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!

Buzzz1325
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 13082018 11:24

the bear
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 13082018 11:28
assume that 1/P is rational.
investigate the product of P and 1/P 
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 13082018 15:08

asdfghjklcupcake
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 14082018 09:09
For b), it's always useful to think of the formula (x  y) (x + y) = x^{2}  y^{2}
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:ShowExample: 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 rationalLast edited by asdfghjklcupcake; 14082018 at 09:17. 
InsertNameHero
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 14082018 09:17
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.Posted on the TSR App. Download from Apple or Google Play 
igotohaggerston
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 14082018 09:21
(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.Posted on the TSR App. Download from Apple or Google Play 
igotohaggerston
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 14082018 09:27
(Original post by asdfghjklcupcake)
For b), it's always useful to think of the formula (x  y) (x y) = x^{2}  y^{2}
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:ShowExample: 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
sqrt(2) is irrational
2sqrt(2) is also irrational but
sqrt(2) + (2sqrt(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.Posted on the TSR App. Download from Apple or Google PlayLast edited by igotohaggerston; 14082018 at 09:34. 
InsertNameHero
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 14082018 09:30
(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
2sqrt(2) is also irrational but
2 + (2sqrt(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.Posted on the TSR App. Download from Apple or Google Play 
igotohaggerston
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 14082018 09:32
(Original post by InsertNameHero)
2+(2sqrt(2))= 4sqrt(2) ≠ 2 and is irrationalPosted on the TSR App. Download from Apple or Google Play 
Muttley79
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 14082018 09:43
(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. 
InsertNameHero
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 14082018 09:54
(Original post by Muttley79)
Not correct  please note in future that 'solutions' should not be posted in the maths forum.Posted on the TSR App. Download from Apple or Google Play 
Muttley79
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 14082018 09:57
(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? 
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 14082018 10:01
(Original post by Muttley79)
Not correct  please note in future that 'solutions' should not be posted in the maths forum. 
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 14082018 10:13
(Original post by Zacken)
Apart from not stating the triviality that 1/P isn't 0, it is correct. 
Farhan.Hanif93
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 14082018 11:11
(Original post by Muttley79)
Read the posts by The Bear and igotohaggerston.
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 nonzero 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. 
Muttley79
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 14082018 11:14
(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 nonzero 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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