State And Prove Thate Multiplication Theorum Of Expection

Can you prove this math theorem?

Given: 2 variables, A and B A and B have area that incorporates all advantageous integers A + B = consistent if B > A than as A will improve and procedures B, A*B will improve to boot if A >B than as B will improve and procedures A, A*B will improve to boot answer: for the reason that A + B = consistent, the comparable might desire to be authentic for the fabricated from those 2 variables: A*B = consistent as long because of fact the smaller variety, wether or no longer it is A or B, is increasing to return closer to the bigger value and not any incorrect way around, the bigger lowering to fulfill the smaller, than the product will constantly get greater because of certainty that the multiplication of two variables will produce a product that will improve if a sort of variables will improve. In mathematical words: A*B = consistent........enable A be the smaller variety and B be the bigger (A + a million)*B = A*B + B = consistent + B, consequently the product has greater via B whilst including a million to A in case you have been to attempt and make those numbers closer jointly via including a million to A and subtracting a million from B then it is what happens: (A + a million)*(B - a million) = A*B - A + B = consistent - A + B, consequently because of fact A is smaller than B the consistent will nevertheless be a greater value then it substitute into earlier, so the evidence holds authentic. On yet another notice nevertheless, if B is subtracted via a million and not something happens to A then: A*(B - a million) = A*B - A = consistent - A, this might shrink the size of the consistent and consequently the product would be smaller even nevertheless A and B are coming closer jointly in length. So for this evidence to hold authentic, for any shrink in the bigger variable, there might desire to be an equivalent or greater desirable improve in the smaller variable. So what ethan stated is powerful except you state the undertaking above.

Mathematical theorems and properties?

It's actually a property of exponents. You have to realize not everything in math has a name. We would probably just call that "a property of exponents".

The property states that:
b^x * b^y = b^(x + y)

Converting to scientific notation:
0.36 = 3.6 x 10^-1

0.36 x 10
= (3.6 x 10^-1) x 10
By associativity of multiplication:
= 3.6 x (10^-1 x 10)
= 3.6 x (10^-1 x 10^1)
By above property of exponents:
= 3.6 x 10^(-1 + 1)
= 3.6 x 10^0
= 3.6 x 1
= 3.6

Can someone prove that the multiplication of two prime numbers is always a non-prime number? :)?

Suppose we have two numbers, A and B, and their product is C.
So, A*B = C.

Assume that C is a prime number.

Since A and B are also prime numbers, then they must be whole numbers.

Then C has four distinct factors: 1, A, B, and itself

This contradicts with the definition of a prime number, which states that a prime number has exactly two distinct whole number factors, 1 and itself.

Therefore our initial assumption is false, and we can conclude that C is a non-prime number.

QED

State and Prove the Addition Theorem for Probability.?

The addition rule is a result used to determine the probability that event A or event B occurs or both occur.


The result is often written as follows, using set notation:
P(A U B) = P(A) + P(B) - P(P int B)

where:
P(A) = probability that event A occurs
P(B) = probability that event B occurs
P(A U B) = probability that event A or event B occurs
P(A int B) = probability that event A and event B both occur

Proof:

For mutually exclusive events, that is events which cannot occur together: P(A int B) = 0
The addition rule therefore reduces to
P(A U B)= P(A) + P(B)

For independent events, that is events which have no influence on each other:
P(A int B) = P(A).P(B)
The addition rule therefore reduces to
P(AUB)=P(A)+P(B)-P(A).P(B)

In both cases the rules stands true.

This can also be proved with venn diagram.

I hope I answered your question.

All the best

Prove this two theorems?

Say (x0,y0) is extremum. Pick a direction and choose (x0+a1,y0+a2) (a1 and a2 are small) around it.
By mean value thm,
f(x0+a1,y0+a2)-f(x0,y0) = ∇f(x0+b1,y0+b2)'[a1 a2]'
for some (b1,b2)=k*(a1,a2) where ∇ is the gradient. Note that 0∇f(x0+b1,y0+b2)-∇f(x0,y0) = ∇²f(x0+h1,y0+h2)[b1 b2]'
for some (h1,h2)=l*(b1,b2) and ∇² is for the second gradient. Note similarly 0Combine all to get;
f(x0+a1,y0+a2)-f(x0,y0) ≈ k*[a1 a2]*∇²f(x0,y0)*[a1 a2]'
for some 1>k>0.
So left side is greater than zero for all [a1 a2] pairs (or directions) if ∇²f(x0,y0) is positive definite, less than zero if negative definite. Positive (negative) definite implies all eigenvalues are positive (negative). So the determinant is positive in both cases. So if determinant is positive, it is either minimum or maximum, depending on the sign of eigenvalues. It is a minimum if they are positive, it is a maximum if they are negative. I have no idea where your (i) and (ii) conditions come from. They seem to have no information about the signs of eigenvalues.

However, by the characteristic polynomial, we can replace them with
(i) a local maximum if f_xx(x0,y0)+f_yy(x0,y0)<0
(i) a local minimum if f_xx(x0,y0)+f_yy(x0,y0)>0

If determinant is negative, it implies one eigenvalue is positive while another is negative. Using eigenvalue decomposition, you can show in one direction, the function is increasing, in another it is decreasing. So, (x0,y0) is a saddle point.

How does one use Femat's Little Theorem to simplifiy a modular such as 4^6 (mod 7)?

The integer 7 is prime and 4 does not have 7 as a factor. By FLT, 4^(7-1) (mod 7) = 1 (mod 7).


A power of an integer may often be simplified by using the following rule:

........If a^k = a mod b, then a mod b = a^(k^2) mod b = a^(k^3) mod b = . . .

where a and k are non negative integers and b is a positive ingeger.

Suppose you were asked to simplify 7^10 (mod 14). Notice that we can't use FLT because 14 is not prime. However, because 7^2 = 7 (mod 14), we have 7^10 = 7^(8 + 2) = 7^2 = 7 mod 14.


Now try 7^58 mod 14.

7^58 = 7^(32 + 16 + 8 +2) = 7^(4) = 7 (mod 14)