Math Is Fun Forum

sumpm1

Member

Registered: 2007-03-05

Posts: 42

Real Analysis Help

Hey guys. I need some help on the following questions. Usually, for a given problem, like proving something is odd or even, there is a trick, or a pattern that you are looking for. I am not sure what to look for when trying to prove that something IRRATIONAL.

1. Prove that √2 + √3 is irrational
2. prove that √(n-1) + √(n+1) is irrational for every positive integer.
3. Prove that there is no rational number r such that (2^r)=3

Thanks

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Real Analysis Help

You can use proof by contradiction for all 3 questions.  Start by assuming that sqrt(2) + sqrt(3) is rational, which means it is equal to some fration a/b where a and b are coprime integers.

The left side of this equation is clearly rational.  If you are allowed to assert without further proof that sqrt(6) is irrational then you have your contradiction, otherwise go ahead and show that sqrt(6) is irrational to get the contradiction.  You can use this same line of work for the second question.

Do the same thing for the third question. Let r = a/b where a and b are coprime integers and find a contradiction.

Last edited by TheDude (2009-01-23 04:28:43)

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JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Real Analysis Help

I think we can safely assume that

is irrational, since the proof of it is so well known. (

is a bit iffy.)

So, assume

were rational.

Then

would be rational.

And then

would be rational – contradiction.

This bypases the problem with the irrationality of

. Sometimes lateral thinking can be much more effective than mere application of old methods.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Real Analysis Help

For #2, however, I don’t think you can escape the necessity of assuming (or otherwise proving) that

is irrational for all square-free positive integers

.

Suppose

were rational.

Note that at least one of

and

must be square free. They can’t both be perfect squares because perfect squares never differ by exactly 2.

If

is square free, then

If

is square free, then

We have a contradiction either way.

sumpm1

Member

Registered: 2007-03-05

Posts: 42

Re: Real Analysis Help

Thanks guys, I appreciate the help that I get here. And Jane, you are a genius and never cease to amaze me, I can tell you really love this stuff! I was just working through #2 and was trying the method that TheDude suggested and did get stuck because the contradiction in the proof of

is irrational arises because we assume

and

in

have no common divisor greater than 1. At least that is how it is done in my text. But this does not occur in this case, and we do not arrive at a case where we can prove that

or

is any multiple of a number. So proving

is irrational for all square-free positive integers

is very useful.

I may seem quite dense here, but where is the contradiction in each case? Is it just because we assumed that

is not a perfect square, and then found the square root of

? I guess I am just hung up on the fact that I considered only 4,9,16,25... to be perfect squares; integers. I guess I never considered

or

to be perfect squares!

Thanks

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Real Analysis Help

The "no common divisor" method works for radical 2, but not much else.  Instead, follow the same proof for radical 2, stopping when you get integers on each side of the equation.  Now use the fundamental theorem of arithmetic to say that the two sides of the equation can never be equal.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Real Analysis Help

This is how I would prove it. Let

be a square-free integer. Then

and we can write it as the product of powers of distinct primes:

   

At least one of the

’s must be odd since

is not a perfect square. Say

is odd.

Suppose

for some positive integers

which are coprime.

Then

.

. And since

and

are coprime,

.

Thus we see that

divides the RHS of

an even number of times and the LHS of

an odd number of times. This is an impossibllity.

Hence

is irrational.

Last edited by JaneFairfax (2009-01-25 01:40:29)

sumpm1

Member

Registered: 2007-03-05

Posts: 42

Re: Real Analysis Help

The "no common divisor" method works for radical 2, but not much else.  Instead, follow the same proof for radical 2, stopping when you get integers on each side of the equation.  Now use the fundamental theorem of arithmetic to say that the two sides of the equation can never be equal.

Hey Ricky. I was able to use the same proof for proving

is irrational in #1. First I thought it would be showing that both sides were even, but instead was able to show that both sides must be multiples of 6 to get the contradiction. I will look at the fundamental theorem of arithmetic.

Thanks guys

jayk

Guest

Re: Real Analysis Help

I think we can safely assume that

is irrational, since the proof of it is so well known. (

is a bit iffy.)

So, assume

were rational.

Then

would be rational.

And then

would be rational – contradiction.

This bypases the problem with the irrationality of

. Sometimes lateral thinking can be much more effective than mere application of old methods.

how did you get

?

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Real Analysis Help

how did you get

?

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Wrap it in bacon

jay17

Member

Registered: 2009-06-21

Posts: 4

Re: Real Analysis Help

to prove that  is irrational,

let x = √2 +√3

square both sides,
x² = 2 + + 3
x² - 5 =  2√6

square again both sides,
x^4 - 10x² + 1 = 0

by the rational root theorem any rational root of this polynomial is either 1 or -1. +- 1,  contradiction

thanks avon for the correction!

Last edited by jay17 (2009-06-25 01:16:13)

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Real Analysis Help

x^4 - 10x² + 1 = 0

you can see that the only rational roots are +- 1,  contradiction

I would interpret this to mean that 1 and -1 are roots of this polynomial, which is clearly untrue.

It would be better to say something like, by the rational root theorem any rational root of this polynomial is either 1 or -1.