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Math Help - If a+√ b=c+√ d, prove that a=c and b=d.

  1. #1
    jwu
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    If a+√ b=c+√ d, prove that a=c and b=d.

    Suppose a, b, c, d are all rational number where √ b and √ d exist and are irrational. if If a+√ b=c+√ d, prove that a=c and b=d.

    I have been thinking for a long time but I don't know where to start. The only thing i can think of is to try using proof by contradiction. I am really confused, please help.
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    Hey jwu.

    Consider a + SQRT(b) = c + SQRT(d) then a - c = SQRT(b) - SQRT(d). Now if a and c are rational then a - c is also rational but if SQRT(b) - SQRT(d) is irrational (two numbers added or subtracted are irrational) then the only time when this can happen is when both sides are equal to 0.

    Hence you get the result.
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    yes, but that proof lies subsumed under a proof that if b ≠ d, and √b, √d are irrational, √b - √d is irrational, which you have not proven.
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    When is the only case when both sides are rational? Also you have both numbers b and d being rational which you need to use.

    The rationality means you can't get things like say (SQRT(2) + 1) - SQRT(2): this is a not a general statement but rather a statement where b and d are forced to be rational which has consequences on what the sums and differences are.
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    I respectfully suggest that a rigorous proof of the following statement hasn't yet been given:

    Let b and d be rational numbers such that \sqrt{b} and \sqrt{d} are irrational. Then either \sqrt{b} = \sqrt{d} or \sqrt{b} - \sqrt{d} is irrational.

    Proof: Suppose that \sqrt{b} - \sqrt{d} = r \in \mathbb{Q} with r \neq 0 (If r = 0, then \sqrt{b} = \sqrt{d}). Then \sqrt{b} = r + \sqrt{d}. Squaring both sides and solving for \sqrt{d}, we get

    \sqrt{d} = \frac{-r^2 + b -d}{2r}

    which implies that \sqrt{d} \in \mathbb{Q}, contrary to the hypothesis. Therefore, \sqrt{b} - \sqrt{d} is irrational.
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    Your problem is terribly stated. From what set do a, b, c, & d come from? Real numbers? Complex numbers? Rationals?

    One could approach this problem from many different angles depending upon the "context" of the statements.

    I would assume b and d are positive integers, having no common factors and square-free, a & c ordinary integers.

    From there you can draw some elementary conclusions about a, b, c & d.
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    Quote Originally Posted by dave0147 View Post
    Your problem is terribly stated. From what set do a, b, c, & d come from? Real numbers? Complex numbers? Rationals?
    From the first post:

    Suppose a, b, c, d are all rational number where √ b and √ d exist and are irrational.
    Seems properly stated to me.

    One could approach this problem from many different angles depending upon the "context" of the statements.

    I would assume b and d are positive integers, having no common factors and square-free, a & c ordinary integers.

    From there you can draw some elementary conclusions about a, b, c & d.
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    I'm sorry. I'm new here and didn't realize that the problem had been better stated in an earlier post.

    I will present my solution once I have it in order.

    Thanks
    dave0147
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    Quote Originally Posted by dave0147 View Post
    I'm sorry. I'm new here and didn't realize that the problem had been better stated in an earlier post.

    I will present my solution once I have it in order.

    Thanks
    dave0147
    Hey, no problem.
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    Quote Originally Posted by Petek View Post
    I respectfully suggest that a rigorous proof of the following statement hasn't yet been given:

    Let b and d be rational numbers such that \sqrt{b} and \sqrt{d} are irrational. Then either \sqrt{b} = \sqrt{d} or \sqrt{b} - \sqrt{d} is irrational.
    yes, this was the point i was trying to make earlier.

    Proof: Suppose that \sqrt{b} - \sqrt{d} = r \in \mathbb{Q} with r \neq 0 (If r = 0, then \sqrt{b} = \sqrt{d}). Then \sqrt{b} = r + \sqrt{d}. Squaring both sides and solving for \sqrt{d}, we get

    \sqrt{d} = \frac{-r^2 + b -d}{2r}

    which implies that \sqrt{d} \in \mathbb{Q}, contrary to the hypothesis. Therefore, \sqrt{b} - \sqrt{d} is irrational.
    short, sweet and to the point.
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    Without loss of generality, one may assume that ALL rational numbers are fractions expressed in lowest terms.
    Hence the set {(a-c), b^(1/2), -d^(1/2)} are linearly independent over the rational numbers (field), Q

    Please excuse the clumsy notation (expressing square roots as ^(1/2)) but I am very new and haven't mastered the learning curve when it comes to using the notation of this platform)

    Thus any linear combination A(a-c) + B(b^(1/2)) + C(-d^(1/2)) cannot equal zero unless each of A, B, & C are zero.
    The original problem is: a+ b^(1/2) = c + d^(1/2) cannot be valid unless either A or (a - c) = 0 --> c = a as advertised! The rest follows easily.

    Could someone please tell me where symbols like sqrt are?
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    Re: If a+√ b=c+√ d, prove that a=c and b=d.

    Quote Originally Posted by dave0147 View Post
    Could someone please tell me where symbols like sqrt are?
    Here are some examples.

    [tex] \cup [/tex] gives  \cup is union
    [tex] \cap [/tex] gives  \cap is intersection
    [tex] \subseteq [/tex] gives  \subseteq is subset
    [tex] x^{\sqrt{x+1}} [/tex] gives  x^{\sqrt{x+1}}
    [tex] \in [/tex] gives  \in is element
    [tex] \emptyset [/tex] gives  \emptyset is emptyset
    [tex] \wedge [/tex] gives  \wedge , and
    [tex] \vee [/tex] gives  \vee , or
    [tex] \sqrt{x+1} [/tex] gives  \sqrt{x+1}
    [tex]x^{-i\theta}[/tex] gives  x^{-i\theta} instead of  x^-i \theta .

    The \boxed{\Sigma} on the toolbar gives the [tex]..[/tex] wrap.
    Last edited by Plato; November 11th 2012 at 02:05 PM.
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