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    Please could I have some help on the following question:

    a.) If P is irrational, show by the method of contradiction that 1/P is also irrational.

    b.) Give an example of two irrational numbers, a and b, where a does not equal b, such that ab is rational

    Many thanks!
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    assume that 1/P is rational.

    investigate the product of P and 1/P
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    (Original post by the bear)
    assume that 1/P is rational.

    investigate the product of P and 1/P
    That's great, thank you!
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    For b), it's always useful to think of the formula (x - y) (x + y) = x2 - y2
    Make x and y irrational. Their sum and difference will be irrational and different from one another, so they can be your a and b.

    Spoiler:
    Show
    Example: x = sqrt(8), y = sqrt(2)
    a = x - y = sqrt(8) - sqrt(2), irrational
    b = x + y = sqrt(8) + sqrt(2), irrational
    ab = (x - y)(x + y) = 8 - 2 = 6, which is rational
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    Assume 1/P is rational.

    i.e. 1/P = a/b; a,b are integers
    => P = b/a which means is rational. This contradicts what we know of P being irrational. Therefore we prove 1/P must be irrational.
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    (Original post by InsertNameHero)
    Assume 1/P is rational.

    i.e. 1/P = a/b; a,b are integers
    => P = b/a which means is rational. This contradicts what we know of P being irrational. Therefore we prove 1/P must be irrational.
    you haven't checked the case where b=0 lol
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    (Original post by asdfghjklcupcake)
    For b), it's always useful to think of the formula (x - y) (x y) = x2 - y2
    Make x and y irrational. Their sum and difference will be irrational and different from one another, so they can be your a and b.

    Spoiler:
    Show
    Example: x = sqrt(8), y = sqrt(2)
    a = x - y = sqrt(8) - sqrt(2), irrational
    b = x y = sqrt(8) sqrt(2), irrational
    ab = (x - y)(x y) = 8 - 2 = 6, which is rational
    It is not necessarily true that if x and y are irrational their sum or their differences are irrational. counter example:
    sqrt(2) is irrational
    2-sqrt(2) is also irrational but
    sqrt(2) + (2-sqrt(2))=2 which is definitely rational.
    hence you must prove sqrt(8) (and minus) sqrt(2) is irrational. A better solution is just sqrt(2) is irrational sqrt(8) is irrational but sqrt(2)*sqrt(8) = sqrt(16)=4 is rational lol.
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    (Original post by igotohaggerston)
    It is not true that if x and y are irrational their sum and differences are irrational. counter example:
    sqrt(2) is irrational
    2-sqrt(2) is also irrational but
    2 + (2-sqrt(2))=2 which is definitely rational.
    hence you must prove sqrt(8) +(and minus) sqrt(2) is irrational. A better solution is just sqrt(2) is irrational sqrt(8) is irrational but sqrt(2)*sqrt(8) = sqrt(16)=4 is rational lol.
    2+(2-sqrt(2))= 4-sqrt(2) ≠ 2 and is irrational
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    (Original post by InsertNameHero)
    2+(2-sqrt(2))= 4-sqrt(2) ≠ 2 and is irrational
    Sorry I meant sqrt(2) + (2-sqrt(2)) I edited the original post
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    (Original post by InsertNameHero)
    Assume 1/P is rational.

    i.e. 1/P = a/b; a,b are integers
    => P = b/a which means is rational. This contradicts what we know of P being irrational. Therefore we prove 1/P must be irrational.
    Not correct - please note in future that 'solutions' should not be posted in the maths forum.
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    (Original post by Muttley79)
    Not correct - please note in future that 'solutions' should not be posted in the maths forum.
    Sorry, can you clarify the mistake I have made?. I assumed the statement isn't true. I then showed that this lead to a contradiction of a fact we knew to be true. Then I concluded that the assumption was incorrect and the statement must be true. I thought this would be correct but can you show me where I have gone wrong?
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    (Original post by InsertNameHero)
    Sorry, can you clarify the mistake I have made?. I assumed the statement isn't true. I then showed that this lead to a contradiction of a fact we knew to be true. Then I concluded that the assumption was incorrect and the statement must be true. I thought this would be correct but can you show me where I have gone wrong?
    Read the posts by The Bear and igotohaggerston.
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    (Original post by Muttley79)
    Not correct - please note in future that 'solutions' should not be posted in the maths forum.
    Apart from not stating the triviality that 1/P isn't 0, it is correct.
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    (Original post by Zacken)
    Apart from not stating the triviality that 1/P isn't 0, it is correct.
    Yes but without that is is not correct - so as it stands it is wrong.
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    (Original post by Muttley79)
    Read the posts by The Bear and igotohaggerston.
    Neither of those posts help InsertNameHero see why his solution is "incorrect", which is arguably much more correct than not so. Simply saying it's not correct is unhelpful and pedantic.

    Igotohaggerston's post asks him to check the case b=0. When assuming that 1/P=a/b is rational, it is implicit that b is non-zero and this does not need to be stated.

    Bear's post, whilst also leading to a correct proof, also skirts over the issue of the product being undefined if 1/P is zero. I wonder if the OP made this distinction in his final solution based on the posts of this thread alone.
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    (Original post by Farhan.Hanif93)
    Neither of those posts help InsertNameHero see why his solution is "incorrect", which is arguably much more correct than not so. Simply saying it's not correct is unhelpful and pedantic.

    Igotohaggerston's post asks him to check the case b=0. When assuming that 1/P=a/b is rational, it is implicit that b is non-zero and this does not need to be stated.

    Bear's post, whilst also leading to a correct proof, also skirts over the issue of the product being undefined if 1/P is zero. I wonder if the OP made this distinction in his final solution based on the posts of this thread alone.
    My point, which you have totally overlooked, is that any solution should NOT have been posted at all. Please amend your post as it is unhelpful.
 
 
 
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