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[Dromikaites]

Quick question, is this math statement true or false?

The equation:

y=(x^2-9) / (x-3) has a vertical asymptote.

My friends say it's FALSE, because once you simplify the equation you get y=x+3, which is a straight line with no asymptote.

I maintain that even if it is a straight line, it will have an open circle at x=3 because it's undefined at that point, therefore making the statement TRUE.

Who's right?

Actually your friend is right because before analyzing an equation (or a curve defined by an equation) you need first to get that equation to its simplest form. In this case to y=x+3.

If you don't do that you end up with lots of "paradoxes" like this one (I hope you know what a complex number is):

i is the square root of -1 (or i=sqr(-1) ), meaning i*i = -1

so on one hand we have: 1 = sqr(1) but because 1 = (-1)* (-1) we have 1 = sqr ((-1) * (-1)) or 1= sqr(-1) * sqr(-1) or 1 = i*i. However i*i = -1 so we've just "proved" that 1= -1 !!!

[Simetrical]

Actually your friend is right because before analyzing an equation (or a curve defined by an equation) you need first to get that equation to its simplest form. In this case to y=x+3.

That is not the same equation. Specifically, it is defined at x = 3, while the provided equation is not. A simplified equation in this case might be something like "y = x + 3 (x ≠ 3)".

If you don't do that you end up with lots of "paradoxes" like this one (I hope you know what a complex number is):

i is the square root of -1 (or i=sqr(-1) ), meaning i*i = -1

so on one hand we have: 1 = sqr(1) but because 1 = (-1)* (-1) we have 1 = sqr ((-1) * (-1)) or 1= sqr(-1) * sqr(-1) or 1 = i*i. However i*i = -1 so we've just "proved" that 1= -1 !!!

The error is not because you failed to simplify. It's because (ab)^c = a^cb^c holds where those quantities are real numbers, not for complex numbers. The step where you go from "1 = sqr ((-1) * (-1))" to "1= sqr(-1) * sqr(-1)" is therefore simply incorrect: that does not hold for complex-valued square roots (or any other complex-valued exponents).

[Dromikaites]

That is not the same equation. Specifically, it is defined at x = 3, while the provided equation is not. A simplified equation in this case might be something like "y = x + 3 (x ≠ 3)".

The "beauty" of symbolic notation (I hope that's the English for y=x+3) is unless really good care is taken one can easily run into "paradoxes". Good care in this case stands for sticking to the rules of manipulating those symbols.

If we set aside the rule that we first eliminate the undetermined forms like (x^2-9)/(x-3) when x=3 then we're open for guessing. For instance you've suggested the form "y=x+3 x≠3" which is OK while I can as legitimately chose "y=(x^2-9)/(x-3) for x≠3 and y=6 for x=3" which plotted on paper/screen looks exactly like y=x+3.

The error is not because you failed to simplify.

Exactly, it's because instead I chose to "complicate". Just like one can chose to "complicate" y=x+3 into y=(x²-9)/(x-3)

It's because (ab)^c = a^cb^c holds where those quantities are real numbers, not for complex numbers. The step where you go from "1 = sqr ((-1) * (-1))" to "1= sqr(-1) * sqr(-1)" is therefore simply incorrect: that does not hold for complex-valued square roots (or any other complex-valued exponents).

I'm sure you know that the rule "(ab)^c=a^cb^c is limited to real numbers" comes from the rigorous mathematical definition of real numbers, complex numbers and from the definition of what it means raising a complex number to a complex power (raising a real number to a real power is just a particular case of the same operation with complex numbers).

The point I want to make is apparently simple expression need actually rigorous handling or otherwise we end up "arguing" or being "fascinated by paradoxes".

In the particular case of the expression which started this discussion y=(x²-9)/(x-3) we have an example of "sloppy" instead of rigorous definition because nothing is said about x=3. In such case my approach would be to avoid the ambiguity of x=3 by means of the "rule" "simplify before performing other manipulations" while yours would be to let the expression as it is and address the ambiguity by placing restrictions on the value of x. In front of an audience we would look like two wizards using two different spells which is not exactly what science is about

[Simetrical]

The conventional way that it's taught, at least in American high schools and colleges, is that if a function like f(x) = (x² − 9)/(x − 3) is given, f is taken to be a function with domain R − { 3 }, i.e., it's just undefined where no value is explicitly given. This is just a convention, but it's the standard convention in low-level math courses, and so would be the "correct" way to view this problem.

In higher math, at least pure math, one tends to be more careful, and specifies the domain explicitly when defining a function for any serious purpose.

[Holger Danske]

Alright then!..
I have a question about the processes that takes place inside our Sun - How the energy is produced by nuclear fusion and how the processes changes as the Sun (Star) gets older?

[Simetrical]

This is a homework problem? For what course? It sounds like it's better suited to its own thread. Science homework problems usually tend to have a single answer that's easy to come up with for anyone with the right background, so it's not a bad idea to cram them all into a thread. This sort of question is much fluffier and really requires lengthy discussion to be properly treated.

[Banned]

Alright then!..
I have a question about the processes that takes place inside our Sun - How the energy is produced by nuclear fusion and how the processes changes as the Sun (Star) gets older?

The process takes place in following steps:

• The combination of two protons gives deuteron, an anti-electron and a neutrino.
  
  ¹H + ¹H → ²H + e⁺ + ν
  
  
• An electron is hit by an anti-electron which results in the emission of light in the form of photons.
  
  e^- + e⁺ → hν
  
  
• A proton is hit by a deuteron which yields Helium and more light.
  
  ²H + ¹H → ³He + hv
  
  
• Two Helium atoms combine to form an alpha particle and two protons.
  
  ³He + ³He → ⁴He + ¹H + ¹H
  
  
• The overall reaction is:
  
  4¹H + 2e^- → 2v + 6hv
  
  
• The total energy released in the process is 26 MeV or 4.1652 x 10^-12 Joules.
  
  
• As far as the talk regarding other reactions is concerned, let's say that they're simply less important in the context, since they produce little to no energy.

As for the question related to change in processes with the lapse of time, there's little change that's noticeable in real time even though the sun is burning matter at the rate of 4 x 10⁹ kilograms per second and releasing energy in the form of heat and light. However, in about a few billion years the mass of Sun will have reduced a lot and its size will have increased by more than 200 times of its present size. It will become a red hot giant, or to be more specific, an asymptotic giant branch star, as a major composition of hydrogen will get converted into Helium and Helium+Helium reactions will become more prominent, resulting in the formation of carbon. Moreover, the temperature of its core will have also increased by many times due to the release of more energy from the next level of fusion reactions.

[Holger Danske]

Thank you so much.
So once the Hydrogen is bruned the Sun(Star) shifts it's fuel comsumption over to Helium. What next?

@Sime
It's a science course (physics)

[chriscase]

It's a science course (physics)

Yea, but it's not really a "problem". This is more along the lines of lecture notes. Did you lose your textbook or something?

[Holger Danske]

No, but I couldn't find the process provided by Marcus in any of them for some strange reason...

(I had it written down once, as my teacher went through it, but guess I was a genius and threw it away)

away, I got the point, and I'll post somewhere else next time.

[chriscase]

I got the point, and I'll post somewhere else next time.

Hey I am not a moderator or a representative of TWC forums in any capacity. I was not telling you not to post here, anyway. It's just hard to substitute actual course notes for you. You really need those to do well, and they are fairly specific to the course and instructor.

[Banned]

So once the Hydrogen is bruned the Sun(Star) shifts it's fuel comsumption over to Helium. What next?

When about 10% of the hydrogen present in the core of a star is consumed, it starts developing a layer of Hydrogen around the core where Helium is present in very low amount due its accumulation in the core. Fusion reactions involving Helium begin inside the core. The equilibrium previously maintained between gravitational force directed towards the center and pressure due to fusion reactions directed outwards, is disturbed because of the production of less amount of energy from fusion reactions. The gravitational energy as a result becomes more dominant and the core starts collapsing because of it. But as we know that the fusion of Helium results in the formation of carbon and oxygen, the explosion of the core does not take place. The carbon accumulated in the core prevents it from exploding and rather the outer layers of the star expand due to immense heat becoming red hot like an iron bar in a temperature of around 2273 K. The star becomes very large in size and starts engulfing the nearby planets in orbit. However, some orbits may shift to great distances because of stellar wind resulting from the loss of mass. If we take into consideration sun and earth, then there's a slight possibility that at some point of time in future the sun will gobble the earth in no time. Anyway, many things will have changed by that time, so you need not worry. After the asymptotic giant branch phase, or in other words, the red phase, the star turns into a white dwarf because of shrinking of the core and lack of radiation. Then it starts cooling slowly and gradually, finally turning into a black dwarf which is rich in carbon and oxygen. Such is the life of a main sequence star like our life-giving sun.

[Ephialtes The Dog]

I've got a biology question that I just can't seem to find the answer to.

Here it is: "How would you demonstrate that 2 particles have opposite electrical charges?"

Any help would be greatly appreciated.

[roy34543]

Isnt that basically...a magnet?

[Jango]

Opposite charges are attracted to each other, if one is positive and one is negative wouldn't they move towards each other? I do believe so!

[chriscase]

If you put them in an electric field, they will be affected by it differently. Not sure whether you need to also specify a means of detecting how the particles move, their momentum, or what...

[Simetrical]

To be picky, the response should be "Using what equipment?" The most reliable way might be to get a strong electric field, drop them both in, and see what happens. If they get pulled in opposite directions, they have opposite charge. If in the same direction, they have the same charge. If one doesn't move at all, it's neutrally charged.

[Bernem]

I've got a biology question that I just can't seem to find the answer to.

Here it is: "How would you demonstrate that 2 particles have opposite electrical charges?"

Any help would be greatly appreciated.

what exactly do want want to demonstrate? which biological molecules do you want to examine? proteins? DNA? The application of an electric field is apropriate to show charge differences of these things!

[chriscase]

To be picky, the response should be "Using what equipment?" The most reliable way might be to get a strong electric field, drop them both in, and see what happens. If they get pulled in opposite directions, they have opposite charge. If in the same direction, they have the same charge. If one doesn't move at all, it's neutrally charged.

They would only move "in opposite directions" along the force lines of the field. If the force lines are curved, the particles would not appear to be moving in opposite directions to an outside observer.

[Simetrical]

When I said "drop them both in" the strong electric field, I of course meant the same electric field, i.e., drop them in places with the same electric field vector. They would then move in instantaneously opposite directions when released.