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sauravshakya

  • one year ago

Prove OR Disprove Every irrational number can be expressed as (x)^y where x and y are rational numbers.

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  1. terenzreignz
    • one year ago
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    Am I allowed to assume that pi is transcendental?

  2. oldrin.bataku
    • one year ago
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    Any transcendental number disproves this.

  3. terenzreignz
    • one year ago
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    Suppose pi (which is irrational) can be expressed as (x)^y where x and y are rational. Then (x)^y can be expressed as \[\huge b^{\frac{p}{q}}\] Where p and q are integers and b is rational Then pi is a solution of \[\huge x - b^{\frac{p}{q}}=0\] \[\huge x = b^{\frac{p}{q}}\] \[\huge x^{q} = b^{p}\] \[\huge x^{q} - b^{p} = 0\]effectively contradicting the transcendental-ness of pi XD

  4. wio
    • one year ago
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    I'm guessing the point is to prove the existence transcendental numbers.

  5. wio
    • one year ago
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    Or rather, to do the proof without assuming they exist.

  6. wio
    • one year ago
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    Let's say \(x = m/n\) and \(y=a/b\) \[ x^{y} = \left( \frac{m}{n}\right)^{a/b} = \left( \frac{m^a}{n^a}\right)^{1/b} = \sqrt[b]{x^a} \]We know \(x^a\) is still rational so the question then just becomes "every irrational number can be expressed as an integer root of a rational number"

  7. wio
    • one year ago
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    But this also means "Every irrational number q, when taken to some integer number n, results in a rational number"

  8. wio
    • one year ago
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    Probably should have noted earlier that if \(y\) is negative, then we can take the reciprocal of \(x\) which is still rational.

  9. wio
    • one year ago
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    Anyway, I'm not sure where to continue from here, but I think: "Every irrational number q, when taken to some integer number n, results in a rational number" Is a bit easier of a thing to disprove. That's what my gut says.

  10. wio
    • one year ago
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    We could note that: \[ \Large \sum_{n=1}^{\infty} \frac{1}{n^{2}} \]Converges, and thus is an irrational number...and try to show that there is no way putting it to any integer power would give you a rational number?? I think...

  11. wio
    • one year ago
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    Oh wait! I got an idea, how about \[ \sqrt[\sqrt{2}]{2} \]

  12. wio
    • one year ago
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    This is obviously an irrational number

  13. oldrin.bataku
    • one year ago
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    I don't see how that is relevant.

  14. wio
    • one year ago
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    Which thing?

  15. wio
    • one year ago
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    Anyway, if we take it to the \(n\)th power: \[ \sqrt[\sqrt{2}]{2}^n= \sqrt{2}^{n/\sqrt{2}} \]No matter what we set \(n\) to (it must be an integer), we are going to get and irational number to an irrational power. hmmmm

  16. wio
    • one year ago
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    @oldrin.bataku What do you think then?

  17. wio
    • one year ago
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    Without the for sight of transcendental numbers, how would you disprove it?

  18. wio
    • one year ago
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    Hmmm, I think maybe a good way is to do a proof by contradiction. Suppose \[ \sqrt{2}^\sqrt{2} = x^y \]

  19. wio
    • one year ago
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    We know the inverse of \(y\) is also rational, let's say \(z = 1/y\)

  20. wio
    • one year ago
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    \[ \Large x = \sqrt{2}^{ \sqrt{2}z} \]

  21. wio
    • one year ago
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    \[ \Large x = 2^{\sqrt{2}z/2} \]

  22. wio
    • one year ago
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    \[ \Large x^2 = 2^{\sqrt{2}z} \]

  23. wio
    • one year ago
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    Now if only I can show that \[ \Large 2^{\sqrt{2}z} \]is not rational....

  24. wio
    • one year ago
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    I know that a rational number times an irrational number is irrational, but an integer to an irrational power...?

  25. oldrin.bataku
    • one year ago
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    @wio the question states \(x,y\) are *rational*, so \(\sqrt2\) is no good here.

  26. wio
    • one year ago
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    'proof by contradiction'

  27. oldrin.bataku
    • one year ago
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    @wio ... that doesn't disprove the claim however, at all. The only relevance \(\sqrt2\) has is that it's an algebraic irrational number that works for our claim, i.e. \(2^\frac12\) where \(2,\frac12\) are rational.

  28. wio
    • one year ago
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    We know that \(x^2\) is rational. If \(2^{\sqrt{2}z}\) isn't rational then it's a contradiction, that's where I was going.

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