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Please explain to me why this simple problem is so

  1. May 30, 2010 #1

    I just wanted to know if someone could explain to me why the following is true:

    (x+y)^2 =/= x^2 + y^2

    So, in words, why doesn't the some of two numbers added together and then squared equal the sum of the numbers squared individually and then added together?

    I obviously know how to show they are not equal by simply explanding the left side, but I'm looking for a really fundamental answer (i.e. is there some axiom which accounts for this?).

  2. jcsd
  3. May 30, 2010 #2


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    Take any number, say x=2 and y=3, now plug into your equality. Obviously they aren't the same numbers so how could both left and right sides be equal to each other then?
  4. May 30, 2010 #3
    Thanks for the response, but thats exactly the type of answer I didn't want (sorry if I wasn't clear). What you said shows both sides aren't equal but it doesn't tell me anything about WHY they aren't equal.
  5. May 30, 2010 #4


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    I'm still a bit confused about exactly what type of answer you want, but what about the explanation :

    "Order of operations" is important in mathematics. That is, in general you cannot exchange the order of operations in a mathematical expression and always expect the result to be unchanged.
  6. May 30, 2010 #5
    That answer might be leading in the right direction...Maybe it will help if I frame the question with reference to the order of operations:

    WHY does squaring two numbers after they have first been added together produce a result different to when they are squared individually first and then added together?
  7. May 30, 2010 #6


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    Sorry I thought you were just looking for some answer other than "expand the Left side and you'll see".

    If the operation of expanding is clearly defined, then you shouldn't have any doubts.

    [tex]a(x+y)=ax+ay[/tex], yes?

    Well, let a=x+y, Now we have [tex](x+y)(x+y)=(x+y)x+(x+y)y=x^2+2xy+y^2=(x+y)^2\neq x^2+y^2[/tex]

    Are you convinced? Else the only other way I can answer your question is with "just because".
  8. May 30, 2010 #7


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    or even take a geometrical approach to the problem:

    Have some length x, and then add some length y to the end of it so now you have x+y length. Now square this (actually make it a square of x+y length and width). Notice how this square can be split up into 4 sections, x2, y2 and the last two are xy each. Obviously now (x+y)2>x2+y2 since we're dealing with positive lengths x and y. Except of course in the obvious case that x=0 or y=0, [tex](x+y)^2\neq x^2+y^2[/tex]
  9. May 30, 2010 #8
    You can just view this as distribution.
    (x+y)(x+y) = y(x+y) + x(x+y) = xy + y^2 + x^2 + xy = x^2 + 2xy + y^2 so if your looking for something "fundamental" as to why you would expand the left side as you do, then it's just really distribution law.

    Edit: oops someone got to it before I did. Mentallic is correct.
  10. May 30, 2010 #9


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    I hope so :bugeye:
  11. May 30, 2010 #10
    All of maths "just is" because it is.
    Maybe that's why it's hard to explain elementary arithmetic/algebra.

    Or stuff like "why isn't (uv)' equal to u' v'?" where ' denotes d/dx. It just...isn't, because we know how to derive the correct answer and everything is a counterexample; and why should it be? Argh lol.

    Sometimes you get asked things like "Why is -x^2 not (-x)^2" and the only reasons you can give are basically 'just apply stuff you know', or 'assume it is, then show x=0 or a contradiction'. However, no one (except someone who already gets it anyway) will listen to an answer like the second one.
  12. May 30, 2010 #11
  13. May 30, 2010 #12
    The answer is very simple: First tell me why you think you can do it your way!

    Maths works like this: It is very well defined what is are allowed to do. It's about 20 rules of algebra you learned at school. And anything else is definitely wrong!
    In fact any manipulation you ever did in school can be traced back to about 10 laws for basic arithmetics. The rules look like
    Some of these you might find even too trivial, but I assure you that everything you learned can be derived from these rules (provided you define notation).

    It's the single most serious mistake to make up new rules. See it this way:
    If there are 10 real rules and you know 9 of then, then you will get 90% of the answers correct and 10% you won't find - at least you know which ones.

    If however you know the 10 rules and one additional erroneous 11th rules, then you get all results wrong, because your wrong rule spoils everything like a virus.

    So I'd rather want to know my student one rule too little, than one rule too much.
  14. May 30, 2010 #13


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    The first figure in this link should provide the answer you need. call the side length of the blue square a, while b is the side length of the red square. So the blue square has area a2 the area of the red square is b2, the larger square has side length of (a+b) and area of (a+b)2. You should be able to see that the sum of a2 + b2 is not the same as (a+b)2. Note that the difference is 2 rectangles with area ab. That gives the total area of the big square as a2+ b2 + 2ab. This is the exact result of algebracally expanding (a+b)2.
  15. May 30, 2010 #14
    Thanks everyone for the input. I was thinking along the lines of the distributive law, which Mentallic later verified, and I guess it does seem to be pretty much as fundamental as you can go. But, of course, the question of why the distributive law is so can then be raised...

    Thanks freireib too, that was an excellent summary of all the posts so far.

    Gerenuk, the statement of "first tell me why you think you can do it your way", really gets at why I raised this question in the first place. The problem I've always had with maths is that when asked a question like the one being analysed, I can only come up with answers which rely on performing the memorised arithmetic processes/substitution etc., rather than really providing the fundamental explanation.

    Does anyone know if there is a field dedicated to studying the very fundamental nature of maths and how the axioms were established?
  16. May 30, 2010 #15
    And also, just one more question:

    If the first question I posed can be shown to be true by reducing the binomial square down to an application of the distributive law, what can the following be reduced down to:

    √(a + b) =/= √a + √b

    So its pretty much the same question but instead of being squared the root is taken...

    Really appreciate the help
  17. May 31, 2010 #16


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    It's really just an application of the same idea here in this thread. If x,y>0 (which is a given since we're dealing with the square roots of these numbers and only want to work in the domain with which this problem exists), then if x=y, then x2=y2 since, by one of those laws (whcih I don't know the name of), if x=y then ax=ay, and let a=x=y to get x2=y2.

    Now let x=(a+b)1/2 and y=a1/2+b1/2
  18. May 31, 2010 #17
    If [itex]x[/itex] and [itex]y[/itex] are positive real numbers and you accept the axioms of Euclidean geometry, then this arguemt might appeal to you. See this figure:

    [PLAIN]http://img413.imageshack.us/img413/1774/binomial.png [Broken]

    [itex](x + y)^{2}[/itex] is the area of the large square. [itex]x^{2}[/itex] is the area of the square in the lower left corner and [itex]y^{2}[/itex] is the area of the square in the upper right corner. You see that their sum is smaller than the area of the big square. You need to include the areas of the two rectangles (which are identical) [itex]2 x y[/itex].
    Last edited by a moderator: May 4, 2017
  19. May 31, 2010 #18
    OK I know the answer has been given but it took me ages to draw this!!


    Anyone know of a better drawing package than MSpaint?

    Edit looks like dickfore does!!
  20. May 31, 2010 #19
    Actually, I used MSPaint. But, there are further options than the free-form line or brush line. Also, there is an option for typing characters.
  21. May 31, 2010 #20
    Thanks!! Yours does look slightly better than mine!!

    I will try to do better in future!

    It will lose the 'back of a fag packet' look though!!
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