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anonymous

  • 4 years ago

Does (x+y)^n=x^n+y^n? If not, give a counterexample.

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  1. anonymous
    • 4 years ago
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    try replacing n with a variable. What variable makes it true? What variable makes it not true? I always start with 2.

  2. JamesJ
    • 4 years ago
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    Right. Try something simple, such as n = 2, and x = y = 1.

  3. Zarkon
    • 4 years ago
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    "Freshman's dream"

  4. anonymous
    • 4 years ago
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    If I replace n with 2, I come out with \[x^2+xy+y^2 = x^2 + y^2 \]. That is not true?

  5. JamesJ
    • 4 years ago
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    No. Suppose x=y=1. Are the two sides equal?

  6. anonymous
    • 4 years ago
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    Do you mean x+y?

  7. JamesJ
    • 4 years ago
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    and btw, \[ (x+y)^2 = x^2 + 2xy + y^2 \] No: I mean x = 1 and y =1.

  8. Zarkon
    • 4 years ago
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    http://en.wikipedia.org/wiki/Freshman%27s_dream

  9. JamesJ
    • 4 years ago
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    But without expanding it out: if x = 1 and y = 1, then x+y = 2 and thus \[ (x+y)^2 = 2^2 = 4 \] But \[ x^2 + y^2 = 1^2 + 1^2 = 1 + 1 = 2 \] Hence for n = 2 and x=y=1, is it true that \[ (x+y)^n = x^2+y^n \] ?

  10. JamesJ
    • 4 years ago
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    **correction: \[(x+y)^n = x^n + y^n \]

  11. anonymous
    • 4 years ago
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    No it's not true. Did we use the Power of a Power Postulate?

  12. JamesJ
    • 4 years ago
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    No. Nothing so fancy was necessary.

  13. anonymous
    • 4 years ago
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    It asks me to list what Postulate was implemented, and what do they mean by counterexample in this case?

  14. JamesJ
    • 4 years ago
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    You'll have to figure out what postulates were used here, given the list you have, as there's no standard list I can use to answer that for you. A counter-example of a proposition in mathematics (or logic more generally) is an explicit example of when the result does not hold and hence the proposition is not true in general. Let P be the following proposition: For all real numbers x and y and all integers n \[ (x+y)^n = x^n + y^n \ \ \ \ -- (*) \] A counter-example to this proposition P is x=y=1 and n=2, because for those values of the variables, the equation (*) does not hold. Hence the proposition P is false in general. I.e., it is NOT the case that for all real numbers x and y, and all integers n that \[ (x+y)^n = x^n + y^n \]

  15. anonymous
    • 4 years ago
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    That clears things up a lot! Thank you very very much.

  16. JamesJ
    • 4 years ago
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    Sure thing

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