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Math Help - About even and uneven numbers and about rational and irrational numbers

  1. #1
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    About even and uneven numbers and about rational and irrational numbers

    ok I have some problems that I need help with:
    if a^2 is even, then prove that a is even too
    if a^2 is uneven, then prove that a is uneven too

    And another problem:
    prove that:
    rational + irrational number = irrational number
    PLS HELP ME!!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by antrewry View Post
    ok I have some problems that I need help with:
    if a^2 is even, then prove that a is even too
    use the contrapostive

    it suffices to show if a is not even, then a^2 is not even.

    so assume a is not even, so that a = 2n + 1 for some integer n, then .....

    if a^2 is uneven, then prove that a is uneven too
    same approach as above

    by the way, "odd" means "uneven" and it sounds nicer.

    And another problem:
    prove that:
    rational + irrational number = irrational number
    PLS HELP ME!!!
    Let \frac ab be a rational number and c be an irrational number. what can you say about their sum?
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  3. #3
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    Proof:
    Let x be an even interger. Then by definition of an even number we can write:
    x=2m
    where m is an interger.
    Therefore x^2= (2m)^2
    =4m^2
    =2(m^2)
    Let P =2m^2. Since the product of integers is an interger, p is an itnerger. Thus x^2 =2p and x^2 is an even integer.
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  4. #4
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    And another problem:
    prove that:
    rational + irrational number = irrational number
    PLS HELP ME!!!
    Let \frac ab be a rational number and c be an irrational number. what can you say about their sum?
    I think it would be funny to assume that \frac ab+c=\frac de, where \frac de is a rationnal. (a,b,d,e are non-zero integers)

    Now, is it logical to say : \underbrace{c}_{\text{irrational}}=\frac de-\frac ab=\frac{bd-ae}{be} ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by reagan3nc View Post
    Proof:
    Let x be an even interger. Then by definition of an even number we can write:
    x=2m
    where m is an interger.
    Therefore x^2= (2m)^2
    =4m^2
    =2(m^2)
    Let P =2m^2. Since the product of integers is an interger, p is an itnerger. Thus x^2 =2p and x^2 is an even integer.
    you should state what question you are answering

    and this approach was already given. give the poster some time to see if they can figure things out
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  6. #6
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    OK not exactly

    Quote Originally Posted by reagan3nc View Post
    Proof:
    Let x be an even interger. Then by definition of an even number we can write:
    x=2m
    where m is an interger.
    Therefore x^2= (2m)^2
    =4m^2
    =2(m^2)
    Let P =2m^2. Since the product of integers is an interger, p is an itnerger. Thus x^2 =2p and x^2 is an even integer.
    Not exactly what I asked, I want to prove that IF a^2 is even THEN a is even too, not the opposite. I approciate the help and efford though. Thank you very much!
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  7. #7
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    Thumbs up

    Quote Originally Posted by Jhevon View Post
    use the contrapostive

    it suffices to show if a is not even, then a^2 is not even.

    so assume a is not even, so that a = 2n + 1 for some integer n, then .....

    same approach as above

    by the way, "odd" means "uneven" and it sounds nicer.

    Let \frac ab be a rational number and c be an irrational number. what can you say about their sum?
    Very helpful, I found the solutions through your guidance. Thank you very much. And I will use "odd" next time if you say so
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  8. #8
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    Talking Thank you

    Quote Originally Posted by Moo View Post
    I think it would be funny to assume that \frac ab+c=\frac de, where \frac de is a rationnal. (a,b,d,e are non-zero integers)

    Now, is it logical to say : \underbrace{c}_{\text{irrational}}=\frac de-\frac ab=\frac{bd-ae}{be} ?

    Thanks for your help!!!
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by antrewry View Post
    Not exactly what I asked, I want to prove that IF a^2 is even THEN a is even too, not the opposite. I approciate the help and efford though. Thank you very much!
    this is not addressing that question (if it is, the poster made a mistake). this actually proves "If x^2 is odd, then x is odd" by the contrapositive. that's why i asked the poster to state what problem he/she was addressing.
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