brightsky's Maths Thread

[brightsky]

okay so this question might sound stupid, but how do you prove, formally, that the sum of two irrational numbers is irrational?

[Phy124]

okay so this question might sound stupid, but how do you prove, formally, that the sum of two irrational numbers is irrational?

More than likely flawed logic here, but I'm curious as to whether you could have cases where the statement is false?

Say you had and

Then the sum of and which are both irrational would give a rational result?

[Alwin]

Yeah, I'm with Phy124 brightsky.

The typical counter example I use is:



Clearly the LHS is the sum of two irrational numbers, but the RHS is a rational number. Thus, this contradiction by example demonstrates that the sum of two irrational numbers need not be irrational



PS: what was the proof for the formula of a tangent plane again?
Something about two vector's dot product is zero so then the equation is df/dx(x-x₁) + df/dy(y-y₁) - (z-z₁) = 0,  where z = f(x,y)................. or something like that?

[brightsky]

lol. i'm retarded. thanks guys!

there's no real proof, alwin. as you said, just dot product of normal vector and some vector on the plane. so we know that the normal vector will be parallel to the gradient vector of f. so we have n = (fx, fy, fz) and we know that (x1, y1, z1) lies on the plane. so (fx,fy,fz).(x-x1,y-y1,z-z1) = 0. in this case fz = -1.

[rife168]

This tangent plane business relies on a cool trick that took me ages to understand just from the notes last year.
We know that the gradient vector is orthogonal to any level curve, so we take the function z=f(x,y) and put all the variables x,y,z on one side and then the constants on the other side.
We then have something resembling g(x,y,z)=c so we treat the 3d surface as a 'level curve' of some function in 3 variables (a 4d surface essentially). We then know that the gradient vector will be orthogonal to the level curve g(x,y,z)=c, but this function is exactly z=f(x,y), just rewritten. So the gradient vector of g is normal to the surface so the plane defined by that vector is tangent to the surface at any point.

Soz, pretty crappy explanation, I'm on my phone.

[brightsky]

question regarding logistics: for umep exams, do we use our VCAA number or do we use our university ID number (do we even have one?)?

thanks!

[stolenclay]

I think one of the early emails from enrolment said we need to have our university student card to do the exam, so we probably use our university ID number (which is on the card), if any.

It should be a 6 digit number, and if you kept your emails from when you applied for UMEP, it should be in the email with the subject University of Melbourne Extension Program Application Receipt Confirmation, labelled as Extension Program application number after the salutation. (Hopefully that's it; for me it is.)

[Alwin]

Since this is kinda become the unofficial UMEP maths thread, good luck to everyone for tomorrow
Hope to see you guys there and meet up!


And with the ID issue, our numbers are on the library cards we got I'm bringing that and school ID tomorrow

[brightsky]

ahh okay thanks guys!

best of luck tomorrow!!

[brightsky]

quick question: say we have a function f(x,y) and we wanted to find the direction of zero change in F at a random point (x0,y0). how do we go about doing that? we have the gradient vector ∇f . say ∇f = (1,2). then why is +-1/sqrt(5)(2,-1) the ONLY directions in which there is zero change in F. so by definition zero change in F means directional derivative is 0. but D_uf = ∇f . u(hat). so surely there are infinite solutions for u(hat), given that the only condition it needs to satisfy is when it is dotted with ∇f it gives a value of 0. is there another condition that u(hat) needs to satisfy? i'm missing something...

edit: never mind, i'm retarded...

[TrueTears]

Seeing as though most of the other questions have been answered with fairly detailed replies, I just want to address this.

a few questions:

2. does the phrase '3 linearly independent vectors' make sense? I've always been under the impression that linearly dependent/independent were adjectives that modified SETS rather than vectors. would it be more correct to say 'a linearly independent set of 3 vectors', as in 'a linearly independent set of 3 vectors is required to span the whole of R^3'?

thanks in advance!

You are completely right (from a completely rigorous point of view). The phrase "linearly dependent" or "linearly independent" are used to describe the "structure" of a set of vectors. However, phrases are often used very "colloquially" and the meaning from '3 linearly independent vectors' is quite clear, just not 100% technical. But I'm sure we can live with that


Seeing there's all this talk about linear independence, spanning sets, basis... etc, the following theorem utilizes several of the aforementioned ideas, can anyone prove it?

If is a (real) matrix, and is a (real) column vector, show that if , then lies in the column space of . Note: is the augmented matrix where is inserted as the rightmost column of .

[brightsky]

ahh I see I see, thanks TT!

for your question, I can present a vague, informal proof, but not sure how to write it up formally. so we the augmented matrix A|b. we perform row operations to turn it into reduced row echelon form. the stuff on the LHS of the | would be identical to the reduced row echelon form of A. so the columns in A corresponding to the columns on the LHS of the | which contain pivots form a basis for the column space of A. now if rank(A) = rank(augmented matrix), then that means that b can be written as a linear combination of the vectors in a basis for the column space of A and so b is in the column space of A. if rank(A) < rank(augmented matrix), then b cannot be written as a linear combination of the vectors in the basis for the column space of A and so b is not in the column space of b. is this right? (I need to learn the art of writing up proofs in fancy maths notation though...any help would be much appreciated!)

I have another question: does the formula A^2 = tr(A)*A - det(A)*I apply when A is a square matrix of higher dimensions than 2*2. if A is a 2*2 matrix, then it is easy to verify, using the cayley hamilton theorem, that the formula works. but I have a suspicion that this formula does not work for 3*3, 4*4, etc. matrices....

thanks!

[kamil9876]

I have another question: does the formula A^2 = tr(A)*A - det(A)*I apply when A is a square matrix of higher dimensions than 2*2. if A is a 2*2 matrix, then it is easy to verify, using the cayley hamilton theorem, that the formula works. but I have a suspicion that this formula does not work for 3*3, 4*4, etc. matrices....

You are right, it does not. Take for instance.




Then so the RHS is . But is a non-zero matrix.

[brightsky]

thanks kamil!

also this might sound silly, but is there a foolproof way of identifying and sketching quadrics. I was confronted with a surface of equation 2x^2 - 2y^2 - 4z^2 = 1, and asked to identify it, but the equation doesn't seem to fit any of the possible cases: http://en.wikipedia.org/wiki/Quadric. perhaps some rearrangement is in order?

help much appreciated!

[Alwin]

thanks kamil!

also this might sound silly, but is there a foolproof way of identifying and sketching quadrics. I was confronted with a surface of equation 2x^2 - 2y^2 - 4z^2 = 1, and asked to identify it, but the equation doesn't seem to fit any of the possible cases: http://en.wikipedia.org/wiki/Quadric. perhaps some rearrangement is in order?

help much appreciated!

I can kinda of see how it it could be like that without the need of the sketch, but only sort of haha. It's like the wikipedia example with two (-) and one (+) when you get 1 on the RHS, but on different x, y, z terms. imho it's like hyperbolas in spesh where the (-) can be on the x^2 term or on the y^2 term and we recognise that signs of the terms tells us to draw an up-down graph or left-right graph... but in 3D this time.

Here's a W|A plot of it btw:
http://www.wolframalpha.com/input/?i=2x%5E2+-+2y%5E2+-+4z%5E2+%3D+1