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Proofs
Proof means sufficient evidence to establish the truth of something. It is obtained from deductive reasoning, rather than from empirical arguments. Proof must show that a statement is true in all cases, without a single exception.
1,294 Questions
If A and B are independent events then are A and B' independent?
if P(A)>0 then P(B'|A)=1-P(B|A)
so P(A intersect B')=P(A)P(B'|A)=P(A)[1-P(B|A)]
=P(A)[1-P(B)]
=P(A)P(B')
the definition of independent events is if P(A intersect B')=P(A)P(B')
that is the proof
If L1=1 and L2=2, we would just get the Fibonacci sequence. Recall that the Fibonacci sequence is recursive and given by: f(0)=1, f(1)=1, and f(n)=f(n-1)+f(n-2) for integer n>1. Thus, we have f(2)=f(0)+f(1)=1+1=2. If L1=1 and L2=2 then we would have L1=f(1) and L2=f(2). Since the Lucas numbers are generated recursively just like the Fibonacci numbers, i.e. Ln=Ln-1+Ln-2 for n>2, we would have L3=L1+L2=f(1)+f(2)=f(3), L4=f(4), etc. You can use complete induction to show this for all n:
As we have already said, if L1=1 and L2=2, then we have L1=f(1) and L2=f(2). We now proceed to induction. Suppose for some m greater than or equal to 2 we have Ln=f(n) for n less than or equal to m. Then for m+1 we have, by definition, Lm+1=Lm+Lm-1. By the induction hypothesis, Lm+Lm-1=f(m)+f(m-1), but this is just f(m+1) by the definition of Fibonnaci numbers, i.e. Lm+1=f(m+1). So it follows that Ln=f(n) for all n if we let L1=1 and L2=2.
How do you prove that a group of order 3 is cyclic?
Any group must have an identity element e. As it has order 3, it must have two other elements, a and b. Now, clearly, ab = e, for if ab = b, then a = e:
abb-1 = bb-1, so ae = e, or a = e.
This contradicts the givens, so ab != b. Similarly, ab != a, leaving only possibility: ab = e. Multiplying by a-1, b = a-1. So our group has three elements: e, a, a-1.
What is a2? It cannot be a, because that would imply a = e, a contradiction of the givens. Nor can it be e, because then a = a-1, and these were shown to be distinct. One possibility remains: a2 = a-1.
That means that a3 = e, and the powers of a are: a0 = e, a, a2 = a-1, a3 = e, a4 = a, etc. Thus, the cyclic group generated by a is given by: = {e, a, a-1}.
QED.
Let g be any element other than the identity. Consider
QED
In fact here is the proof that any group of order p where p is a prime number is cyclic. It follow precisely from the proof given for order 3.
Let p be any prime number and let the order of a group G be p. We denote this as
|G|=p. We know G has more than one element, so let g be an element of the group and g is not the identity element in G. We also know
|
|
QED
Notation...Order of an element
We will use the fact that if p prime, a divides p, then a = p or a = 1.
Then if p + q = r, for primes p, q, r, then one of p,q,r is even, or all three are (consider mod 2). p = q = r = 2 clearly doesn't work, and p + q = 2 doesn't work for primes p,q >= 3. So without loss of generality p = 2, then r = q+2. r is also the difference of two primes, r = s - t. Again considering mod 2, knowing that r is odd, one of s or t is even (and so equal to 2). If s = 2 then r is negative, so t = 2, and we have q + 2 = r = t - 2, so t = q + 4.
So we have q, r = q + 2 and t = q + 4 all prime. By considering q mod 3, one of them has a factor 3. If a prime has a factor 3, it is equal to 3. So q = 3, as q + 2 = 3 or q + 4 = 3 mean q is not prime. So, r = q + 2 = 5. Therefore, 5 is the only prime that can be represented as both the sum of two primes and the difference of two primes: 5 = 2 + 3 = 7 - 2. Since it is the only one, it is the greatest.
Proof of Bernoulli's Theorem
To prove Bernoulli's theorem, we make the following assumptions:
1. The liquid is incompressible.
2. The liquid is non-viscous.
3. The flow is steady and the velocity of the liquid is less than the critical velocity for the liquid.
Imagine an incompressible and non-viscous liquid to be flowing through a pipe of varying cross-sectional area as shown in Fig. The liquid enters the pipe with a normal velocity v11 and at a height h1 above the reference level (earth's surface). It leaves the pipe with a normal velocity v2 at the narrow end B of cross-sectional area a2 and at a height h2 above the earth's surface. at its wide end A of cross-sectional area a
If r is the density of the incompressible liquid, then in accordance with the equation of continuity, the mass m of the liquid crossing any section of the pipe is given by
a1 v1 = a2 v2 = m (say) or a1 v1 = a2 v2 = m/ ..... (ii)
Let P1 and P2 be the values of the pressure due to the liquid at the ends A and B respectively. If the liquid moves from the end A to B under the action of pressure difference P1 - P2, then in accordance with energy conservation principle, the work done by the pressure energy of the liquid must appear as the increase in potential and kinetic energies of the liquid.
The pressure energy exerts a force P1 a1 on the liquid at the end A. The liquid covers a distance v1 in one second at the end A and therefore
Work done per second on the liquid at the end A = 1 a1 v1
The liquid reaches the end B against pressure P2 i.e. against a force P2 a2. At the end B, the liquid covers a distance v2 in one second and therefore
Work done per second by the liquid at the end B = 2 a2 v2
Hence, net work done by the pressure energy in moving the liquid from the end A to B in one second = 1 a1 v1 - 2 a2 v2
Using the equation (ii), we have
Net work done by the pressure energy per second = 1 m/ - p2 m/ = (1 - 2) m/ ... (iii)
When the mass m of the liquid flows in one second from the end A to B, its height increases from h1 to h2.
Therefore, increase in potential energy of the liquid per second
= m g h2 - m g h1 = m g (h2 - h1) … (iv)
Further, when the mass m of the liquid flows in one second from the end A to B, its velocity increases from v1 to v2.
Therefore, increase in kinetic energy of the liquid per second = 1/2 mv22 - 1/2 mv12 = 1/2 m(v22 - v12) ... (v)
According to work-energy conservation principle,
Work done by the pressure energy per second = increase in potential energy per second + increase in kinetic energy per second
Therefore (P1 - P2) = m/? = mg (h2 - h1) + 1/2 m(v22 - v02) or P1/? - P2/? = gh2 - gh1 + 1/2 v22 - 1/2 v12 or P1/? + gh1 + 1/2 v12 = P2/? + gh2 + 1/2 v22 ..... (vi)
or P/? + gh + 1/2 v2 = constant
It proves the Bernoulli's theorem. This is the most convenient form of Bernoulli's equation. However, it can be expressed in some other forms as explained below:
Multiplying both sides of the equation (vi) by ?, we have
P1 + ?gh1 + 1/2 ?v12 = P2 + ?gh2 + 1/2 ?v22 ... (vii)
In this equation, each term has got dimension of pressure.
Again, dividing both sides of the equation (vi) by g, we have
P1/?g + h1 + v12/2g = P2/?g + h2 + v22/2g ..... (viii)
In this equation, each term has got dimensions of length. The terms P/?g, h and v2/2g are called pressure head, elevation (or gravitational) head and velocity head respectively.
When the liquid flows through a horizontal pipe (h1 = h2), then the equation (vi) becomes
P1/ + 1/2 v12 + P2/ + 1/2 v22
How do you do radicals in math?
radicals are quite simple. if say u have √27, then u find the largest square number that you can multiply to get the total. In this case u would have √9*3. Take the square root of 9, which is 3. The you have 3√3. this is now in simplest form.
If the greatest common factor/divisor of A and B is 1 then they are coprime - they do not share any prime factors. Multiplying both through by C means, obviously, that each number now divides by C. In fact, C is their greatest common divisor, since AC and BC do not have further common factors after C is taken out. Hence the GCF of AC and BC is not merely a factor of C - it is C. (The question makes sense only if A, B and C are integers.)
How do you prove the quadratic formula?
The essence of the proof is simply to complete the square for a generalised quadratic equation. Like this:
ax2 + bx + c = 0
Take 'a' outside:
a[x2 + bx/a + c/a] = 0
Divide through by 'a':
x2 + bx/a + c/a = 0
Complete the square:
(x + b/2a)2 - b2/4a2 + c/a = 0
Rearrange to find x:
(x + b/2a)2 = b2/4a2 - c/a
x + b/2a = (+/-)sqrt[b2/4a2 - c/a]
x = -b/2a (+/-) sqrt[b2/4a2 - c/a]
Finally, fiddle around so that (1/2a) can be taken out as a common factor:
x = -b/2a (+/-) sqrt[b2/4a2 - 4ac/4a2]
x = -b/2a (+/-) sqrt[(1/4a2)(b2 - 4ac)]
x = -b/2a (+/-) sqrt(1/4a2)sqrt(b2 - 4ac)
x = -b/2a (+/-) (1/2a)sqrt(b2 - 4ac)
x = [ -b (+/-) sqrt(b2 - 4ac) ] / 2a.
What is the formula to find the radius?
It depends on the information that you do have. If you know the diameter, the circumference or the area the equations are relatively straightforward.
What is the circumcenter of a triangle?
Circumcenter - the center of the circle that circumscribes the triangle, ie. goes through all its vertices.
How is maths useful in day-to-day life?
Math helps you determine speed, helps you calculate how much food to get, helps you make a recipe, lets you know how much you weigh. It tells you how much money you have, how much money you make in a job, what size clothes you wear, and when your favorite show is on. It lets you cook properly, tells you how much taxes to pay, and what time of the year it is. Math is a little important.
Two of the most famous badlands are the Makoshika State Park in Montana and the Badlands National Park in South Dakota. There is also The Big Muddy Badlands in Saskatchewan, Canada and Valle de la Luna in Argentina.
What the 'Oyabun Kobun' relation mean?
Oyabun-Kobun literally translates as "father/child relationship". It is a relationship in which both parents and children show respect to each other, but mostly focuses on the respect of the child towards his/her parents.
What is the formula for finding the radius?
The formula for finding the radius of a circle is circumference divided by double of pi. If area is given instead of circumference, the radius may be found by dividing the area by pi, then finding the square root of the resultant quotient. It is also equal to half of the diameter, or the diameter divided by 2.
They are dense in that between any two decimals you have another one. A much better way to think of it is that a decimal must be either rational or irrational. For example, .34 is certainly rational, but the decimal expansion of square root of 2 goes on forever and is irrational. So a decimal, being either one, must be dense in the reals.
