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What else can I help you with?
What is an example of using the word finite?
There is only a finite number of opportunities in your life.
How do we prove that a finite group G of order p prime is cyclic using Lagrange?
Lagrange theorem states that the order of any subgroup of a group G must divide order of the group G. If order p of the group G is prime the only divisors are 1 and p, therefore the only subgroups of G are {e} and G itself. Take any a not equal e. Then the set of all integer powers of a is by definition a cyclic subgroup of G, but the only subgroup of G with more then 1 element is G itself, therefore G is cyclic. QED.
How do you prove that order of a group G is finite only if G is finite and vice versa?
(1). G is is finite implies o(G) is finite.Let G be a finite group of order n and let e be the identity element in G. Then the elements of G may be written as e, g1, g2, ... gn-1. We prove that the order of each element is finite, thereby proving that G is finite implies that each element in G has finite order. Let gkbe an element in G which does not have a finite order. Since (gk)r is in G for each value of r = 0, 1, 2, ... then we conclude that we may find p, q positive integers such that (gk)p = (gk)q . Without loss of generality we may assume that p> q. Hence(gk)p-q = e. Thus p - q is the order of gk in G and is finite.(2). o(G) is finite implies G is finite.This follows from the definition of order of a group, that is, the order of a group is the number of members which the underlying set contains. In defining the order we are hence assuming that G is finite. Otherwise we cannot speak about quantity.Hope that this helps.
What are infinite set that has a finite complement?
An infinite set with a finite complement is a set that contains infinitely many elements, while the elements not in the set (the complement) are limited to a finite number. For example, the set of all natural numbers excludes a finite number of integers, such as only the number 0. This means that the complement, which in this case would be {0}, is finite, while the set of natural numbers itself is infinite. Thus, such sets are often used in various mathematical contexts, especially in topology and set theory.
Is 64 rational or irrational?
Rational
G is finite where the number of subgroups in g is finite?
Actually a stronger statement can be made:A group G is finite if and only if the number of its subgroups is finiteLet G be a group. If G is finite there is only a finite number of subsets of G, so clearlya finite number of subgroups.Now suppose G is infinite , let'ssuppose one element has infinite order. The this element generates an infinite cyclicgroup which in turn contains infinitely many subgroups.Now suppose all the subgroups have finite order Take some element of G and let it generate a finite group H. Now take another element of G not in H and let it generate a finite group I. Keep doing this by next picking an element of G not H or I. You can continue this way.
What is an example of using the word finite?
There is only a finite number of opportunities in your life.
Who proved that there are an infinite amount of phone numbers?
It was presumably proven when it was discovered that there were infinitely many counting numbers. However, whoever it was, did not consider the mathematical possibility with practicality. The universe has a finite life. Within that our solar system is finite. People, in their turn, have finite lives. In a finite life you can only "dial" a finite number of digits. therefore, you can only call a number if it has a finite number of digits. For any finite number of digits, there are only a finite amount of phone numbers. So, having infinitely many telephone numbers is no use if you need to wait an infinite amount of time (longer than you'll live) for the first person to call you!
How do we prove that a finite group G of order p prime is cyclic using Lagrange?
Lagrange theorem states that the order of any subgroup of a group G must divide order of the group G. If order p of the group G is prime the only divisors are 1 and p, therefore the only subgroups of G are {e} and G itself. Take any a not equal e. Then the set of all integer powers of a is by definition a cyclic subgroup of G, but the only subgroup of G with more then 1 element is G itself, therefore G is cyclic. QED.
Does the Catholic church believe that there are only a finite number of souls?
Every human being born on this planet has a soul. So the number always remains a finite number. So the belief is true.
What are the limitations of finite automata?
The defining characteristic of FA is that they have only a finite number of states. Hence, a finite automata can only "count" (that is, maintain a counter, where different states correspond to different values of the counter) a finite number of input scenarios.There is no finite automaton that recognizes these strings:The set of binary strings consisting of an equal number of 1's and 0'sThe set of strings over '(' and ')' that have "balanced" parenthesesThe 'pumping lemma' can be used to prove that no such FA exists for these examples.
Let Sn be the group of permutations on n symbols Then 1. S4 has no subgroup isomorphic to S3 2. S4 has only one subgroup isomorphic to S3 3. S4 has exactly 3 distinct subgroups isomorphic to S3?
It has 4 subgroups isomorphic to S3. If you hold each of the 4 elements fixed and permute the remaining three, you get each of the 4 subgroups isomorphic to S3.
How do you prove that order of a group G is finite only if G is finite and vice versa?
(1). G is is finite implies o(G) is finite.Let G be a finite group of order n and let e be the identity element in G. Then the elements of G may be written as e, g1, g2, ... gn-1. We prove that the order of each element is finite, thereby proving that G is finite implies that each element in G has finite order. Let gkbe an element in G which does not have a finite order. Since (gk)r is in G for each value of r = 0, 1, 2, ... then we conclude that we may find p, q positive integers such that (gk)p = (gk)q . Without loss of generality we may assume that p> q. Hence(gk)p-q = e. Thus p - q is the order of gk in G and is finite.(2). o(G) is finite implies G is finite.This follows from the definition of order of a group, that is, the order of a group is the number of members which the underlying set contains. In defining the order we are hence assuming that G is finite. Otherwise we cannot speak about quantity.Hope that this helps.
Is countries of the world finite or infinite?
Well, honey, the number of countries in the world is finite. Last time I checked, there were 195 countries recognized by the United Nations. So unless aliens start setting up embassies, we're not gonna see an infinite number of countries anytime soon.
Is income discrete or continuous?
Income is discrete. People can have only a finite number of possible income values.
A random variable is said to be discrete if?
It can take only a finite number of values. These need not be integer values.
Why can't you have a money multiplier of inifinity?
The money multiplier is the reciprocal of the reserve requirement, which can only be a finite number.
