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    Prove that any irrational number can be a root of at most one cubic equation of the form x^3+ax =b where a and b are rational.

    How do I get started?
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    (Original post by rdj201)
    Prove that any irrational number can be a root of at most one cubic equation of the form x^3+ax =b where a and b are rational.

    How do I get started?
    Proof by contradiction. Assume there is an irrational number that is the root of two equations of that form, and show this leads to a contradiction.
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    (Original post by ghostwalker)
    Proof by contradiction. Assume there is an irrational number that is the root of two equations of that form, and show this leads to a contradiction.
    Thank you...still struggling at the moment, can I have a bit more help...
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    (Original post by rdj201)
    Thank you...still struggling at the moment, can I have a bit more help...
    Based on the suggested assumption, form and solve two simultaneous eqns.
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    (Original post by ghostwalker)
    Proof by contradiction. Assume there is an irrational number that is the root of two equations of that form, and show this leads to a contradiction.
    PRSOM
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    (Original post by atsruser)
    Based on the suggested assumption, form and solve two simultaneous eqns.
    .

    Is this sufficient...
    If n is an irrational number.
    If assume it works for two equations x^3 + ax - b = 0 and x^3 + cx - d = 0.
    Then
    f(n) = n^3 + an - b = 0
    and
    f(n) = n^3 + cn - d = 0

    So n^3 + an - b = n^3 + cn - d
    an - b = cn - d
    an - cn = b - d
    n = (b-d)/(a-c)
    If a, b, c and d are rational numbers, then n must be rational.
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    (Original post by rdj201)
    .

    Is this sufficient...
    If n is an irrational number.
    If assume it works for two equations x^3 + ax - b = 0 and x^3 + cx - d = 0.
    Then
    f(n) = n^3 + an - b = 0
    and
    f(n) = n^3 + cn - d = 0

    So n^3 + an - b = n^3 + cn - d
    an - b = cn - d
    an - cn = b - d
    n = (b-d)/(a-c)
    If a, b, c and d are rational numbers, then n must be rational.
    In essence, yes. I would tighten up the phraseology a bit. Also you need to deal with the case a=c, since you've dividing by a-c.

    Here's my take:

    Spoiler:
    Show



    Let n be an irrational number for which there exist two distinct equations x^3 + ax - b = 0 and x^3 + cx - d = 0, where n is a root of each of them.

    We then have:

    n^3 + an - b = 0..............(1)
    n^3 + cn - d = 0..............(2)

    (I avoid f(n) since this is one specific function. If you want to use that then you'd have f(n) and g(n))

    Subtracting we get:

    (a-c)n -(b-d)=0

    If a=c, then (0)n- (b-d) = 0 and it follows that b-d = 0 and b=d, and they are the same equation.

    So, a not = c and we then have n=(b-d)/(a-c)

    Since all the terms on the right are rational, it follows that n is rational.

    This is a contradiction, since n is irrational.

    Therefore no such n can exist and any irrational number can the root of at most one equation of the form x^3 + cx - d = 0


    QED.


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    (Original post by ghostwalker)
    In essence, yes. I would tighten up the phraseology a bit. Also you need to deal with the case a=c, since you've dividing by a-c.

    Here's my take:

    Spoiler:
    Show




    Let n be an irrational number for which there exist two distinct equations x^3 + ax - b = 0 and x^3 + cx - d = 0, where n is a root of each of them.

    We then have:

    n^3 + an - b = 0..............(1)
    n^3 + cn - d = 0..............(2)

    (I avoid f(n) since this is one specific function. If you want to use that then you'd have f(n) and g(n))

    Subtracting we get:

    (a-c)n -(b-d)=0

    If a=c, then (0)n- (b-d) = 0 and it follows that b-d = 0 and b=d, and they are the same equation.

    So, a not = c and we then have n=(b-d)/(a-c)

    Since all the terms on the right are rational, it follows that n is rational.

    This is a contradiction, since n is irrational.

    Therefore no such n can exist and any irrational number can the root of at most one equation of the form x^3 + cx - d = 0


    QED.



    Thank you :-)
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    (Original post by rdj201)
    Thank you :-)
    Your welcome. Your answer was spot on - I was only dotting i's and crossing t's.

    Can't recall seeing a proof by contradiction being required at A-level before - could just be my memory though.
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    (Original post by ghostwalker)
    Your welcome. Your answer was spot on - I was only dotting i's and crossing t's.

    Can't recall seeing a proof by contradiction being required at A-level before - could just be my memory though.
    It is a suggestion for the new A Level
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    (Original post by ghostwalker)
    Your welcome. Your answer was spot on - I was only dotting i's and crossing t's.

    Can't recall seeing a proof by contradiction being required at A-level before - could just be my memory though.
    The combination of the words "proof" and "A level" switched my brain off - I couldn't see what they were getting at, using standard A level methods.
 
 
 
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