Extremely difficult to disprove “some” because you have to eliminate all competing claims. And the only way we can do that is to find and show every man on Earth is not silly.

Categorical statements that begin with “some” must be treated differently from categorical statements that begin with “all or no”. Some is often ambiguous in ordinary usage. Does it mean “a few,” “at least a few,” “at least one but not all,” “at least one and maybe all,” “at least a few but not all,” “lots,” “many”? To avoid such confusions, logicians always use some with the same consistent meaning. In logic, “some” always means “at least one.”

For example, “Some dogs are animals,” means “At least one dog is an animal” (which is true), not “At least one dog is an animal, but not all” (which is false).

(1) Rebuild the structure of the argument:

If men are humans, ... (premise)
and some human tends to be silly, ... (premise)
Then, some men are silly. ... (conclusion)

(2) The first main way to attack an argument is to challenge one of its premises. Another way is to show that the claim to be refuted implies something that is ridiculous or absurd in ways that are independent of any particular counterexample. This mode of refutation is called a reductio ad absurdum, which means a reduction to absurdity.

If dogs are humans, then some dogs are silly,
and dogs are not humans,
Therefore, dogs are not silly.

A1 is the uniform cross-sectional area along the entire the coil of wire, which is a circle (πr²).
The area A1 is used to calculate the resistivity.

A2 is the area of the plane of the coil, which is the rectangular shape (2 cm × 2 cm).
The area A2 is used to calculate the torque acting on the coil of wire.

I'm sure you have already known that charged particles whose charges have the same sign repel one another, and particles whose charges have different signs attract. Therefore, the same thing can be used to explain the electric force F1.

For the magnetic force F2, there must be some kind of law governs the behavior or motion of charged particle. Well, this law is called Lorentz force. If fact, Lorentz force applies to both applications involving charged particles moving in a electric field and magnetic field. The total force F (called the Lorentz force) acting on the charge is given by

F = FE + FB = qE + (qv × B)

where the bolded are vectors. E is electric field. B is magnetic field. v is the velocity of the charged particle.

Electric force :: FE = qE

Magnetic force :: FB = qv × B

In math, you have learned that cross product on two vectors in 3-D space results in a vector which is perpendicular to both and therefore normal to the plane containing them. In the Lorentz force, the magnetic force is perpendicular to both v and B. In the following diagram, it is clearly shown oppositely directed magnetic forces are exerted on two oppositely charged particles moving at the same velocity in a magnetic field:



A general rule of thumb, you can use a MODIFIED Fleming's right-hand rule for positively (+) charged particles, and Fleming's left-hand rule for negatively (−) charged particles. For details, see Post #228.



Since the electron in your problem is a negatively charged particle, you may apply modified Fleming's left-hand rule, and you will discover the magnetic force F2 is directed downward.

This post has been edited by Critical_Fallacy: Dec 10 2013, 10:28 PM

Notice that in both cases, L = 1, therefore, the both ratio tests failed. They are pretty much worthless and we would need to resort to a different test to determine the convergence of the series. Like I mentioned previously, there are no hard and fast rules about which test to apply to a given series, but it is not wise to apply a list of the tests in a specific order until one finally works. But there are some strategies we can apply to decide which test is appropriate.

In Case 1 and 2, both can be tested and compared using the p-series Σ{1/(n^p)} because the denominator is a polynomial function. In a nutshell, p-series says, if p ≤ 1, it is divergent, and if p > 1, it is convergent.

It does not stop there. In fact, you have only proved Σ{(n^n)/(e^n)} is divergent, which is true, using the Root Test.

You did not prove anything about the convergence of Σ{(n^n)/n!}.

To continue, since you know Σ{(n^n)/(e^n)} is divergent, all you need to do now is to prove

(n^n)/n! > (n^n)/(e^n)

because Series Comparison Test says if the term of a given series Σ{(n^n)/n!} is larger than the corresponding term of another series Σ{(n^n)/(e^n)}, which is known to be divergent, then the given series must be divergent too.

Please remind maximR to copy down this to his note book ::

Strategy for Testing Series ::
---------------------------------------
1. If the series is of the form Σ{1/(n^p)}, it is a p-series, which we know to be convergent if p > 1 and divergent if p ≤ 1. See Example 1 of the Tutorial #4.

2. If the series has the form Σ{ar^(n−1)} or Σ{ar^n}, it is a geometric series, which converges if |r| < 1 and diverges if |r| ≥ 1. Some preliminary algebraic manipulation may be required to bring the series into this form.

3. If you can intuitively see that lim {an} > 0, as n → ∞, then the Term Test should be used for checking the Divergence.

4. If the series is of the form Σ{(−1)^(n−1)*bn} or Σ{(−1)^n*bn}, then the Alternating Series Test is an obvious possibility.

5. Series that involve factorials or other products (including a constant raised to the nth power) are often conveniently tested using the Ratio Test.

6. If {an} is of the form {(bn)^n}, then the Root Test may be useful.

7. If {an} = f(n) where [1,∞] ∫ f(x) dx can be easily evaluated, then the Integral Test is effective (assuming the hypotheses of this test are satisfied).

This post has been edited by Critical_Fallacy: Dec 10 2013, 11:29 AM

Up to this point, x² − x = 0 is a quadratic equation. And you don't need me to tell you how to solve it, right?

Thanks v1n0d for the explanation!

Again, up to this point, x + y = y, it is very obvious that x = 0 in order to satisfy the equation.

From this point onward, "Since x = y: ∴ y + y = y" writing down the equation is blatant ignorance.

Let me give you an example. We know a = b, and we also know a = 0.

Mathematically, it is true that

0 + 0 + 0 + 0 = 0

Knowing that a = 0, Can I write

a + a + a + a = a

and therefore, 4a = a?

Yes I can. But can cancel the a on both sides? To find what is a, we do a little algebraic manipulation

4a − a = 0

3a = 0

In the end, we still find a = 0.

Remember that we DO NOT simply cancel or divide out the unknown a!

This post has been edited by Critical_Fallacy: Dec 10 2013, 11:44 AM

Modus tollens is one of a mathematician's finest weapons, because it is a logically reliable pattern of deductive reasoning, and it is is absolutely guaranteed to have a true conclusion if the premises are also true. We can symbolize this reasoning pattern in the following way:

If p, then q. {p → q}

Since not q, {∵ ¬q}

Therefore, not p. {∴ ¬p}

**********************
Show that √2 is irrational.
**********************
If √2 were rational, then √2 could be expressed as a fraction m/n in irreducible terms, where m and n are integers having no common factors. {p → q}

So, 2 = m²/n², which implies m² = 2n². Clearly, m² is even. Because the squares of even numbers are even (2a)² = 2(2a²), that in turn implies that m must be even.

Let m = 2k. So m² = 4k² = 2n². It follows that n² = 2k² is even. By the same token, we see that n is even.

This shows m/n, a ratio of two even integers must be a reducible fraction because they have a common divisor 2.

Since m/n is not is reduced form {¬q}, therefore √2 is not rational {¬p}.

I seriously do not know STPM physics is that hard. If you have troubles with imagining the Fleming's hands rule, maybe this will help you. I just figured out that it is very convenient to assign the middle fingers ┌∩┐()┌∩┐ as magnetic field B.

For a negatively (−) charged particle: use Fleming's LEFT hand rule

Left Thumb = Magnetic Force, F
Left Index Finger = Direction of motion, v
Left Middle Finger = Magnetic Field, B

For a positively (+) charged particle: use Fleming's RIGHT hand rule

Right Thumb = Magnetic Force, F
Right Index Finger = Direction of motion, v
Right Middle Finger = Magnetic Field, B

Procedure:
1. Begin with pointing your middle finger according to the field B, either point in or point out.
2. Once the field B is fixed, start to rotate your hand until your index finger is pointing to the Direction of motion, v.
3. Voila! You can determine the Force direction by looking at your thumb direction.

This post has been edited by Critical_Fallacy: Dec 10 2013, 10:07 PM

It's OK. I also have a hard time learning other subjects as well. We just cover for each other and the world will be a better place to live. Please check out the updated post in #228. It should be very EASY now.

This post has been edited by Critical_Fallacy: Dec 10 2013, 10:12 PM

The Fleming's hands rule is just a suggestion. That's why I stated "modified". In fact, why don't we call them Critical's LEFT hand & RIGHT hand rules? Just stick to #228.

It was meant for comic relief. As long as you find the modified rules convenient and useful, just name it whatever we want. Do you understand how to find the Lorentz Force now?

What trigonometric identities are you using? I don't get it. Perhaps you could show some of your workings here, and then let me have a look.

Using Double-Angle Formulas for sin and cos, this gives:

Oh! I just discovered that STPM Syllabus suffers for a lack of Trigonometry. Foundation topics in Math are very important for students who are starting courses in Applied Math and Engineering Math from diverse backgrounds. Refer to Fundamental Identities, csc θ = 1 / sin θ. Remember to print out this Trigonometry Reference Sheet and keep by your side.

The following topics are lacking in STPM syllabus. To prepare for Calculus, you have to master them before going to college.

1. Functions (Trigonometric identities, Hyperbolic functions)
2. Complex numbers in exponential form
3. Differentiation of Hyperbolic functions
4. Taylor Series (General form of Maclaurin Series)
5. Partial differentiation
6. Integration by Reduction formula (useful for functions that can't be integrated directly)
7. Advanced integration applications (mechanics)
8. Polar coordinate systems
9. 2nd-order Differential equations

Thanks for clarifying that! In fact, I learned the Law of Sines, and Law of Cosines in my classic SPM.

How are you doing on the mechanics?

Conventionally, mathematicians will tell you that there is no number that you can multiply by 0 to get a non-zero number. Hence, there is NO solution, and so any non-zero number divided by 0 is undefined or rather meaningless.

However, to physicists, it remains an unsolved problem in Black Hole Astrophysics. At the center of a black hole, the singularity point has zero volume (V = 0 m³) and infinite density (ρ = m/V = ∞). If you are a fan of Albert Einstein, you probably know that he once quoted, “Black holes are where God divided by zero.”

There are controversies about the existence of black holes, but astrophysicists generally agree that black holes exist. Moreover, there is good observational evidence from X-ray observations and from the Hubble Space Telescope that there are massive black holes (with masses more than a million times that of the Sun) exist in the centers of some galaxies.

Technically you are correct because you have proven that any attempt at a definition for "Division by 0" leads to a contradiction.

In order to show maximR why 0/0 is meaningless, now we let a = 0, and from a = b × c, this gives

0 / 0 = b

The equation tells us that we have to find a number b such that

0 = b × 0

The problem occurs when we could argue that b is 1, or 2, or any other number.

Therefore, it means "meaningless", because the result of 0/0 could be anything, you name it!