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Math Help - 2 Proof Questions

  1. #1
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    2 Proof Questions

    Anyone got a step by step approach to solve this, it aint that hard just checkin my work

    Prove (There exists a real number x)(For all real numbers y)(4+6y-y^2<x)

    and the 2nd one is

    Prove sqrt(2) +sqrt(5) is irrational

    the way i did this was by contradiction and i proved that sqrt(10) is irrational because via contradiction i said sqrt(10) i srational then for integers m, n swrt(10) = m/n and m and n are both not even integers, but as i worked it i showed that m and n were both even therefore it cantbe rational, was my approach correct?
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by ruprotein View Post
    Anyone got a step by step approach to solve this, it aint that hard just checkin my work

    Prove (There exists a real number x)(For all real numbers y)(4+6y-y^2<x)

    and the 2nd one is

    Prove sqrt(2) +sqrt(5) is irrational

    the way i did this was by contradiction and i proved that sqrt(10) is irrational because via contradiction i said sqrt(10) i srational then for integers m, n swrt(10) = m/n and m and n are both not even integers, but as i worked it i showed that m and n were both even therefore it cantbe rational, was my approach correct?
    for the first, 4+6y-y^2 is a real number for every real number y and you can always a real number greater than a given real number.
    hence there exists infinite x for every y

    for the second, your approach is absolutely correct.

    Keep Smiling
    Malay
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  3. #3
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    thanks

    Can you just say that for the 1st part though im sure you have to say it in more detail, i said somethinlike what u said i received 4 out of 12 pts out of it and i dunno why

    What about this

    Step 1: Assume y is an arbirary real number
    Step 2: Then 4+6y^2-y^2 is a real number [Axiom (For all x)(x+y element of real #'s)
    Step 3: Assume x is some real number... dunno how to finish it off
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by ruprotein View Post
    thanks

    Can you just say that for the 1st part though im sure you have to say it in more detail, i said somethinlike what u said i received 4 out of 12 pts out of it and i dunno why

    What about this

    Step 1: Assume y is an arbirary real number
    Step 2: Then 4+6y^2-y^2 is a real number [Axiom (For all x)(x+y element of real #'s)
    Step 3: Assume x is some real number... dunno how to finish it off
    It may be like this
    let a is a positive real number
    let x= 4+6y^2-y^2 + a=sum of two real numbers.
    now x is a real numbers and for any postive real value of a, it is greater than 4+6y^2-y^2

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  5. #5
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    Quote Originally Posted by malaygoel View Post
    for the first, 4+6y-y^2 is a real number for every real number y and you can always a real number greater than a given real number.
    hence there exists infinite x for every y
    You are supposed to show that there is a x \in \mathbb{R} for which 4+6y-y^2 < x for all y \in \mathbb{R}.

    RonL
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  6. #6
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    Quote Originally Posted by ruprotein View Post
    Prove sqrt(2) +sqrt(5) is irrational

    the way i did this was by contradiction and i proved that sqrt(10) is irrational because via contradiction i said sqrt(10) i srational then for integers m, n swrt(10) = m/n and m and n are both not even integers, but as i worked it i showed that m and n were both even therefore it cantbe rational, was my approach correct?
    Why are you working with \sqrt{10}?

    \sqrt{2}+ \sqrt{5} \ne \sqrt{10}

    so if this approach is to work you need to show how \sqrt{10} is related to \sqrt{2}+ \sqrt{5}.

    RonL
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  7. #7
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    (sqrt(2)+sqrt(5))^.5 = 7 + 2*sqrt(2)*sqrt(5) = 7 + 2*sqrt(10)
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  8. #8
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    Quote Originally Posted by ruprotein View Post
    (sqrt(2)+sqrt(5))^.5 = 7 + 2*sqrt(2)*sqrt(5) = 7 + 2*sqrt(10)
    Hello,

    this transformation isn't correct:

    \sqrt{\left( \sqrt{2}+\sqrt{5} \right)^2}=\sqrt{ 7+2 \cdot \sqrt{10} }

    But does this really help?
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  9. #9
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    Quote Originally Posted by ruprotein View Post
    Anyone got a step by step approach to solve this, it aint that hard just checkin my work

    Prove (There exists a real number x)(For all real numbers y)(4+6y-y^2<x)
    Show that there exists an x \in \mathbb{R} such that for all y \in \mathbb{R}:\ \ 4+6y-y^2 < x.

    Suppose otherwise, then for any X \in \mathbb{R} there exists a k \in \mathbb{R},\ k>X such that:

    4+6y-y^2=k,

    or:

    y^2-6y+(k-4)=0

    The quadratic formula can be used to find those y's that satisfy this, and these are:

    y=\frac{6\pm\sqrt{36-4(k-4)}}{2}

    This has real roots only if the discriminant 36-4(k-4) \ge 0 or k \le 13.

    So if X=14, there is no k having the properties assumed, which is a contradiction, hence there exists an x such that:

    4+6y-y^2 < x

    for all y, and x=14 is one such.

    RonL
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  10. #10
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    Hello, ruprotein!

    In #2, m and n are relative prime.
    . . It does not involve even/odd properties.


    Prove \sqrt{2} + \sqrt{5} is irrational.

    First, we will prove that \sqrt{10} is irrational.

    Assume \sqrt{10} is rational.
    Then: \sqrt{10} \:=\:\frac{a}{b} for relatively prime integers a and b.
    . . That is, \frac{a}{b} is already reduced to lowest terms.

    Square both sides: . 10 \:=\:\frac{a^2}{b^2}\;\;\Rightarrow\;\;10b^2\:=\:a  ^2 [1]
    Since a^2 is a multiple of 10, then a is a multiple of 10: . a \,=\,10p

    Substitute into [1]: . 10b^2\:=\:(10p)^2\;\;\Rightarrow\;\;10b^2\:=\:100p  ^2\;\;\Rightarrow\;\;b^2\:=\:10p^2
    Since b^2 is a multiple of 10, then b is a multiple of 10: . b\,=\,10q

    But this contradicts the fact that a and b are relatively prime.
    . . Therefore, \sqrt{10} is irrational.


    Assume \sqrt{2} + \sqrt{5} is rational.

    Then: . \sqrt{2} + \sqrt{5} \:=\:\frac{m}{n} for relatively prime integers m and n.
    . . That is, \frac{m}{n} is already reduced to lowest terms.

    Square both sides: . \left(\sqrt{2} + \sqrt{5}\right)^2 \:=\:\left(\frac{m}{n}\right)^2
    . . and we have: . 2 + 2\sqrt{10} + 5 \:=\:\frac{m^2}{n^2}\;\;\Rightarrow\;\;\sqrt{10} \:=\:\frac{\frac{m^2}{n^2} - 7}{2}
    Hence: . \sqrt{10}\:= \:\frac{m^2-7n^2}{2n^2}

    The right side is rational. .It is the product, difference and quotient of integers.
    . . But we have shown that the left side is irrational.
    We have reached a contradiction.

    Hence, our assumption was incorrect.
    . . Therefore: . \sqrt{2} + \sqrt{5} is irrational.

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  11. #11
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    The set of point,
    4+6y-y^2
    Has a maximum value.
    (An upside down parabola).
    Thus, if you choose a number above the maximum point then you are safe.
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  12. #12
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    [QUOTE=Soroban;33016]Hello, ruprotein!

    In #2, m and n are relative prime.
    . . It does not involve even/odd properties.


    [size=3]
    First, we will prove that \sqrt{10} is irrational.

    Assume \sqrt{10} is rational.
    Then: \sqrt{10} \:=\:\frac{a}{b} for relatively prime integers a and b.
    . . That is, \frac{a}{b} is already reduced to lowest terms.

    Square both sides: . 10 \:=\:\frac{a^2}{b^2}\;\;\Rightarrow\;\;10b^2\:=\:a  ^2 [1]
    Since a^2 is a multiple of 10, then a is a multiple of 10: . a \,=\,10p

    Substitute into [1]: . 10p)^2\;\;\Rightarrow\;\;10b^2\:=\:100p^2\;\;\Righ tarrow\;\;b^2\:=\:10p^2" alt="10b^2\:=\10p)^2\;\;\Rightarrow\;\;10b^2\:=\:100p^2\;\;\Righ tarrow\;\;b^2\:=\:10p^2" />
    Since b^2 is a multiple of 10, then b is a multiple of 10: . b\,=\,10q

    But this contradicts the fact that a and b are relatively prime.
    . . Therefore, \sqrt{10} is irrational.


    =========================

    Here is the other proof of sqrt(10) is irrational:

    Theorem. If nth root of a is not an integer , then in fact nth root of a is irrational.

    sqrt(9) < sqrt(10)< sqrt(16)
    3 < sqrt(10) < 4

    So sqrt(10) is not an integer, hence it is irrational.
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  13. #13
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    Quote Originally Posted by Jenny20 View Post

    Here is the other proof of sqrt(10) is irrational:

    Theorem. If nth root of a is not an integer , then in fact nth root of a is irrational.

    sqrt(9) < sqrt(10)< sqrt(16)
    3 < sqrt(10) < 4

    So sqrt(10) is not an integer, hence it is irrational.
    I think the user is asked not to use that method. Otherwise it is too easy.
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  14. #14
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    i see.
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  15. #15
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    Quote Originally Posted by ThePerfectHacker View Post
    The set of point,
    4+6y-y^2
    Has a maximum value.
    (An upside down parabola).
    Thus, if you choose a number above the maximum point then you are safe.
    Pointing at a diagram does not constitute a proof. You will have the
    Bourbaki police after you at this rate.

    RonL
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