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anonymous

  • one year ago

I need some help with understanding a basic mathematical proof.

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  1. anonymous
    • one year ago
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    Disprove the statement: There is a real number x such that \[x ^{6}+x ^{4}+1 = 2x ^{2}\]

  2. anonymous
    • one year ago
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    Proof: For all real numbers, x, \[x ^{6}+x ^{4}+1\neq 2x ^{2}\].

  3. anonymous
    • one year ago
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    Case 1: suppose x = 0, then x^6 + x^4 + 1 = 1 and 2x^2 = 0, so x^6 + x^4 +1 =/= 2x^2.

  4. anonymous
    • one year ago
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    Case 2 (this is the one I'm having a little trouble understanding) : Suppose \[x \neq0\], then x^2 >0. Since \[x ^{6}=x ^{2^{3}}\], x^6 >0. Therefore x^6+(x^2-1)^2 >x^6, so x^6 +x^4 +1 - 2x >0

  5. anonymous
    • one year ago
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    (x^2-1)^2 is factored from x^6+x^4+1-2x^2...can I end the proof there?

  6. jim_thompson5910
    • one year ago
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    x is nonzero ---> x^2 is positive x^4, x^6, etc are all positive as well

  7. jim_thompson5910
    • one year ago
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    and as you pointed out, x^4-2x^2+1 turns into (x^2-1)^2 so (x^2-1)^2 is always positive when x is nonzero

  8. anonymous
    • one year ago
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    Yes, so x^6+x^4+1 - 2x^2 >0 ...can I say that x^6+x^4+1 >2x^2? Thus x^6+x^4+1 =/= 2x^2

  9. jim_thompson5910
    • one year ago
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    I would do it like this x^6+x^4+1 - 2x^2 >0 x^6+x^4- 2x^2+1 >0 x^6+(x^2- 1)^2 >0 when x is nonzero, x^6 is positive and (x^2- 1)^2 is positive so that's why x^6+(x^2- 1)^2 >0 is true and x^6+(x^2- 1)^2 = 0 will never be true

  10. anonymous
    • one year ago
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    I understand what you did, but how (exactly) does that show x^6+x^4+1 =/= 2x^2?

  11. anonymous
    • one year ago
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    Or is that supposed to be obvious so I don't have to write anything more?

  12. jim_thompson5910
    • one year ago
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    it shows that x^6+(x^2- 1)^2 =/= 0

  13. jim_thompson5910
    • one year ago
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    which consequently means x^6+x^4+1 - 2x^2 =/= 0

  14. anonymous
    • one year ago
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    ahh, yes so bringing 2x^2 to the left side gives x^6+x^4+1 =/= 2x^2

  15. jim_thompson5910
    • one year ago
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    yes

  16. anonymous
    • one year ago
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    thank you so much~

  17. jim_thompson5910
    • one year ago
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    no problem

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