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Math Help - Proof using contradiction!

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    Proof using contradiction!

    Let x, y be real numbers such that x<= y+c for every c>0. Then x<=y. Prove this by contradiction....
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Vedicmaths View Post
    Let x, y be real numbers such that x<= y+c for every c>0. Then x<=y. Prove this by contradiction....
    Assume to the contrary that x <= y + c but x > y

    since x > y we have x - y > 0. Choose c > 0 such that c < x - y (this is of course possible, c = (x - y)/2 is an example).

    Thus we have: x <= y + c < y + x - y = x. So that x < x. A contradiction!
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    Thank you very much Sir..

    but I could not understand one point here:
    "since x > y we have x - y > 0. Choose c > 0 such that c < x - y (this is of course possible, c = (x - y)/2 is an example)."..

    How could you say that c < x - y....example is not very clear!

    Thanks!
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    Quote Originally Posted by Vedicmaths View Post
    Thank you very much Sir..

    but I could not understand one point here:
    "since x > y we have x - y > 0. Choose c > 0 such that c < x - y (this is of course possible, c = (x - y)/2 is an example)."..

    How could you say that c < x - y....example is not very clear!

    Thanks!

    Jhevon
    is asking you to choose a positive c such that x - y > c, and moreover he has given you an example of (x-y)/2 in case you worry how to choose...
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    yeah that makes sense perfectly now...that is what I was wondering for..how do you choose that c = (x - y)/2 and hence c < x - y...but now I understand that c is any number greater than 0 and that is we choosing it to be equal to (x-y)/2...

    Thank you!
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    Lord of certain Rings
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    Quote Originally Posted by Vedicmaths View Post
    yeah that makes sense perfectly now...that is what I was wondering for..how do you choose that c = (x - y)/2 and hence c < x - y...but now I understand that c is any number greater than 0 and that is we choosing it to be equal to (x-y)/2...

    Thank you!
    Yes thats right
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