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Math Help - please help me with these exercises

  1. #1
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    Unhappy please help me with these exercises

    I need your help for these 3 exercises. Ex.1 Considering that x+y=a+b and X2+y2=a2+b2 demonstrate that xn+yn=an+bn which n is a natural number. (n is the exponent.) Ex.2 Demonstrate that the quadratic sum of 2 odd numbers is not a full quadrate number. (It means (a+b)2 Ex.3 Considering that a x b> a+b a>0 and b>0 demonstrate that a+b>4. I hope you understand me and I am sorry for my english.
    Last edited by blertta; January 19th 2008 at 05:15 AM.
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  2. #2
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    Quote Originally Posted by blertta View Post
    I need your help for these 3 exercises. Ex.1 Considering that x+y=a+b and X2+y2=a2+b2 demonstrate that xn+yn=an+bn which n is a natural number. n is the exponent. Ex.2 Demonstrate that the quadratic sum of 2 uneven numbers is not a full quadrate number. Ex.3 Considering that a x b> a+b a>0 and b>0 demonstrate that a+b>4. I hope you understand me and I am sorry for my english.
    To get you started:

    Ex 1: Not sure what you're asking - perhaps you should look at Fermat's Last Theorem ....?

    Ex 2: Look at Euclid's formula for generating a Pythagorean triple .....
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  3. #3
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    Hello, blertta!

    I think I understand the second problem . . .


    2) Demonstrate that the sum of the squares of two odd integers is not a square.

    Consider the square of an even integer: . (2m)^2 \:=\:4m^2
    Consider the square of an odd integer: . (2n+1)^2 \:=\:4(n^2+n) + 1


    \text{The square of an integer is: }\begin{array}{c}\text{[1] a multiple of 4, or} \\ \text{[2] one more than a multiple of 4}\end{array}


    Now consider the sum of the squares of two odd integers:

    . . (2a+1)^2 + (2b+1)^2 \;=\;4a^2 + 4a + 1 + 4b^2 + 4b + 1 \;=\;4(a^2+b^2+a+b) + 2

    This is two more than a multiple of 4 . . . It cannot be a square.

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  4. #4
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    Quote Originally Posted by blertta View Post
    Ex.3 Considering that a x b> a+b a>0 and b>0 demonstrate that a+b>4. I hope you understand me and I am sorry for my english.
    There is a problem with this:

    If a = b = 0.1 then 0.1^2 = 0.01 \not > 0.2.

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    There is a problem with this:

    If a = b = 0.1 then 0.1^2 = 0.01 \not > 0.2.

    -Dan
    I think the intent of the question is:

    IF a b > a + b then a + b > 4.

    In other words, show that a b > a + b \Rightarrow a + b > 4.

    Your counter-example doesn't satisfy the necessary condition. (And yes, I too fiddled around for a bit thinking a counter-example might exist).

    Perhaps the fact that a b > a + b is only true when a b - a > b \Rightarrow a(b - 1) > b \Rightarrow \left\{ \begin{array}{cl} a > \frac{b}{b - 1}, &  b > 1 \\ \\ a < \frac{b}{b - 1}, & 0 < b < 1 \\ \\ \text{False}, & b = 1 \end{array}\right.

    leads somewhere ....

    For b > 1, this gives a + b > \frac{b}{b - 1} + b = b + 1 + \frac{1}{b - 1}. The graph of y = b + 1 + \frac{1}{b - 1} has a minimum turning point at b = 2 (which satisfies b > 1), corresponding to y = 4 ......
    Last edited by mr fantastic; January 19th 2008 at 04:21 PM. Reason: Fixed up the latex spacing, added the b > 1 case at the end and corrected a typo
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