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Math Help - Infinite Number Question

  1. #1
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    Infinite Number Question

    My college algebra teacher was explaining how an isoceles right angle triangle will always have a hypotenuse of an irrational number, where 1 ^2 + 1^2 = square root of 2 = 1.414213562...etc. My observation and question to him was that, while I can kind of understand how a number like Pi is infinite, given that it's based on a circle, I do not see why a triangle with 2 equal sides have a hypotenuse of an irrational number. In my mind, such straight lines of known proportions would be the DEFINITION of rational. He really couldn't explain WHY this is the case.

    Can anyone here offer an explanation, other than "that's just the way it is"? Thanks.
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  2. #2
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    Quote Originally Posted by paulmo View Post
    My college algebra teacher was explaining how an isoceles right angle triangle will always have a hypotenuse of an irrational number
    This is quite simply not true. Perhaps you forgot to add that there is some restriction on the legs of the triangle, e.g., they must be integers?

    If the legs are length a and hypotenuse length x, we get the relation

    2a^2 = x^2

    x = a*sqrt(2)

    Obviously letting a = 1/sqrt(2) will result in a rational hypotenuse.
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  3. #3
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    Quote Originally Posted by undefined View Post
    This is quite simply not true. Perhaps you forgot to add that there is some restriction on the legs of the triangle, e.g., they must be integers?

    If the legs are length a and hypotenuse length x, we get the relation

    2a^2 = x^2

    x = a*sqrt(2)

    Obviously letting a = 1/sqrt(2) will result in a rational hypotenuse.
    ...His prof must have meant something else.
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  4. #4
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    OK, a^2+b^2=c^2
    Take 5^2+5^2=25, which squared = 7.0710678...etc. An irrational number. Works with any number. Try it.

    Explanation? Thanks.
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  5. #5
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    Quote Originally Posted by paulmo View Post
    OK, a^2+b^2=c^2
    Take 5^2+5^2=25, which squared = 7.0710678...etc. An irrational number. Works with any number. Try it.

    Explanation? Thanks.
    Did you even read what I wrote?
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    Quote Originally Posted by paulmo View Post
    OK, a^2+b^2=c^2
    Take 5^2+5^2=25, which squared = 7.0710678...etc. An irrational number. Works with any number. Try it.
    It does not work for any number.
    Let a=b=\dfrac{\sqrt{2}}{2} then a^2+b^2=1.
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    OK, sorry I meant = 50 not 25. Yes I read what you wrote. So the irrational hypotenuse only resolves with INTEGERS? If so, that is probably what prof was referring to.

    Still, my main question stands, that no one has addressed: why, in this situation, with straight lines with KNOWN endings does a hypotenuse resolve as irrational?
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    Quote Originally Posted by paulmo View Post
    OK, sorry I meant = 50 not 25. Yes I read what you wrote. So the irrational hypotenuse only resolves with INTEGERS? If so, that is probably what prof was referring to.
    Still, my main question stands, that no one has addressed: why, in this situation, with straight lines with KNOWN endings does a hypotenuse resolve as irrational?
    As a retired chair of a mathematics department, I wish I could say that you probably misunderstood your college algebra instructor, but sadly I cannot say that.

    That said, this is a true statement: If in an isosceles right triangle the legs have rational length then the hypotenuse has irrational length.
    If a\in \mathbb{Q} then a^2+a^2=2a^2.
    So c=a\sqrt{2}.
    The number \sqrt{2} is irrational so a\sqrt{2} must be irrational.
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  9. #9
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    But why do you object so to "irrational numbers"? The length of a hypotenuse of a right triangle must be a number and there's nothing wrong with irrational numbers- most numbers are irrational (in fact, in a very specific sense, "almost all" number are irrational).

    (Perhaps it has something to do with your titling this thread "Infinite Numbers Question". This whole subject has nothing at all to do with "infinite numbers.)
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  10. #10
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    Hallsofivy, I'm sorry: this post should have been titled "Irrational Numbers Question." And thanks for your points that irrational numbers are OK, even though I have a hard time conceptualizing WHY they are irrational (ie. infinite and not in a repeating sense) in regard to STRAIGHT, CONNECTED LINES

    Plato: can you please interpret your reply as to why you wish I misunderstood my instructor? And please explain what these symbols mean: a\in \mathbb{Q} (your code introducing the Pythagorean Theorem)?

    Thanks.
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