proving inequality with sqr roots (1 Viewer)

[underthesun]

this question is so annoying, someone help me!

prove:

sqrt(5 + sqrt(5)) - sqrt(5 - sqrt(5)) > 1

is there any general tips with square roots that appear like this? thanks in advance

 

[nike33]

lol ill give it a go

sqrt(5 + sqrt(5)) - sqrt(5 - sqrt(5)) > 1

LHS

= sqrt(5 + sqrt(5)) - sqrt(5 - sqrt(5))
= sqrt(10-4sqr(5)) after some minor algebra

now for sqrt(10-4sqr(5)) to be > 1 then
10-4sqr(5) > 1

which is true as 2.25 > sqr(5) (as 81/16 > 5)
(ie 81/16 > 80/16)

 

[underthesun]

okay, can you please expand the "minor algebra" part? Im algebraically, mathematically and logically challenged here .

cheers for the solution though

 

[CM_Tutor]

The minor algebra involves changing LHS to sqrt(LHS²)

So, LHS = sqrt{[(sqrt(5 + sqrt(5)) - sqrt(5 - sqrt(5))]²}
= sqrt{[sqrt(5 + sqrt(5))]² - 2 * sqrt(5 - sqrt(5)) * sqrt(5 + sqrt(5)) + [sqrt(5 - sqrt(5))]²}
= sqrt{5 + sqrt(5) - 2 * sqrt[(5 - sqrt(5)) * (5 + sqrt(5))] + 5 - sqrt(5)}
= sqrt{10 - 2 * sqrt[5² - (sqrt(5))²]}
= sqrt[10 - 2 * sqrt(20)]
= sqrt(10 - 4 * sqrt(5)), as 20 = 4 * 5

 

[underthesun]

ok thanks for that, cheers

just another simple question (I think this thread title should be changed to "basic questions for my algebra exam")

prove that log₁₀3 is irrational.

Now for this one I started with stating its contrary, that log₁₀3 = p/q (whereas p & q rationals). Now, i completely forgot what kind of contradiction I was supposed to achieve..

 

[nike33]

one possible contradiction is for example you find both p and q are even if defining log(_10)3 = p/q where p/q is the simplist fraction ...i think

 

[Xayma]

umm from there you are suppose to prove that p or q cant be an integer (ie one of them is irrational itself) (ie contradicting the first statement)

 

[nike33]

ahh failed attempt 1

 

[CM_Tutor]

Originally posted by nike33
...i think as b/a is defined as rational then 10^(b/a) must be irrational..you may have to prove this --> i think this is true as im pretty sure e(b/a) for example, b,a rational is irrational

Sorry, but this isn't true. Suppose b = 2, a = 1, in which case b / a is rational. 10^b/a = 100 is also rational.

 

[nike33]

ok well is it true for e(b/a) ? then it would be true for ln2

 

[CM_Tutor]

I think that you're thinking of Gelfond's Theorem, which was mentioned recently by KeypadSDM. Here's a reference:

http://mathworld.wolfram.com/GelfondsTheorem.html

It came in a discussion of transcendental numbers in the thread on stupid proofs involving e.

You could also have a look at:

http://mathworld.wolfram.com/IrrationalNumber.html

 

[nike33]

ok attempt 2

log_10 3 = b/a. b,a integers
10^(b/a) = 3
10^b = 3^a

10^b is even... 3^a is odd... which means log_10 3 is irrational??

 

[underthesun]

Originally posted by nike33
ok attempt 2

log_10 3 = b/a. b,a integers
10^(b/a) = 3
10^b = 3^a

10^b is even... 3^a is odd... which means log_10 3 is irrational??

ok great, that one looks pretty good for me

now for next question (bet your wondering when this will end):

let a and n be integers, greater than 1. prove aⁿ - 1 is prime only if a = 2 and n is prime. Is the converse of this statement true? Show that 2ⁿ + 1 is prime only if n is a power of 2.

Dah, discrete maths is killing me. Thanks a lot in advance

 

[CM_Tutor]

Nike33, attempt 2 works, but should be cleaned up a bit to be a proper proof.

Theorem: log₁₀3 is irrational

Proof: By contradiction

Suppose the theorem is false, and so log₁₀3 is rational.
It follows that log₁₀3 = p / q, where p and q are positive integers, q <> 0, and p and q have no common factors other than 1.
So, 3 = 10^p/q
and so 3^q = 10^p
Now, 10^p must be divisible by 10, and hence by 2, as p is an integer and p => 1.
But, since q is also an integer and q => 1, the only factors of 3^q are 1, 3 and powers of 3. Thus, 3^q is not divisible by 2.

This is a contradiction, and so our supposition (that the theorem is flase) is false, and so the theorem is true and log₁₀3 is irrational.

 

[CM_Tutor]

Originally posted by underthesun
let a and n be integers, greater than 1. prove aⁿ - 1 is prime only if a = 2 and n is prime. Is the converse of this statement true? Show that 2ⁿ + 1 is prime only if n is a power of 2.

Some hints:

1. aⁿ - 1 can be factorised if n is an integer and n > 1. From this it is easy to show that aⁿ - 1 is not prime if a <> 2

2. Now, using a = 2, take n to be a number that is not prime, and you should be able to prove that 2ⁿ - 1 is not prime if n is not prime.

This proves the first result. Its converse, that if a = 2 and n is prime, aⁿ - 1 is prime is false. All you have to do is find a counter-example.

I'll leave you to think abou the 2ⁿ + 1 problem.

 

[Xayma]

Originally posted by underthesun
Is the converse of this statement true? Show that 2ⁿ + 1 is prime only if n is a power of 2.

I think someone had a theory about all of these being prime, but only the first 4 are primes. Can't remember exactly if it was this or not, the book is around at my dads.

 

[Affinity]

2^n - 1 are mersenne primes and 2^n +1 are fermat primes.

reminds me of a question (B session question for anyone who knows what that is):

find all primes in the form 2^n + 1 where n is a fibonnaci number.

 

[underthesun]

Hi, just another silly question: prove that log(5)15 is irrational. Tried to go the old log₁₀3 way, but now it all boils down to trying to prove that 3^q != 5^m where q and m are integers. or was I just lead in the wrong direction?

 

[Affinity]

that's obviously true...

apply the fundamental theorem of arithmetic (each interger has a unique prime number factorisation)

 

[underthesun]

oh, thanks for that! this theorem looks like as if it'll be my best friend during that exam time .

oke thanks a lot for the helps people, really greatly appreciated