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Homework Help: Prove that if a² + ab + b² = 0 then a = 0 and b = 0

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  1. Dec 25, 2016 #1

    TheBlackAdder

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    1. The problem statement, all variables and given/known data
    Prove that if a² + ab + b² = 0 then a = 0 and b = 0
    Hint: Recall the factorization of a³-b³. (Another solution will be discussed later when speaking about quadratic equations.)

    2. Relevant equations
    a² + ab + b² is close to a² + 2ab + b² = (a+b)²
    a³-b³=(a-b)(a²+ab+b²)

    3. The attempt at a solution
    a² + ab + b² = 0
    a² + ab + b² + ab = ab
    a² + 2ab + b² = ab
    (a+b)² = ab
    (a+b)² - ab = 0

    Here I figured, to get zero after a subtraction, both terms need to be equal. An addition of two numbers is the same to the multiplication of those two numbers only when a,b = 0 or a,b = 2. (0+0=0*0 and 2+2=2*2)
    But since the first term is squared, (2+2)² - 2*2 = 16 - 4 ≠0. This leaves a,b = 0 as the only option to get a zero.

    A second try was
    a² + ab + b² = 0
    a² + ab + b² + ab = ab
    a² + 2ab + b² = ab
    (a+b)² = ab
    (a+b)² - ab = 0
    (b+b)² - bb = 0 (given that a=b, and same works for b=a)
    (2b)² - b² = 0
    (2b-b)(2b+b) = 0
    b*b = 0
    b = 0

    A third unsuccessful attempt
    a² + ab + b² = 0
    a² + ab + b² + ab = ab
    a² + 2ab + b² = ab
    (a+b)² = ab
    a+b = √(ab)
    a+b = √a√b
    (a+b) - (√a√b) = 0

    The solution:
    a³-b³=(a-b)(a²+ab+b²), if (a²+ab+b²)=0, then a³-b³=0 or
    a³ = b³ which can only happen when a=b. If a=b, then 0=a²+ab+b²=3b² which implies that
    b=0 and a=b=0.

    I have no experience with proofs whatsoever. I understand the solution but I want to know if any of my first three tries have any correctness in them. Or is the solution above the only possible way to prove this equality without any other 'techniques'?
     
  2. jcsd
  3. Dec 25, 2016 #2

    robphy

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    Are there restrictions on a and b? e.g. only integer solutions.
     
  4. Dec 25, 2016 #3

    TheBlackAdder

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    Yes, only integers. No irrational or complex numbers as the books hasn't touched that yet.
     
  5. Dec 25, 2016 #4

    PeroK

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    Your proofs only get so far and then fail on the key step. An alternative would be to start with simply:

    ##a^2 + b^2 = -ab##

    ##a^2 + b^2 = |a||b|##

    Can you justify that second step?

    Then consider the case ##|a| \ge |b|##

    That's perhaps a more direct approach.
     
  6. Dec 25, 2016 #5

    PeroK

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    You should be trying to prove it for any real numbers.
     
  7. Dec 25, 2016 #6

    TheBlackAdder

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    ##a^2 + b^2 = -ab##

    ##a^2 + b^2 = |a||b|##

    The second step can be justified by the fact that any number to an even power will be positive, except for zero.

    ##|a| \ge |b|## With this I think you want to show me that, if one number is bigger than the other, there is no possibility that both numbers are zero, which is the only case that leads to the equation equaling zero.
    So, if only one number would be zero, let's say a=0, then you're left with 0² + b² = |0||b| = 0, which can only mean b=0.
    The other case, where both numbers are the same, results in:
    a²+b² = |a||b|
    a²+a² = |a||a|
    2a² = a²
    2a²-a² = 0
    a² = 0
    a = 0

    Is this what you meant?

    (Another side question: let's say I make the mistake of doing
    2a² = a²
    2 = a²/a² (can't divide by 0)
    2 = 1
    How's this called in mathematical terminology? )
     
  8. Dec 25, 2016 #7
    Another approach would be solving the quadratic for a. Actually you only need to consider the determinant to get the answer.
     
  9. Dec 25, 2016 #8

    TheBlackAdder

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    Can not be used yet.
     
  10. Dec 25, 2016 #9
    Is that a requirement by the author or you simply don't want ?Anyhow book's solution is ideal.
    By the way what you are doing in your other attempts is whats called 'Completing the square' which is used to solve quadratic equations and quadratic formula is also derived by that method.
     
  11. Dec 25, 2016 #10

    PeroK

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    You've a bit of work to do to understand the logic of proofs. Often you are using what you are trying to prove during steps in the process.

    For this proof there are two distinct approaches:

    The direct approach where you assume nothing extra about ##a, b##.

    Proof by contradiction, where you assume one or both of ##a, b## are non zero and show this leads to a contradiction.

    In either case, you need a clear strategy for your proof.

    For this proof, the key thing you should be thinking is: Isn't ##a^2 + b^2## always bigger than ##ab##? If ##a## is bigger than ##b## then ##a^2## on its own is bigger than ##ab##.

    Can you use that idea to guide you through a formal proof?

    By the way, the simplest way to justify that second step was just to notice that the left hand side is non-negative, hence ##-ab## is non-negative, hence ##-ab = |-ab| = |a||b|##.
     
  12. Dec 25, 2016 #11

    TheBlackAdder

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    Isn't that what I said? "The second step can be justified by the fact that any number to an even power will be positive, except for zero."
    Meaning that the left hand side is positive and therefore also the right hand side.

    I'll see what I can do for the formal proof. Thank you.
     
  13. Dec 26, 2016 #12

    TheBlackAdder

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    Okay so.
    a²+ab+b² = 0
    a²+b² = -ab
    a²+b² = |a||b|

    1. If a=b where a,b ≠ 0, then
    a²+b² = |a||b| becomes
    b²+b² = |b||b|
    2b² = b² which is impossible unless when b=0, thus a=0 too.

    2. If a>b where a,b ≠ 0, then
    a²+b² = |a||b| becomes
    a²+b² > |a||b| because a² > |a||b| and therefore a²+b² - |a||b| > 0
    Going back to a²+ab+b² = 0
    If a²+b² is greater than |a||b| then a²+ab+b² shouldn't be able to equal 0 because in either case when ab is positive or negative, it can not cancel a²+b² because the absolute value of ab is not large enough? So a>b is impossible which leaves a=b, which in turn leaves b=0 and a=0.
     
    Last edited: Dec 26, 2016
  14. Dec 26, 2016 #13

    Ray Vickson

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    Who told you that you cannot use the method of solving a quadratic equation?
     
  15. Dec 26, 2016 #14

    SammyS

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    Look at the statement of the problem in OP.

     
  16. Dec 26, 2016 #15

    PeroK

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    I was thinking of a much simpler proof. Assume, without loss of generality, that ##|a| \ge |b|##. Then:

    ##a^2 \ge |a||b|##

    And, unless ##b =0##:

    ##a^2 +b^2 > |a||b|##

    That's the core of the proof. The key here is to use the fact that one of ##a^2## or ##b^2## on its own is larger than |ab|. With equality only when ##a=b##.
     
  17. Dec 26, 2016 #16

    TheBlackAdder

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    "And, unless ##b = 0##"
    Isn't the unless ##b = 0## redundant? Because the expression holds even when b = 0.
    a² + 0² > a * 0. Or maybe you meant to say unless a,b = 0? Or is it added so the complete expression remains?

    "I was thinking of a much simpler proof." Does that mean my proof is correct also? I'm self studying and have never learned how to write proofs yet, so I have nothing to go on. Thanks for your patience.
     
  18. Dec 26, 2016 #17

    PeroK

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    Your proof goes round the houses, as it were. You get to the front door, the you run round the house 2-3 time and climb in a back window.

    Also, you say one equation "becomes" another, that isn't logical.

    You need to step back and decide what you are assuming, where this leads and how it proves the original statement.

    Finally, my statement "unless ##b =0##" was not redundant, as you need the strictly ##>## to rule out ##b \ne 0##.

    Do you want to have another go at a full proof? Here's the outline:

    Assume ##a^2 + ab + b^2 = 0##

    Show that this implies

    ##a^2 + b^2 = |ab| \ ## (1)

    Assume, wlog (without loss of generality) that ##|a| \ge |b|##

    Show that ## b = 0##, otherwise we have a contradiction to equation (1).

    Show that also ##a= 0##.
     
  19. Dec 26, 2016 #18

    lurflurf

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    $$a^2+ab+b^2=\left.\left.\frac{1}{4}\right[3(a+b)^2+(a-b)^2\right]$$
     
  20. Dec 27, 2016 #19

    TheBlackAdder

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    K, I understand my redundancy mistake.
    Isn't 'becomes' logical because it gives no information about the action you're performing?

    I'll have another try at the proof after I've made some other exercises.
     
  21. Dec 28, 2016 #20
    if ##a^2+ab+b^2=0##, assuming a and b are real, then
    ##a^2## is positive
    ##b^2## is positive
    therefore ##ab## MUST be negative, in order for the sum to be 0.

    The original equation can be rewritten as:
    ##(a+b)^2-ab=0##
    Therefore ##a+b=(ab)^{1/2}##
    which the only real solution, if ##ab## IS negative, is ##{a=0,b=0}##
     
  22. Dec 28, 2016 #21

    Ray Vickson

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    No. If ##ab < 0## you cannot take its square root (at least, not in the real numbers), and of course, if ##ab < 0## you cannot have ##a = b = 0## (so you should never, ever write that). However, you can get a contradiction another way:
    (1) You have just shown that ##ab = -a^2 - b^2##, which is ## \leq 0##.
    (2) You can write ##a^2 + b^2 + ab## as ##(a+b)^2 - ab##, so you also have ##ab = (a+b)^2##, which is ## \geq 0##.
    (1) and (2) are compatible only when ##a = b = 0##.
     
    Last edited: Jan 2, 2017
  23. Dec 28, 2016 #22
    I should rephrase some stuff in my post, when I say must be positive or must be negative, I mean, ## ≥ 0 or ≤ 0##
    So, ##ab ≤ 0##
    If ##a+b=(ab)^{1/2}##
    The only real solution is if ##ab = 0##, exactly because there is no real root of a negative number, which means that ##a=0## or ##b=0## or ##a=b=0##
    if ##ab = 0##, then ##a+b=0##, so ##a=-b##
    Since one of them is 0, so must be the other.
     
  24. Dec 28, 2016 #23

    PeroK

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    That's a neat approach. A better way to finish it off might be to say:

    ##a^2 + b^2 = -(a+b)^2##

    And then ##a=b=0##.

    As already pointed out you should be saying non-positive instead of negative, as the latter excludes ##0##.
     
  25. Dec 31, 2016 #24

    LCKurtz

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    You have ##(a+b)^2=ab##. Put that in the original equation ##a^2+ab+b^2 = 0## giving ##a^2 + (a+b)^2 + b^2 = 0## so ## a=b=0##.

    Edit: Only after posting this did I notice there was a second page of replies. But, hey, it's New Year's Eve, so why am I in here posting anyway.
     
  26. Jan 11, 2017 #25
    Well late to the party but I've been thinking about extending the domain of this question to including all numbers including complex numbers. This is my proof for it and I'd like to know if it is valid.

    Given ##a^2+ab+b^2=0## ∀ ##a,b ∈ ℂ##
    ##a^2+b^2=-ab## ⇒(Realise that for this to be true only either ##a,b∈ℝ## or both ##a,b∉ℝ##)[tex] ab =
    \begin{cases}
    ≤0 & \text{if } a,b ∈ ℝ \\
    ≥0 & \text{if } a,b ∉ ℝ
    \end{cases}
    [/tex]
    ##(a+b)^2-2ab=-ab##
    ##(a+b)^2=ab##
    Suppose ##a,b≠0##
    ##⇒(a+b)^2/(ab)=1##
    Since ##(a+b)^2## and ##ab## have to be of the same sign, it is a contradiction with the case statement above and hence a,b cannot not be 0.
     
    Last edited: Jan 11, 2017
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