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shaniehh

  • one year ago

A student is attempting to show that the product of a nonzero rational number and an irrational number is always rational or always irrational.

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  1. shaniehh
    • one year ago
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    Part A: Find the value of \[0.22 \] * \[\sqrt{112}\]and place it in simplified form.

  2. shaniehh
    • one year ago
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    Part B: Is the answer in Part A irrational or rational? Make a conjecture about the product of a nonzero rational number and an irrational number.

  3. anonymous
    • one year ago
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    not really clear what the question is find the value of what?

  4. shaniehh
    • one year ago
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    I need help with part B the answer to A is 2.32826115

  5. anonymous
    • one year ago
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    it just says "make a conjecture" but the truth is that a rational number times an irrational number is always irrational

  6. anonymous
    • one year ago
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    the proof is very simple if you want to see it

  7. shaniehh
    • one year ago
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    yes please

  8. anonymous
    • one year ago
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    a rational number is any number that can be expressed as a fraction (ratio of two integers) an irrational number is one that cannot be expressed that way we need to know this to start

  9. anonymous
    • one year ago
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    perhaps the easiest way to show that the product of an irrational number and a rational number is irrational is to prove it by contradiction, that is, assume that the product IS rational, then get a contradiction

  10. anonymous
    • one year ago
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    so suppose \(a\) is rational, \(b\) is irrational and \(ab=c\) is rational that means that \(b=\frac{c}{a}\) is irrational, but since both \(c\) and \(a\) are rational so is \(\frac{c}{a}\) contradicting the fact that \(b\) if irrational

  11. anonymous
    • one year ago
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    btw if this seems like kicking the can down the road, it is not you can prove for yourself that if \(c, a\) are rational, then so is \(\frac{c}{a}\) by writing them both as the ratio of two integers, invert and multiply to get another ratio of two integers

  12. shaniehh
    • one year ago
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    thank you

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