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tomiko Group Title

prove that sqrt(5) is irrational :)

  • 2 years ago
  • 2 years ago

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  1. Sheng Group Title
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    it's not a perfect square, so it's irrational

    • 2 years ago
  2. tomiko Group Title
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    how to prove this?

    • 2 years ago
  3. Sheng Group Title
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    all square roots of numbers other than perfect squares yields an irrational response, because that's the definition of a perfect square

    • 2 years ago
  4. tomiko Group Title
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    i know this already. but if you were given this question to work out. how would you do it?

    • 2 years ago
  5. asnaseer Group Title
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    One way is to prove by contradiction. i.e. start off by assuming that \(\sqrt{5}\) IS rational and show that this leads to a contradiction - thus proving that it is irrational.

    • 2 years ago
  6. tomiko Group Title
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    if i do this assumption. i can say that \[\sqrt{5} = \frac{ a }{ b }\] then \[5 = \frac{ a ^{2} }{ b ^{2} }\] \[5b ^{2} = a ^{2}\] and then i'm stuck.

    • 2 years ago
  7. asnaseer Group Title
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    all steps are correct so far :)

    • 2 years ago
  8. asnaseer Group Title
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    you also need to state at the start that a and b are co-prime, i.e. they have no common factors so that a/b is a fraction in its simplest form. now think about what the last step means - what does it say about \(a^2\)

    • 2 years ago
  9. asnaseer Group Title
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    yo found:\[a^2=5b^2\]therefore \(a^2\) must be a multiple of ???

    • 2 years ago
  10. tomiko Group Title
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    multiple of 5

    • 2 years ago
  11. asnaseer Group Title
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    correct. now 5 is a prime number, so if \(a^2\) is a multiple of 5, then a must also be a multiple of 5 - agreed?

    • 2 years ago
  12. tomiko Group Title
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    yes...

    • 2 years ago
  13. asnaseer Group Title
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    therefore we can express a as:\[a=5m\]where m is some other integer. therefore:\[a^2=(5m)^2=25m^2\]now substitute this back into your expression that relates a^2 to b^2 and see what that leads to.

    • 2 years ago
  14. tomiko Group Title
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    i get \[25m ^{2} = 5b ^{2}\]

    • 2 years ago
  15. asnaseer Group Title
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    yup - now divide both sides by 5 and see what you can say about \(b^2\)...

    • 2 years ago
  16. tomiko Group Title
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    \[5m ^{2} = b ^{2}\]

    • 2 years ago
  17. asnaseer Group Title
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    ... so \(b^2\) must be a multiple of ???

    • 2 years ago
  18. tomiko Group Title
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    or 5, just like \[a ^{2}\] was

    • 2 years ago
  19. tomiko Group Title
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    of**

    • 2 years ago
  20. asnaseer Group Title
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    yes - and, following the same reasoning as we did for \(a^2\), we conclude that b is also a multiple of 5.

    • 2 years ago
  21. asnaseer Group Title
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    so we have concluded that both a and b are multiples of 5. which means they are NOT co-prime, i.e. they have a common factor of 5. this contradicts our original assumptions, therefore \(\sqrt{5}\) must be irrational.

    • 2 years ago
  22. asnaseer Group Title
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    I hope that makes sense?

    • 2 years ago
  23. tomiko Group Title
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    makes sense if we can say that a rational number is one which when expressed as a fraction has no coprime (except 1)

    • 2 years ago
  24. asnaseer Group Title
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    yes - that is how the proof is supposed to start off as - remember I stated it above as: "you also need to state at the start that a and b are co-prime, i.e. they have no common factors so that a/b is a fraction in its simplest form."

    • 2 years ago
  25. tomiko Group Title
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    makes sense. thanks! goodnight!

    • 2 years ago
  26. asnaseer Group Title
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    yw :) and goodnight to you too.

    • 2 years ago
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