2005 Q8 (1 Viewer)

[Dumsum]

2001 we had irrationality of e.
2003 we had irrationality of pi.

What shall we have in 2005?

 

[acmilan]

prove cubed root of 5 is irrational

 

[Trev]

Uh, if they asked us to prove anything is irrational I wouldn't know where to start

 

[withoutaface]

Irrationality of root 2?
Hmm, nah, that one's too easy.

 

[wanton-wonton]

Irrationality of root 2?
Hmm, nah, that one's too easy.

How would you start off?

 

[Templar]

Prove that pi is transcendental

How would you start off?

Proof by infinite descent (ie strong induction).

Let's suppose sqrt2 is p/q, p, q are integers.
2=p^2/q^2, 2q^2=p^2, hence p is even, p=2k
2q^2=4k^2
q^2=2k^2, hence q is even, q=2m

Hence p/q=k/m, and this can be repeated infinitely. But since this cannot happen with integers (divide infinitely by 2), the original assumption is false.

 

[lucifel]

they have to firstly know what transcendental means. Be interesting if they have like a whole paragraph telling you just exactly what transcendental means. Hmm if they make us prove irrationality of a number, it's not too bad because pi and e are done, and thats all the 'special' numbers we have to deal with. Hmm they could say prove the e^iπ +1 = 0 but that'd be delving into complex analysis. By the way π is meant to be a pi (i can't find a proper pi)

 

[acmilan]

Hmm they could say prove the e^iπ +1 = 0 but that'd be delving into complex analysis. By the way π is meant to be a pi (i can't find a proper pi)

That is extremely easy. Firstly they'd have to tell you that e^i@ = cos@ + isin@ or in some way get you to prove it. From there all you have to do is sub pi in.

 

[MAICHI]

I thought of a good famous one. If you can do this you should be at university already.

2005 Q8

C)

i) Let A, B, C be points on a straight line such that:

A------------B------C

If AC/AB=AB/BC, prove that AC/AB is irrational.

ii) Hence or otherwise prove that (1+root(5))/2 is irrational.

 

[withoutaface]

Similar to Templar's proof:
Suppose that pq⁻¹ has square 2, where p ∈ Z and q ∈ N. We may assume, without loss of generality, that p and q have no common factors apart from ±1. Then p² = 2q² is even, so that p is even. We can write p = 2r, where r ∈ Z. Then q² = 2r² is even, so that q is even, contradicting that assumption that p and q have no common factors apart from ±1.

Also, coincidently, we got given another proof of the irrationality of root 2 in our calculus lecture today.

 

[dawso]

as long as what we are trying to prove is correct, and not actually impossible such as question 7 from my 3unit paper yesterday, ma and a few others actually proved the statement commpletely false.....i hate wen teachers make questions themselves...

 

[withoutaface]

Well the second parts fairly easy, you just take the original identity, rearrange it to get AC*BC=AB²
AB²-AB*BC-BC²=0

AB = (BC +/- root(BC²+4BC²))/2
AB=BC(1+/- sqrt(5))/2
AB/BC = (1+/-sqrt(5))/2

And since we know that AC/AB is irrational, then so is AB/BC and hence (1+sqrt(5))/2

Can't get the first bit though.

EDIT: To get the first bit I work backwards, taking the quadratic for which the solution is (1+sqrt(5))/2

let (1+sqrt(5))/2 = p/q, where p and q have no common factors apart from +/- 1

Then p²/q² - p/q - 1 = 0
p²-pq-q²=0

p(p-q)=q²
p/q=q/(p-q)

now if p/q have no common factors apart from +/- 1, then the fraction p/q is in its lowest terms. But q/(p-q) is in even lower terms, hence contradiction, hence p/q is not rational, work backwards from part 2 to prove part 1, etc

 

[MAICHI]

p(p-q)=q²
p/q=q/(p-q)

now if p/q have no common factors apart from +/- 1, then the fraction p/q is in its lowest terms. But q/(p-q) is in even lower terms, hence contradiction, hence p/q is not rational, work backwards from part 2 to prove part 1, etc

Yeah that's right. Just do this at the start so you go forward:

let a=AC, b=AB, a=f*p, b=f*q. where f is the largest commonfactor.

a/b=b/(a-b)

f(p/q)=f(q/(p-q))

(p/q)=q/(p-q)

 

[acmilan]

Also, coincidently, we got given another proof of the irrationality of root 2 in our calculus lecture today.

With highlarious results.

 

[who_loves_maths]

Originally Posted by MAICHI
I thought of a good famous one. If you can do this you should be at university already.

2005 Q8

C)

i) Let A, B, C be points on a straight line such that:

A------------B------C

If AC/AB=AB/BC, prove that AC/AB is irrational.

ii) Hence or otherwise prove that (1+root(5))/2 is irrational.

to do both part of the question at once:

let AB = a ; BC = b

therefore, from AC/AB = AB/AC, you have (a + b)/a = a/b -----> 1 + b/a = a/b ; where 'b/a' and 'a/b' are reciprocals.

let the ratio a/b =x -----> 1 + 1/x = x -----> 0 = x^2 - x -1

the solution of this quadratic is irrational for x = a/b ; hence the original ratio is irrational and you get the value of this ratio at the same time.


Alternatively (to simply show the irrationality):

1 = x - 1/x -----> 1 = (x - 1/x)^2 -----> 1 + 4 = 5 = (x + 1/x)^2 -----> Sqrt(5) = x + 1/x

but no rational values of 'x' will satisfy (x + 1/x) to give the irrational number Sqrt(5)... hence, x = a/b must have been irrational to start with.

 

[who_loves_maths]

Originally Posted by acmilan
prove cubed root of 5 is irrational

isn't that obvious? like Sqrt(2) is obviously irrational?

you can just use the "sandwich" between squares/cubes technique:

let cuberoot(5) = a/b , where (a, b) are integers, and mutually prime.

hence, 5 = a^3/b^3 , but since 'a' and 'b' are mutually prime, then 'a^3' and 'b^3' MUST also be mutually prime (for obvious factor reasons).

but 5 is an integer, and so is 'a' and 'b', and so since they are mutually prime, then b = 1 is the only solution to 'b'.

hence, a^3 = 5 ; but 'a' is an integer: 1^3 = 1 < 5 , and, 2^3 = 8 > 5

---> ie. 1 < a < 2 ; which is a direct contradiction... hence, the cube root of 5 must have been irrational to begin with.


this 'sandwiching' between integers, im my opinion, is a much more general and accessible proof to the irrationality of common and small roots and cube roots (larger than 1) than other unique methods like using even or odd numbers, etc...

eg. to prove that Sqrt(2) is irrational:

let Sqrt(2) = a/b ; with 'a' and 'b' mutually prime integers.

ie. 2 = a^2/b^2 , now since 'a' is not divisible by 'b', then 'a^2' is also mutually prime to 'b^2'.

but since 2 in an integer, then the only integer solution for 'b' is b =1.

---> a^2 = 2 ; but, 1^2 = 1 < 2 , and, 2^2 = 4 > 2

---> ie. 1 < a < 2 ; which is a direct contradiction since 'a' in an integer ... hence, Sqrt(2) was irrational to begin with.


P.S. if 'a' and 'b' are mutually prime integers, then 'a^n' and 'b^n' are also mutually prime integers for all positive integral 'n'.

 

[withoutaface]

Vince you just use a similar method to what he showed us in lectures.

Define l(x)=n as giving the largest integer n for which x/5ⁿ is an integer.

Then you have the properties, which just follow from the proof this morning:
- l(x) always an integer if x is an integer
- l(xy) = l(x)+l(y)
- l(x/y) = l(x)-l(y)

Now we let p³/q³=5, where p, q are integers with no common factors apart from +/- 1
l(x) both sides
l(p³)-l(q³)=1
3(l(p)-l(q)) =1
l(p)-l(q)=1/3

Now since p, q are integers and hence l(p),l(q) should be integers, so their diffence should be an integer, but it is not, hence one of p or q is not an integer and hence cube root 5 is not rational.

 

[jumb]

prove cubed root of 5 is irrational

It's not irrational, you jack ass. u r sooooo dum

 

[acmilan]

It's not irrational, you jack ass. u r sooooo dum

perhaps, but you are a mormon!

@ cube root of 5, i knew how to prove it Just all i could think of at the time and i was looking at next weeks tutorial questions at the same time

 

[jumb]

perhaps, but you are a mormon!

@ cube root of 5, i knew how to prove it Just all i could think of at the time and i was looking at next weeks tutorial questions at the same time

ur dumer then this