Math discussion/help

[isosceles]

If my name didn't clue you in, I like math.

Soo... I thought I would try to spread the fun . I have a feeling that nobody around here gives a shit, but what the heck. If you have any interesting math concepts/problems/questions, post them here.


Also, I would be glad to try to explain anything or help with problems, but I will not actually do all your homework for you!



I will start off with this (note - I got this "proof" from another forum):

-1 = -1

1/-1 = -1/1

sqrt( 1 / -1 ) = sqrt( -1 / 1)

sqrt( 1 )/sqrt( -1 ) = sqrt( -1 )/ sqrt ( 1 )

Now cross multiply...

sqrt( 1 ) * sqrt( 1 ) = sqrt ( -1 ) * sqrt( -1 )

1 = -1

or even: 

2 = 0

Let's see who knows exactly what is wrong with this .

 

[enjoi888]

im in a very tuff math course in college so i shall b posting in this thread quite often.

and for ur problem pls say wat sqrt means and also try to b more clearer( its tuff to understand wat ur asking)

cheers

 

[Hobo]

Maths isnt my best subject lol. What year is that maths at?

 

[isosceles]

im in a very tuff math course in college so i shall b posting in this thread quite often.

and for ur problem pls say wat sqrt means and also try to b more clearer( its tuff to understand wat ur asking)

cheers

Alright, good to hear.

As for your questions: I should have labeled the problem I posted as a proof. It is just a series of steps going from one conclusion to another. Obviously 2 does not equal 0; I just put that there to get everyone thinking. Your job is to find out what's wrong with those steps.

SQRT is square root. It's just difficult to get the square root notation without going to great lengths.

Maths isnt my best subject lol. What year is that maths at?

Well, the little "proof" I posted is just fairly simple algebra I suppose. I just put it there to spark a some interest/discussion.

 

[Macroman]

That looks like some sort of code, Or LISP..

 

[Typhoon]

Now you started posting ? I had maths homework on proportion that I was stuck with. I manged anyway. What kind of level is your maths ? University level ?

 

[Shredderbeam]

What you do to one side, you must do to the other. Multiplying one side by 1, and the other by -1 may yield the same result, but it sets the problem up for your answer later on.

 

[Sin666]

The sqrt( 1 / -1 ) = sqrt( -1 / 1) (or, sqrt (-1) = sqrt (-1) and therefore i = i), cannot be rewritten as sqrt( 1 )/sqrt( -1 ) = sqrt( -1 )/ sqrt ( 1 ), which is instead 1/i = i/1, and therefore 1/i = i. That is a fallacy.

 

[Shredderbeam]

i^2 is the same thing as -1. >.>

That is, if you believe in imaginary numbers.

 

[Sin666]

I know, too long ago Fixed my reasoning.

 

[isosceles]

Now you started posting ? I had maths homework on proportion that I was stuck with. I manged anyway. What kind of level is your maths ? University level ?

Sorry for my bad timing . Maybe I can help in the future. I am taking calculus II.

The sqrt( 1 / -1 ) = sqrt( -1 / 1) (or, sqrt (-1) = sqrt (-1) and therefore i = i), cannot be rewritten as sqrt( 1 )/sqrt( -1 ) = sqrt( -1 )/ sqrt ( 1 ), which is instead 1/i = i/1

I had to keep changing my response as you kept changing your answer

This time, I believe you have it! The problem with this proof is that a principle is applied incorrectly. sqrt( x/y ) only equals sqrt( x )/ sqrt( y ) when x and y are BOTH positive real numbers. Good job .

I'll try to come up with something else to post a little later perhaps.

 

[Tr1gg4]

alt 251 √
taking maths(need A to get in) further maths (need A* to get in) Physics (need A to get in) and History (need B to get in) for Alevels

 

[enjoi888]

(4/y)-5=(5/2y)

i think ur supposed to mult all by lcm but its tuff help

need done in hour

i came up with y=negative 1/8
am i right?

 

[isosceles]

(4/y)-5=(5/2y)

i think ur supposed to mult all by lcm but its tuff help

need done in hour

i came up with y=negative 1/8
am i right?

Is that (4/y)-5=(5/(2y)) or (4/y)-5=((5/2)y)? I'm not sure if you mean that the y is in the denominator or if it is being multiplied by the fraction 5/2.

If it is in the denominator:
(4/y)-5 = 5/(2y)

Multiply both sides by 2y:
2y(4/y - 5) = 5

Use distributive property:
2y*4/y - 2y*5 = 5

Cancel and simplify:
8 - 10y = 5

Subtract 8 from both sides:
-10y = -3

y = 3/10 or .3

I have to go for a few minutes, if you meant for the y to not be in the denominator on the right side of the equation, tell me and I will post a solution for that.

 

[PurePwnage]

I'm in Geometry in 9th grade, woo!

 

[rs_revolution]

luckily i am done with all my math classes.. hehe. but i can only hope the rest of you enjoy urs more.

 

[jebckr]

I didn't read peoples post. If someone hasn't said it, you can't take the square root of a negative, unless you get into imaginary numbers and whatnot.

Anyhow, are you able to help with calculus as I'm thinking I'll have some trouble with it in a few weeks.

 

[Billyy]

Right now, in trig, I just finished doing functions. Those are a bitch, it's like they give you a function:

f(x) = x^2

Show me the proof of why that above function is Even.

I'll post the answer later when I have my trig notes.

There are a few steps involved in doing this problem, show them.

Btw, I'm a junior, and I am in Functions/Statistics & Trigonometry.

 

[Shredderbeam]

Right now, in trig, I just finished doing functions. Those are a bitch, it's like they give you a function:

f(x) = x^2

Show me the proof of why that above function is Even.

I'll post the answer later when I have my trig notes.

There are a few steps involved in doing this problem, show them.

Btw, I'm a junior, and I am in Functions/Statistics & Trigonometry.

A function is even if it is symmetrical with respect to the Y-axis. x^2 produces a parabolic curve that is vertically symmetrical, so yes, it is even.

If you just graph it for a few values (1, -1, 2, -2, etc.), you'll see how.

 

[Billyy]

Well I meant without graphing the equation, there is a way to solve it algebraically.