitive. It's a real
| area that can be drawn like any other. The sign is an
| operator describing what function to visualize, not a
| property of the measured area. So thinking of it in that
| way eliminates any need for the term "negative area."
|
| But, intuition is subjective, so you may need to adjust
| the terminology to fit the visualization.
| FartyMcFarter wrote:
| It is intuitive on some higher level, now that I know
| about signs, operators, negative numbers and all that.
| But when talking about visual information (i.e. "visual
| proof"), it isn't intuitive in that context.
|
| In addition to that, for all I know there could be some
| pitfalls involved with negative areas which I'm not aware
| of. Even if there aren't any pitfalls, this isn't
| immediately obvious to someone who isn't familiar with
| the concept of negative area.
|
| If I'm willing (or forced) to think in such abstract
| terms, I would much prefer an algebraic proof to this
| visual proof.
| ses1984 wrote:
| Your intuition is irrelevant if they're mathematically
| equivalent.
| jacobolus wrote:
| There are no serious pitfalls with oriented areas. Adding
| them to your arsenal of geometric proof tools will
| greatly simplify many proofs. Not having such a concept
| makes ancient geometry books much more complicated than
| they need to be, often requiring lots of detailed case
| analysis where the separate cases are essentially the
| same, just oriented opposite ways.
| wruza wrote:
| It's intuitive, just in more dimensions. People have
| different ideas and abilities to imagine/think
| dimensions, but on top of that we rarely train them to do
| that.
|
| I think that build up to tensor fields should be in every
| school program. If you can't think of a field, you're
| mathematically disabled and too many basic ideas about
| real world are inaccessible to you. This limits the
| ability to vote on a set of topics and participate in
| non-local decisions that involve systemic understanding.
| Same for formal logic and statistics.
|
| Once familiarized with that, you can easily start
| thinking of nonlinearly signed areas, complex areas and
| areas simultaneously positive and negative by an
| attribute.
| hansvm wrote:
| The concept of negative area still feels like it'd get
| messy in a hurry. For a square pillar, the side lengths
| should be the same, suddenly giving you imaginary lengths
| just for the eventual area subtraction to work out. For a
| negative volume though, you need cubic roots of unity for
| the side lengths, throwing off your area calculations.
| Has anyone actually put together a system where the sort
| of concept you're describing is cohesive?
| mtizim wrote:
| You're in for a nice trip, the concept is called
| Geometric Algebra:
|
| https://youtu.be/60z_hpEAtD8?si=HHs_9m0IJ43nfI3S (~50m
| video)
|
| TLDW: Yes, the concept is there, makes much more sense
| than a cross product (which is just an oriented area) and
| generalizes really nicely.
|
| Alternatively, read:
| https://en.m.wikipedia.org/wiki/Bivector
| mithametacs wrote:
| It seems like an elegant addition to the understanding.
|
| Maybe if you viewed an animation where `a` starts larger
| than `b`, and steps till it's smaller.
|
| Then you could see where the negativity happens. And seeing
| is nice.
| thaumasiotes wrote:
| If you're bothered by the idea of negative-area rectangles,
| there's no need to justify the assumption that b < a,
| because that's the only way you can assign any meaning to
| either side of the equation.
| FartyMcFarter wrote:
| The equation isn't about rectangles at all, so I
| disagree. The rectangles only exist for the purposes of
| this proof. Any resulting complications are on the proof,
| not the reader's burden.
| kleiba wrote:
| _> at least it assumes that b < a_
|
| But you can fairly assume that wlog.
| im3w1l wrote:
| No
| umanwizard wrote:
| Why not? Either b < a, b = a or b > a. If b = a both sides
| are zero and the result is trivial. If b > a just swap the
| names of a and b and multiply both sides by -1.
| im3w1l wrote:
| With this additional justification I accept it, yes. But
| (imo) now you are doing algebra. And then you might as
| well prove it just by distributing the RHS of the
| original expression.
| dagss wrote:
| This additional justification is so trivial it's just
| left out for brevity. In general a lot of statements are
| accepted even if not each and every detail is spelled
| out.
| davrosthedalek wrote:
| Sorry, but no. First, "just switch a and b" is clearly
| wrong, and we had multiple people propose that (Without
| the adjustment of sign). So not that trivial, I would
| argue. At least not more trivial than the underlying
| equality one wants to prove in the first place.
| im3w1l wrote:
| Let me fully spell out the proof that generality is not
| lost
|
| Assume that b > a.
|
| 1. Swap the names a and b.
|
| b^2-a^2 ?= (a+b)(b-a)
|
| 2. Multiply both sides by minus one
|
| a^2-b^2 ?= -(a+b)(b-a)
|
| 3. Absorb the minus into the second factor
|
| a^2-b^2 ?= (a+b)(a-b)
|
| This is the same as the original equality we wanted to
| prove.
|
| Now compare that to the purely algebraic proof for the
| whole theorem.
|
| 1. Distribute over the first parenthesis
|
| (a+b)(b-a) = a(a-b) + b(a-b)
|
| 2. Distribute again
|
| (a+b)(b-a) = a^2-ab + ba-b^2
|
| 3. Cancel equal terms
|
| (a+b)(b-a) = a^2 - b^2
|
| The proof that generality is not lost is of similar
| length to a full proof of the theorem!
|
| So if you think the former can be skipped, then you must
| accept a proof of the whole theorem that simply reads
| "Follows from trivial algebraic manipulation"
| wrsh07 wrote:
| Isn't this resorting to algebra to fix the visual proof?
| How do you fix it visually?
| olddustytrail wrote:
| Turn your monitor sideways?
| bmacho wrote:
| You do it 3 times, a=b, a<b, a>b.
|
| .. now that I think about it, the "visual proof" only
| 'proves' the statement for a specific 'a' and 'b'.
| Probably there is a proof, that can handle all 'a' and
| 'b' pairs at once.
| stephencanon wrote:
| The purely geometric proof works just fine without the
| case analysis if you use the oriented area.
| cowsandmilk wrote:
| What is a is positive and b is negative? How do you
| extend the visual proof?
| jacobolus wrote:
| You can do this if you extend your concepts of length and
| area to signed quantities. You have to be clear to
| explain how the setup of the drawing works, but instead
| of cutting away a square of side _b_ out of the corner of
| the square of side _a_ , you might end up appending a
| negatively-signed square of side - _b_.
| lizzas wrote:
| I am not convinced since in this world you can have the
| following rects:
|
| 1 x 1
|
| 1 x -1
|
| -1 x -1
|
| Is just the middle one a cut out?
|
| I reckon you need the axiom from the field to do
| generalize this to negative numbers.
|
| https://en.m.wikipedia.org/wiki/Field_(mathematics)
| chisquared wrote:
| There's no need for this extension:
| https://news.ycombinator.com/item?id=42425681
| chisquared wrote:
| If b is negative, define c = -b, and use the linked
| result to show that
|
| a^2 - b^2 = a^2 - (-c)^2 = a^2 - c^2 = (a + c)(a - c) =
| (a + b)(a - b)
| callingbull wrote:
| > just swap the names
|
| Then you've just skipped the case when a^2 - b^2 is
| negative. The diagram does not prove that case and
| swapping the names still doesn't prove it.
| shreyshnaccount wrote:
| if you just extend the metaphor in the diagram, and
| imagine a negative length to just refer to direction,
| sure it does :)
|
| personally, I love visual proofs because they can
| communicate an idea efficiently, sure they have their
| pitfalls, but its less about the actual mechanism of the
| proof and more about the core idea that lets me
| appreciate how its working- and visual proofs add a
| pseudo-physical intuition that helps me appreciate it.
| callingbull wrote:
| Trying the proof with a < b, with the b square from the
| bottom-right as in the diagram, I get a region to the top
| and left, and moving a piece (differently to the diagram)
| I get (a + b)(b - a) as a positive area for that region,
| and then flip the sign because it's negative.
| chisquared wrote:
| > Then you've just skipped the case when a^2 - b^2 is
| negative.
|
| Not really. If b > a, then swap them to conclude that b^2
| - a^2 = (b + a)(b - a), which is what the visual proof
| demonstrates.
|
| Your conclusion is equivalent to saying that a^2 - b^2 =
| (a + b)(a - b).
| karmakaze wrote:
| For those thinking this doesn't hold, note that "wlog" means
| without loss of generality (having numerous examples in
| proofs).
| lizzas wrote:
| A good example of the shortening of proofs. Like "Observe
| that... {unproven thing}"
|
| It is not necessarily bad but explains why every step isn't
| always proven. To avoid tedium and long papers.
| colejohnson66 wrote:
| Yes, but that's because it relies on infinite repetitions of
| the perimeter adjustment. Rearranging some boxes here does not
| involve infinity.
| amelius wrote:
| Here's a similar one:
|
| https://en.wikipedia.org/wiki/Missing_square_puzzle
| blablabla123 wrote:
| I vaguely remember from an Analysis lecture when Mathematical
| rigor wasn't such a thing, the professor mentioned a famous
| Mathematician (maybe Euler?) who had a notebook with hundreds
| of proofs where many of the visual ones didn't turn out right.
| Definitely nice for intuition though since this is oftentimes
| lacking
| TheRealPomax wrote:
| No, it assumes "one side is smaller than the other": the labels
| are arbitrary, so there iss no case where b is larger than a as
| we can just swap the labels and get a>b again. Even though you
| might think there are two cases that we need to look at because
| we're using two letters, there's actually only on case to
| demonstrate because we're demonstrating a property of the
| inequality.
|
| The only "but what if..." would be if a=b, which has no
| geometric proof, but also doesn't need one because "zero = zero
| times anything" is (by definition) true for fields.
| davrosthedalek wrote:
| The problem is not symmetric between a and b (because a-b
| isn't). You cannot swap them. (And even if, this swapping is
| not part of the "proof", so in any case the proof is
| incomplete.)
| xlii wrote:
| IMO GP is about the labelling: Instead of a2-b2 use
| longer2-shorter2 (and on visual representation one side
| will always be longer, as per GPs explanation).
|
| Diagram doesn't show proof for "shorter2-longer2". I
| believe showing negative area would spark more controversy
| (imagine that negative area is painted orange, visual area
| would still be positive).
|
| Shorter2-longer2 gives you same absolute value with reverse
| signed, so it feels symmetric to me (can't remember what
| the formal definition is), i.e: 1. a2 -
| b2 = ( a-b)(a+b) # * -1 2. -(-a2 + b2) =
| -(-a+b)(a+b) 3. -(b2 - a2) = -( b-a)(b+a) # / -1
| 4. b2 - a2 = ( b-a)(b+a)
|
| Edit: https://imgur.com/a/olETWfr - crude image with labels
| swap as it seems it's bringing some controversy.
| TheRealPomax wrote:
| It's fully symmetric. Basic arithmetic covers the following
| identities: 1. (a2 - b2) = -(b2 - a2),
| because of even powers 2. (a - b) = -(b - a)
|
| So the following two statements are the same statement:
| 3. (a2 - b2) = (a - b)(a + b) 4. -(b2 - a2) = -(b -
| a)(b + a)
|
| Let's assume this only holds for a>b (because we're content
| that the geometric proof shows that): 3a.
| (a2 - b2) = (a - b)(a + b), a > b
|
| But we don't know if it also holds for b>a... after all,
| how would you show a negative areas? What does that even
| mean Turns out: it doesn't matter, the b>a relation reduces
| to the same formulae as the a>b relation, so the geometric
| proof covers both. To see why, some more elementary
| algebra: we can invert both sides of (4), provided we also
| invert the relation between a and b, so this:
| 4a. -(b2 - a2) = -(b - a)(b + a), b > a
|
| Is the same as this: 4b. (b2 - a2) = (b -
| a)(b + a), a > b, by inversion
|
| Of course, algebra doesn't care about which labels you use,
| as long as the identities and relations between them are
| preserved, so we can swap "a" for "b" and "b" for "a" in
| both the identity and relation in (4b) to get:
| 4c. (a2 - b2) = (a - b)(a + b), b > a
|
| And we found a symmetry that we (maybe) didn't realize was
| there: 3a. (a2 - b2) = (a - b)(a + b), a >
| b 4c. (a2 - b2) = (a - b)(a + b), b > a
|
| Same formula, inverse relation. It turns out that it
| doesn't matter whether we start with a>b or b>a, they
| reduce to the same expression, thanks to those even powers,
| and a geometric proof for one is by definition a proof for
| the other.
| davrosthedalek wrote:
| That's the point. The geometric proof requires that you
| show the applicability for the b>a case with algebra. If
| you don't, it's not complete. And if you do, you can just
| show everything with algebra in the first place, and
| shorter, and also for a,b element C (instead of R), at
| the same time.
| TheRealPomax wrote:
| I'd have to disagree - it is perfectly fine to show a
| visual proof and simply _state_ that we can reduce b >a
| to a>b and this proof therefore covers both, optionally
| with a little "I don't believe you, show me the math" pop
| out.
|
| The visual proof is the neat part that people literally
| can't think of unless you show it to them, after which
| things might suddenly click for them. The algebraic proof
| is boring AF and doesn't make for a good maths hook ;)
| petermcneeley wrote:
| Blasphemy! The value of pi is 3
| https://religions.wiki/index.php/Biblical_value_of_pi
| ComplexSystems wrote:
| There is no problem with the proof except the assumption that
| the value in the limit is the same as the value at infinity. If
| we simply define pi(n) as a function from N U {inf}, which
| gives the value that "pi" takes at the nth step of the process,
| and pi(inf) as the value that it actually takes for the circle,
| then we simply have a function where lim n->inf pi(n) [?]
| pi(lim n->inf). For all finite n, it equals 4, and then at
| infinity it equals 3.1415... .
|
| There are ways to reformulate the above so that "infinity"
| isn't involved but this is the clearest way to think of it. It
| isn't much different than the Kronecker delta function
| delta(t), which is 1 at t=0 and 0 elsewhere. We have lim t->0
| delta(t) [?] delta(lim t->0 t).
| saghm wrote:
| My geometry teacher in 9th grade was adamant that we couldn't
| assume lengths and angles in diagrams except where explicitly
| labeled. In particular, he said that we should never assume
| that a diagram is drawn to scale; if the diagram happened to
| depict a quadrilateral as a square, unless something stated
| that it was a square (or we had sufficient information to
| determine that it was), we should treat it as an unidentified
| quadrilateral and proceed only with the actual information
| provided or else he would "take more points off than the
| question was worth" on tests. On one of our quizzes, he did
| provide a diagram that looked like a kite (two pairs of
| adjacent sides each with the same length but different length
| than the other pair) but listed the angles in a way that could
| only work for a parallelogram that was not a kite, and he made
| good on his word to take off extra points for people who
| misidentified it as a kite.
| trescenzi wrote:
| As one of the bad at geometry good at algebra people this blows
| my mind. I cannot even begin to comprehend how this shows, even
| for these specific boxes, the math works. But I can very clearly
| feel the relatedness of multiplication which makes the algebra
| work.
|
| That's not to say the example is bad, or good, more to marvel at
| how differently people think.
| umanwizard wrote:
| Which part are you struggling to understand?
| Jabbles wrote:
| In which one of the five images would you say your
| comprehension fails?
| dbsmith83 wrote:
| In the first image on the left, you can see the large square
| has length and width of a, which would have an area of a*a, or
| a^2. There is then a little square inside with length and width
| b, for an area of b^2. Essentially, the little square is
| getting removed from the big one (a^2 - b^2). In the last image
| on the right, you can see that the length of one side is (a-b)
| and the top side is (a+b), which would mean the area is equal
| to the product of (a-b)(a+b). This means that a^2 - b^2 = (a +
| b)(a - b). The intermediate steps just show how to move the
| area around visually
| trescenzi wrote:
| I think this helps the most. The piece that doesn't "just
| click" for me is that area = product of multiplication.
| Manipulating symbols generally just feels more natural to me.
| lupire wrote:
| What is area? Do you find the the basic concept of area (or
| the relationship between different dimensions in general)
| "doesn't click"?
| umanwizard wrote:
| Imagine you have 100 square tiles that are each 1cm x 1cm.
| You can make 20 groups of 5 tiles each with them -- this is
| what it means to say 20x5=100.
|
| Now take each of your groups and arrange its 5 tiles into a
| vertical line. Each line is now 5cm long and 1cm wide.
|
| Arrange the lines side by side. You now have a rectangle
| whose height is 5cm and whose width is 20cm. You already
| know its area 100 cm^2, because you made it out of 100
| tiles that were 1 cm^2 each. And now you can see that its
| area also corresponds to the multiple of its side lengths.
| jameshart wrote:
| It's worth rehearsing that instinct. The insight that areas
| are products is a really useful one for deepening your
| intuition of what integration is doing, as well as for
| enhancing your ability to interpret graphs and charts.
|
| Like, if you have a graph showing power consumption over
| time, it's great to be able to mentally recognize that,
| say, if the time units are hours and the power units are
| Watts, that the area under the graph will be counted in
| Watt hours; that a rectangle one hour wide by one Watt tall
| is one Watt hour, and so on.
| dagss wrote:
| Not sure if it helps, but trying: In a sense area isn't
| anything else than multiplication. It isn't like you have
| a) multiplication and b) area and then prove that a=b.
|
| Rather, area IS multiplication.
|
| The unit of "square meter" quite literally means "meter
| multiplied by meter".
| quietbritishjim wrote:
| I agree and I'd say that area is almost, in some
| intuitive way, the more basic thing and multiplication
| follows from that (although I know that's not
| mathematically true). The definition of multiplication
| for natural numbers is repeated addition (e.g. 3 x 5 is
| defined to be 5 + 5 + 5). Many people would see that as
| the count of a 3 by 5 grid of objects, and that's
| certainly how we'd explain the commutativity of
| multiplication in school. If those individual objects
| happen to be unit squares then you have area.
| nuancebydefault wrote:
| Any 100x100 puzzle holds 10000 pieces. Never counted them?
| /i
| hi41 wrote:
| This is so beautiful! I could have never imagined this. I learnt
| this formula by rote when I was in school. Didn't realize that it
| had a geometric equivalent. Same thing with differentiation and
| integration. Couldn't understand. Learnt that too by rote. Is
| there a geometric equivalent for most formulas if not all? Is
| there a website?
| palsecam wrote:
| There is https://en.wikipedia.org/wiki/Proof_without_words /
| https://commons.wikimedia.org/wiki/Category:Proof_without_wo...
| zozbot234 wrote:
| There is some geometric equivalent for any formula that can be
| built using some arbitrary combination of the four operations
| of arithmetic, plus square roots. If you allow for either 3D
| geometry or origami folds, you can extend this allowing for
| cube roots too. Note that the geometric equivalent is not going
| to be necessarily trivial or intuitive in most cases, though.
| wrsh07 wrote:
| I have really enjoyed reading some of the better explained
| pages. I wouldn't recommend most people start on calculus on
| that site, but since you requested it specifically here's the
| overview: https://betterexplained.com/guides/calculus/
| mturmon wrote:
| The notion of "completing the square" is often done as an
| algebraic trick ("just add and subtract this magic term").
|
| But it can also be viewed as translating the quadratic
| expression along the "X-axis" so that (at its new origin) it is
| left/right symmetric.
|
| That is,
|
| Q(x) = ax^2 + b x + c
|
| With the right substitution, x' = (x - B), the linear term
| vanishes. So when you re-write in terms of "x", you get:
|
| Q (x) = a (x - B)^2 + C
|
| So the intuition is that the linear term in the original
| quadratic is the thing that shifts the "symmetry axis" of the
| quadratic.
|
| I have found this helpful when "X" is a vector and you have a
| quadratic form. In this case, the coordinate shift centers the
| quadratic "bowl" about some point in R^n.
|
| *
|
| The chain rule for differentiation is another one with simple
| geometry but cumbersome notation. It's like: we know[*] that
|
| f(x) = g(h(j(k(x))))
|
| must have a linear approximation about some point x0. The only
| possible thing it could be is the product of all the little
| local curve slopes of k, j, h, and g, at the "correct" point in
| each.
|
| Thinking about little slopes also clarifies derivatives like
|
| f(x) = g(x^3, x^2)
|
| where g is an arbitrary function of two variables.
|
| [*] Because we read it in baby Rudin, ofc
| layer8 wrote:
| The geometric "proof" isn't actually equivalent, because it
| assumes a > b, and doesn't generalize geometrically to b > a.
| The algebraic proof, on the other hand, generalizes at least to
| commutative rings.
|
| Geometric "proofs" like this are neat, but are no real
| substitute for the algebraic ones. I'd argue that in cases like
| the present one they also don't provide any deeper insights.
| You're just moving geometric shapes around instead of algebraic
| symbols. They might give you the feeling that the theorem isn't
| as arbitrary as you thought, but it isn't arbitrary in algebra
| either.
|
| I'm putting "proof" in quotes here because there are many
| examples of incorrect geometric "proofs", and there is
| generally no formal _geometric_ way to verify their
| correctness.
| zozbot234 wrote:
| > there is generally no formal _geometric_ way to verify
| their correctness.
|
| There are formal models of synthetic (i.e. axiom-and-proof
| based) Euclidean geometry where proofs can in fact be
| verified. This is accomplished by rigorously defining the set
| of allowed "moves" in the proof and their semantics, much
| like one would define allowed steps in an algebraic
| computation.
| layer8 wrote:
| Yes, but once you work with those formal models, you're
| really doing algebra and not intuitive geometrics anymore.
| kbutler wrote:
| This is best considered an illustration, rather than a proof, as
| it fails for any of the following:
|
| a=0 and b is non-zero
|
| b>a (result is negative. If you swap the variables, you have to
| imagine the area is a negative area)
|
| either variable is negative
| dbsmith83 wrote:
| I think you could make it work for negative numbers. You would
| just shade the negative area differently to show it. For
| example, if you subtract 1000 sq ft from 100 sq ft,
| geometrically it means you are trying to remove more area than
| you have. So, to represent this, draw a rectangle of 100 sq ft.
| Conceptually "extend" this area by subtracting 1000 sq ft. The
| extra 900 sq ft that you try to subtract but can't "exists" as
| a negative representation. This could be represented by
| flipping the surplus 900 sq ft into a negative axis or shading
| it differently to denote that it's a deficit rather than an
| actual, positive area.
| xpe wrote:
| A koan in four parts:
|
| 1. What constitutes a proof?
|
| 2. In what context is a "proof" embodied?
|
| 3. Why is scribbling lines on a paper (that look like math to
| humans) 'more' of a proof than visual diagrams, if at all?
|
| 4. If you came across a proof that was persuasive to alien
| intelligences -- and led them to conclude true things were true
| and false things were false -- but, alas, you did not
| understand it, does that make it less of a proof?
| kbutler wrote:
| 1. What constitutes a proof?
|
| An irrefutable demonstration of a conclusion, possibly via
| sequence of steps or combination of elements.
|
| 2. In what context is a "proof" embodied?
|
| Do you mean what the range or domain of the proof are? Not
| sure on the "embodied". I think you mean the communication of
| the proof and the expected base knowledge to understand the
| proof.
|
| 3. Why is scribbling lines on a paper (that look like math to
| humans) 'more' of a proof than visual diagrams, if at all?
|
| You seem to be focusing on the representation of the proof in
| a particular notation, rather than the actual logic of the
| proof.
|
| The graphical demonstration leads to false conclusions. For
| example, if a=0, it implies that a^2 - b^2 is 0 (or it
| requires some unfamiliar graphical representation of negative
| areas)
|
| 4. If you came across a proof that was persuasive to alien
| intelligences -- and led them to conclude true things were
| true and false things were false -- but, alas, you did not
| understand it, does that make it less of a proof?
|
| Again, the representation is not the proof, it is a means to
| record or communicate the proof.
|
| If the representation implies that false things are true
| (e.g., if a==0), then it is not a proof.
| Ma8ee wrote:
| > Result is negative. If you swap the variables, you have to
| imagine the area is a negative area
|
| And why is that a problem? Just reverse the equation before and
| after the proof.
| loadingcmd wrote:
| A good reminder. This was taught in the elementary school along
| with (a + b)^2,
| loadingcmd wrote:
| Not sure how to edit, I think it may have been middle school,
| is been quite a few years :)
| sahmeepee wrote:
| A similar method is handy for some mental arithmetic involving
| squares, e.g. it's easy to calculate 10052 because it's 10002
| plus two added blocks of 5 x 1000, plus a small 52 block, so
| 1,010,025. Going the other way, 9952 is 10002 minus those same
| two 5 x 1000 blocks, plus 52, so 990,025.
|
| (Edited due to formatting fail!)
| jgrahamc wrote:
| There are other fun tricks for squares like this:
| https://blog.jgc.org/2010/03/squaring-two-digit-numbers-in-y...
| sahmeepee wrote:
| Thanks - the comments are worth reading too
| jfengel wrote:
| Futility Closet used to have a charming and fascinating podcast.
| I miss it. I'm glad that he's still writing the blog.
| papercrane wrote:
| I really enjoyed their podcast. I haven't found anything quite
| like it to replace it.
| fghorow wrote:
| s/proof/demonstration/ and be done with the arguments.
|
| Intuitive but formally wrong is still good for intuition.
| xpe wrote:
| Such a visual argument isn't "formally wrong" because it isn't
| trying or claiming to be a formal proof at all. "Formally
| wrong" would mean that formal mathematics is used incorrectly.
| layer8 wrote:
| No, "formally wrong" means that it fails formal verification.
| xpe wrote:
| >> "Formally wrong" would mean that formal mathematics is
| used incorrectly.
|
| > No, "formally wrong" means that it fails formal
| verification.
|
| The word "formal" has meaning in mathematics independent of
| formal verification. The latter builds upon the former.
| Agree?
|
| > In the context of hardware and software systems, formal
| verification is the act of proving or disproving the
| correctness of a system with respect to a certain formal
| specification or property, using formal methods of
| mathematics. - Wikipedia
| https://en.wikipedia.org/wiki/Formal_verification
|
| Here is how I'm using the terms:
|
| - "Formal" in mathematics refers to rigorous logical
| reasoning with precise definitions and deductive steps.
| This is the older, more general meaning.
|
| - "Formal verification" emerged later as a specific term in
| computer science referring to automated/mechanical
| verification of system properties. This is now the standard
| meaning in software/hardware contexts.
|
| So, putting these terms into use... If a human verifies a
| formal proof "by hand", I think it is fair to say that does
| not comprise "formal verification". On the other hand, if
| an automated system verifies the proof, then I would say
| "formal verification" has happened. Perhaps you will agree
| this is how many, if not most, experts use the terms.
|
| Are we mostly playing language games -- or is there a key
| insight you think I don't understand?
| layer8 wrote:
| You can perform a formal verification of an informal
| proof, like for example the ones in
| https://youtu.be/VYQVlVoWoPY linked elsethread. Of
| course, you have to come up with _some_ formalization of
| what the informal proof is trying to claim, but it
| doesn't mean that what you're proving wrong must be a
| formal proof to start with.
| xpe wrote:
| Yes, I understand what you mean. We don't disagree on the
| substance.
| layer8 wrote:
| It therefore makes sense to me (and represents actual
| usage of those terms) to say that some informal proof is
| formally wrong, meaning that translating its reasoning
| into a formal representation will reveal its
| incorrectness.
| xpe wrote:
| Analogously, one wouldn't call a mathematical proof "visually
| wrong" simply because it lacked visual diagrams.
| lupire wrote:
| Quod erat _demonstrandum_.
| quantum_state wrote:
| would think the point of algebra is to mechanize quantitative
| reasoning ... trying to cast algebra operations in geometry may
| not be the most productive ...
| taeric wrote:
| A fun book I have is Proofs Without Words
| (https://www.amazon.com/gp/aw/d/1470451867). Tons of neat
| diagrams like this.
| lupire wrote:
| Modern version:
|
| https://m.youtube.com/@MathVisualProofs
| exe34 wrote:
| Is there by any chance a book that explains what the
| illustrations mean? I had a quick look and it looks like I
| might need to know the theorems to begin with?
| taeric wrote:
| Proofs is not really accurate. Just visualizing some
| equations. I suspect this is part of some intuitions.
| jrmg wrote:
| Be careful: with visual 'proofs' you can end up believing
| something like this:
| https://en.wikipedia.org/wiki/Missing_square_puzzle.
| gklitz wrote:
| What do you mean "end up believing" are you insinuating that
| the shown example isn't true? It very much is true. The reason
| it's true might initially be confusion to you as a viewer
| because you have a difficult time telling the difference
| between 3/8 and 2/5 and assume the triangles have identical
| slopes, but the visual proof very much and truthfully shows
| that this is not he case.
| dagss wrote:
| That appears to be constructed in order to deceive though.
|
| Someone who was thinking about a problem and drawing something
| would always with a drawing like that either intend the angle
| to be the same, or otherwise highlight the fact that one
| triangles is 8/3 and the other is 5/2 so that the slope is
| obviously not the same.
|
| Good visual proofs simply use lines and figures to talk about
| actual algebra instead of symbols; but every outcome is still
| in a sense algebraic -- like the one linked and the popular
| about Pythagoras. Once you pull our your ruler and measure you
| are obviously lost. Every result should be algebraic, not
| visual, but it's fine to express the algebra in figures instead
| of letters.
| konschubert wrote:
| Here is a visual proof for the Pythagorean theorem:
|
| https://www.dbai.tuwien.ac.at/proj/pf2html/proofs/pythagoras...
|
| I find this much more "useful" since the Pythagorean theorem
| isn't immediately intuitive to me.
|
| As for the proof in the original post, it seems really redundant
| to me. it follows from a (b+c) = ab + ac.
|
| And while building intuition for this distributive property of
| multiplication is _extremely essential_ when teaching maths, I
| feel that the intuition for why this is true is better built
| without leaning on geometry.
| littlestymaar wrote:
| I don't feel like it's more redundant that Pythagorean theorem
| though, as we can say that the later directly follows from the
| definition of dot product...
| konschubert wrote:
| I guess my intuition for vector algebra is much weaker...
| nuancebydefault wrote:
| So you prove something in the 2d space via a 3d space
| intermezzo? Not very intuitive to me. Distribution on the
| other hand, can be explained by counting a handful of the
| same objects.
| r0uv3n wrote:
| The dot product exists in any dimension
| quietbritishjim wrote:
| Indeed I was even taught it in 2d before 3d (and higher).
|
| Even Pythagoras applies to any dimension, although
| admittedly it doesn't quite fit its usual statement in
| terms of triangles for higher dimensions: if a vector v
| has components (v1, v2, ...) then its length squared
| equals v12 + v22 + ...
| quietbritishjim wrote:
| How does Pythagoras's theorem follow from the definition of
| the dot product?
|
| Do you mean that x.y = x1y1 + x2y2 and x.x = |x|2, so it
| follows directly from that? If you define the dot product to
| be the first of those identities then you need Pythagoras's
| theorem to prove the second, so your argument is circular.
|
| (Or you can prove that x.y = |x| |y| cos th but that's even
| further removed from the component-wise definition than
| Pythagoras's theorem. Or you can define the dot product that
| way, but then you still have to prove the component-wise
| formula from it.)
| littlestymaar wrote:
| > but then you still have to prove the component-wise
| formula from it
|
| In an orthogonal basis this is trivial because cos(Pi/2) =
| 0 though...
| quietbritishjim wrote:
| It may seem trivial, but to use that to prove the
| component-wise formula for general vectors you're
| assuming distributivity of the dot product over addition
| of one of its arguments. But if you're starting with the
| x.y = |x| |y| cos th definition, how do you prove that
| (without first going via the component wise definition
| that you're still in the process of proving)? You end up
| needing trigonometric angle formulae that are at least as
| hard to prove as Pythagoras's theorem.
|
| Sorry, but you can't bypass proving Pythagoras's theorem
| by definition of the dot product or anything else.
| gregfjohnson wrote:
| This same identity can be used to provide geometric intuition as
| to why i*i must equal -1. This is shown in the diagrams at the
| bottom of http://gregfjohnson.com/complex/.
| ndsipa_pomu wrote:
| I enjoy some of the Mathologer YouTube videos and they often show
| some great visual proofs:
|
| https://www.youtube.com/watch?v=DjI1NICfjOk (Fermat's sum of two
| squares)
|
| https://www.youtube.com/watch?v=rr1fzjvqztY (Ptolemy's theorem)
|
| https://www.youtube.com/watch?v=yk6wbvNPZW0 (Irrational numbers)
| dvh wrote:
| But weren't the multiplication and addition rules set such that
| this visual proof holds true? Not the other way around.
| larodi wrote:
| this again opens the question what does multiplication actually
| mean. if you approach it geometrically the metric is
| something^squared, if you approach it as numbers - its not that
| straightforward on the numbers line.
| nuancebydefault wrote:
| Also it leans on the premise a x b (orthogonally measured)
| equals surface and that cutting pieces off surfaces leads to
| subtraction. Are those premises definitions or not, is a valid
| question.
| pn3k0 wrote:
| oh my heart! this is very pleasant
| flufluflufluffy wrote:
| I love this, giving an intuition about something that is usually
| taught by rote memorization. I also love how it makes all the
| math nerds and pedants uneasy xD Man like, of course you need to
| be careful. You need to be careful about unintentional
| assumptions in logical proofs too. It's a cool little creative
| visualization.
| sebtron wrote:
| If you like this, there is a whole book full of visual proofs
| [1]. See also wikipedia [2].
|
| A few years ago I re-drew a bunch of these in latex with my PhD
| advisor and another colleague [3]. We planned to print them as
| posters and hang them for a Pi day event that unfortunately never
| happened because the pandemic broke out.
|
| [1] https://www.amazon.com/Proofs-without-Words-Exercises-
| Classr...
|
| [2] https://en.m.wikipedia.org/wiki/Proof_without_words
|
| [3] https://www.antonellaperucca.net/didactics/proof-without-
| wor...
| kelnos wrote:
| I'm a little disappointed at the focus here on how it's hard or
| impossible to visualize when b<a or when either a or b is
| negative.
|
| That's not the point. To me, it's useful to already know that an
| algebraic proof of that equation exists, but to see it work out
| visually. I don't need to see it worked out visually for every
| single possible value for this to be helpful for understanding.
|
| It also nicely illustrates how algebra and geometry are linked.
| And that multiplication is geometrically taking you from 1
| dimension to 2.
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