Rational and Irrational numbers

**Geometor** · May 22nd 2007, 08:47 AM

Now, don't think this is just an ordinary question! Here goes:

a and b are irrational numbers
a + b is rational
a x b is rational

Work out possible values for a and b where they are positive irrational numbers.

I am sure that I have gotten down all the details, any help will be much appreciated, thank you!

**Jhevon** · May 22nd 2007, 09:01 AM

Originally Posted by Geometor

Now, don't think this is just an ordinary question! Here goes:

a and b are irrational numbers
a + b is rational
a x b is rational

Work out possible values for a and b where they are positive irrational numbers.

I am sure that I have gotten down all the details, any help will be much appreciated, thank you!

a = (1 + sqrt(2))
b = -sqrt(2)
then a and b are irrational numbers and...

a + b = (1 + sqrt(2)) + (-sqrt(2)) = 1 which is rational

a = sqrt(2)
b = 1/sqrt(2)
then a and b are both irrational and...

a*b = sqrt(2)*1/sqrt(2) = 1 which is rational

i should be using this oppurtunity to try out Latex, i don't know how to use it

EDIT: o sorry, i forgot a and b must be positive, so the first example is incorrect, i'll come up with another one

how about
a = 10 - sqrt(5) + sqrt(2)
b = sqrt(5) - sqrt(2)

then a + b = 10 which is rational and both a and b are positive irrational numbers

**Geometor** · May 22nd 2007, 11:39 AM

Originally Posted by Jhevon

a = (1 + sqrt(2))
b = -sqrt(2)
then a and b are irrational numbers and...

a + b = (1 + sqrt(2)) + (-sqrt(2)) = 1 which is rational

a = sqrt(2)
b = 1/sqrt(2)
then a and b are both irrational and...

a*b = sqrt(2)*1/sqrt(2) = 1 which is rational

i should be using this oppurtunity to try out Latex, i don't know how to use it

EDIT: o sorry, i forgot a and b must be positive, so the first example is incorrect, i'll come up with another one

how about
a = 10 - sqrt(5) + sqrt(2)
b = sqrt(5) - sqrt(2)

then a + b = 10 which is rational and both a and b are positive irrational numbers

Many thanks but I don't think that is correct since
10 - sqrt(5) + sqrt(2) times sqrt(5) - sqrt(2) is irrational :S
thanks for trying, lol this is a really complex problem ay?

is this a trick question and it might be impossible?

**ThePerfectHacker** · May 22nd 2007, 11:46 AM

Originally Posted by Geometor

Now, don't think this is just an ordinary question! Here goes:

a and b are irrational numbers
a + b is rational
a x b is rational

Work out possible values for a and b where they are positive irrational numbers.

$a=1-\sqrt{2} \mbox{ and }b=1+\sqrt{2}$

**Geometor** · May 22nd 2007, 11:49 AM

Originally Posted by ThePerfectHacker

$a=1-\sqrt{2} \mbox{ and }b=1+\sqrt{2}$

erm... the question requires "positive" numbers and I believe that 1- sq. root 2 is negative?

thanks for helping anyway

**Jhevon** · May 22nd 2007, 11:51 AM

Originally Posted by Geometor

Many thanks but I don't think that is correct since
10 - sqrt(5) + sqrt(2) times sqrt(5) - sqrt(2) is irrational :S
thanks for trying, lol this is a really complex problem ay?

is this a trick question and it might be impossible?

o, we have to come up with a pair of numbers that satisfy both condtions at the same time? i thought we could use different examples for each.

EDIT: TPH came up with an example

**ThePerfectHacker** · May 22nd 2007, 11:52 AM

Originally Posted by Geometor

erm... the question requires "positive" numbers and I believe that 1- sq. root 2 is negative?

thanks for helping anyway

So then change it a little bit,

$a = 2 - \sqrt{2} \mbox{ and } b = 2 + \sqrt{2}$

What is so hard?

**CaptainBlack** · May 22nd 2007, 12:06 PM

Originally Posted by Geometor

Now, don't think this is just an ordinary question! Here goes:

a and b are irrational numbers
a + b is rational
a x b is rational

Work out possible values for a and b where they are positive irrational numbers.

I am sure that I have gotten down all the details, any help will be much appreciated, thank you!

Because their sum is rational $a$ and $b$ are of the form:

$a = A + x$ ,

$b = B - x$

where $A$ and $B$ are rational and x is irrational.

Then:

$ a\ b = AB + (-A+B)x - x^2 = C $

where C is rational.

Now we have demanded that A, B and C be rational, but we may as well
demand that they be integers (the derivation of rational solutions from
integer solutions is fairly elementary and left to the reader).

Anyway we have:

$ x^2 + (A-B)x + (C- AB) = 0 $

and $x$ is irrational. However:

$ x=\frac{-(A+B) \pm \sqrt{(A-B)^2 - 4(C-AB)}}{2}=\frac{-(A+B) \pm \sqrt{(A+B)^2 - 4C}}{2} $

which is irrational only if $q= (A+B)^2 - 4C$ is not a perfect
square.

So here is our algorithm for finding solutions:

Choose an integer $K$ , and an integer $C$ such that $K^2 - 4C$ is not a perfect square,
then choose $A$ and $B$ so that $A+B=K$ , then:

$ x=\frac{-(A+B) \pm \sqrt{(A+B)^2 - 4C}}{2} $

is an irrational number such that if:

$a=A+x$
$b=B-x$

then $a$ and $b$ are irrational and $a+b$ is an integer as is $a \times b$

Of course this does not guarantee that both $a$ and $b$ are positive.

RonL

**Geometor** · May 22nd 2007, 12:12 PM

thanks for the help! sorry, can only thank one person a day I believe?
I find this very confusing though :S

**CaptainBlack** · May 22nd 2007, 12:14 PM

Originally Posted by Geometor

thanks for the help! sorry, can only thank one person a day I believe?
I find this very confusing though :S

No you can thank as many as you want.

RonL

**Jhevon** · May 22nd 2007, 12:15 PM

Originally Posted by Geometor

thanks for the help! sorry, can only thank one person a day I believe?

how come?

**Geometor** · May 23rd 2007, 08:50 AM

Originally Posted by Jhevon

how come?

oh woops, nevermind it's just that I can't thank on the same post so I got mixed up

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