Let S^2 and R^2 be the sample variance from two independent samples .pdf
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Let S^2 and R^2 be the sample variance from two independent samples of size Ns=8 and Nr=7. Identify the distribution of the ratio S^2/R^2 Solution Assuming these samples came from the same normal distribution, the distribution of S^2/R^2 will be F(s-1,r-1).
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Let T R^2 - R^2 be the linear transformation given by T (x.pdf
Let T: R^2 -> R^2 be the linear transformation given by
T*
(|x|
|y|) =
|2x-y |
|-8x+4y|.
Find a basis for the range of T
Solution
basis = number of rows = 2
range ==> range of r^2, where r is any real number
so range is[0,).
let S1be a countable set,S2 a set that is not countable, and S1S2. .pdf
let S1be a countable set,S2 a set that is not countable, and S1?S2. Show that S2 must then
contain an infinite number of elements that are not inS1.
part (b.) show that in fact S2 - S1cannot be countable
Solution
a)
S1 is countable implies S1 is one-one to Natural numbers
S2 is countable basically there exists numbers which are more then that of natural numbers
implies S2 must then contain an infinite number of elements that are not inS1
b)
S1 = (S1 - S2) ? (S1 ? S2) (Draw the Venn diagram to see this) (S1 ? S2) ? S2, hence it is
countable. We have shown that the union of countable sets is countable. If we assume that S1-S2
is countable, this will mean that S1 is countable. Thus we come to a contradiction. Hence S1-S2
is uncountable.
let S1 and S2 be nonempty subspaces of a vector space V. Show sp(S1).pdf
let S1 and S2 be nonempty subspaces of a vector space V. Show sp(S1) + sp(S2) = sp(S1 U S2)
Solution
let K be from Span(S1 U S2) then,
K = a1V1 + ... + anVn ... + b1U1 + ... + bmUm +...
where Vi\'s are from S1 and Ui\'s from S2 for some ai\'s and bi\'s from the field
now all you have to do is group all the Vi\'s and group all the Ui\'s
K = (a1V1 + ... + anVn ... ) + (b1U1 + ... + bmUm +... )
and it is clearly seen that
K = W1 + X1 where W1 is from Span(S1) and X1 is from Span(S2)
therefore K belongs in Span(S1) + Span(S2)
now for the reverse just go exactly backwards providing necessary explanations..
Let S be a subspace. Prove that S (intersection symbol) S(orthogonal.pdf
Let S be a subspace. Prove that S (intersection symbol) S(orthogonal/perpindicular symbol) =
{0}
Solution
Let S be a subspace. Prove that S ? S? = {0}....WHAT ARE THE ? ....PLEASE REPOST THE
QUESTION USING WORDS TO DESCRIBE ..IF YOU ARE NOT ABLE TO TYPE THE
SYMBOLS...
Let S and T be finite sets with n and m elements, respectively. .pdf
Let S and T be finite sets with n and m elements, respectively.
(a.) If m is greater than or equal to n, how many injective (one-to-one) functions F: S---> can be
defined?
(b.) If m=n, how many surjective (onto) functions f: S---> T can be defined?
Solution
n*m n^2.
Let S be a non empty subset of the real numbers. Assume that L is th.pdf
Let S be a non empty subset of the real numbers. Assume that L is the greatest lower bound of S
and L is not in S, prove that L is an accumulation point of S.
Solution
Let S be a bounded set and let GLB(S) be its greatest lower bound.
THAT IS THERE IS NUMBER L , SUCH THAT NO ELEMENT IN THE SET S IS
SMALLER THAN L..
FO EX ..IF S = [1,1/2 , 1/3,1/4...] ETC WE CAN SAY L= -1 , SINCE NO ELEMENT IN THE
SET IS LESS THAN L=-1 ..WE CAN EVEN SAY L=-0.5 AS A GLB , SINCE THAT IS ALSO
TRUE ..
Assume that GLB(S) is not an element of S...OK... IT CAN BE FROM THE DEFINITION OF
GLB , AS WE HAVE SHOWN ABOVE ...
Prove that GLB(S) is an acumulation point of S.
SO LET US TAKE ANY SEQUENCE [ S1,S2,...Sn] FROM SET S ....
SINCE S IS A BOUNDED SET AS GIVEN IN SOME SPACE SAY RN , THIS SEQUENCE
SHALL BE CONTAINED IN A BOUNDED CLOSED BALL & SO IS A COMPACT SUB
SET OF THE SPACE RN...
THE SEQUENCE THUS HAS A CONVERGING SUBSEQUENCE IN RN AND THE LIMIT
OF THAT CONVERGENCE IS THE ACCUMULATION POINT OF A ...
Let S be a bounded set and let GLB(S) be its greatest lower bound. A.pdf
Let S be a bounded set and let GLB(S) be its greatest lower bound. Assume that GLB(S) is not
an element of S.
Prove that GLB(S) is an acumulation point of S.
Solution
Let S be a bounded set and let GLB(S) be its greatest lower bound.
THAT IS THERE IS NUMBER L , SUCH THAT NO ELEMENT IN THE SET S IS
SMALLER THAN L..
FO EX ..IF S = [1,1/2 , 1/3,1/4...] ETC WE CAN SAY L= -1 , SINCE NO ELEMENT IN THE
SET IS LESS THAN L=-1 ..WE CAN EVEN SAY L=-0.5 AS A GLB , SINCE THAT IS ALSO
TRUE ..
Assume that GLB(S) is not an element of S...OK... IT CAN BE FROM THE DEFINITION OF
GLB , AS WE HAVE SHOWN ABOVE ...
Prove that GLB(S) is an acumulation point of S.
SO LET US TAKE ANY SEQUENCE [ S1,S2,...Sn] FROM SET S ....
SINCE S IS A BOUNDED SET AS GIVEN IN SOME SPACE SAY RN , THIS SEQUENCE
SHALL BE CONTAINED IN A BOUNDED CLOSED BALL & SO IS A COMPACT SUB
SET OF THE SPACE RN...
THE SEQUENCE THUS HAS A CONVERGING SUBSEQUENCE IN RN AND THE LIMIT
OF THAT CONVERGENCE IS THE ACCUMULATION POINT OF A ...
Let S be a nonempty bounded subset of R. Prove that sup S is unique..pdf
Let S be a nonempty bounded subset of R. Prove that sup S is unique.
Solution
The supremum is the least upper bound. That is, it is an upper bound (so everything in S is less
than or equal to it) AND it is the LEAST upper bound (so any other upper bound is larger).
We know that S is bounded, so there is some M such that every element of S is less than or equal
to M. So an upper bound exists. Now, the sup is the smallest of all the upper bounds.
Suppose we had two sups (by contradiction). That is suppose M and N were both the sup of S.
One of them is the smaller of the two (if they are the same number, then we don\'t actually have
two sups). Call the smaller one M and the larger one N. If both are supremums of S, then they
are both upper bounds of S. But now we have a problem: N is a sup, but there is an upper bound
(namely M) that is smaller than N. If N were a sup, it can\'t have any upper bounds that are
smaller. So N can\'t really be a sup of S, and we have a contradiction.
Thus, we must only have one supremum. So the supremum of S is unique..
Recommended
Let T R^2 - R^2 be the linear transformation given by T (x.pdf
Let T: R^2 -> R^2 be the linear transformation given by
T*
(|x|
|y|) =
|2x-y |
|-8x+4y|.
Find a basis for the range of T
Solution
basis = number of rows = 2
range ==> range of r^2, where r is any real number
so range is[0,).
let S1be a countable set,S2 a set that is not countable, and S1S2. .pdf
let S1be a countable set,S2 a set that is not countable, and S1?S2. Show that S2 must then
contain an infinite number of elements that are not inS1.
part (b.) show that in fact S2 - S1cannot be countable
Solution
a)
S1 is countable implies S1 is one-one to Natural numbers
S2 is countable basically there exists numbers which are more then that of natural numbers
implies S2 must then contain an infinite number of elements that are not inS1
b)
S1 = (S1 - S2) ? (S1 ? S2) (Draw the Venn diagram to see this) (S1 ? S2) ? S2, hence it is
countable. We have shown that the union of countable sets is countable. If we assume that S1-S2
is countable, this will mean that S1 is countable. Thus we come to a contradiction. Hence S1-S2
is uncountable.
let S1 and S2 be nonempty subspaces of a vector space V. Show sp(S1).pdf
let S1 and S2 be nonempty subspaces of a vector space V. Show sp(S1) + sp(S2) = sp(S1 U S2)
Solution
let K be from Span(S1 U S2) then,
K = a1V1 + ... + anVn ... + b1U1 + ... + bmUm +...
where Vi\'s are from S1 and Ui\'s from S2 for some ai\'s and bi\'s from the field
now all you have to do is group all the Vi\'s and group all the Ui\'s
K = (a1V1 + ... + anVn ... ) + (b1U1 + ... + bmUm +... )
and it is clearly seen that
K = W1 + X1 where W1 is from Span(S1) and X1 is from Span(S2)
therefore K belongs in Span(S1) + Span(S2)
now for the reverse just go exactly backwards providing necessary explanations..
Let S be a subspace. Prove that S (intersection symbol) S(orthogonal.pdf
Let S be a subspace. Prove that S (intersection symbol) S(orthogonal/perpindicular symbol) =
{0}
Solution
Let S be a subspace. Prove that S ? S? = {0}....WHAT ARE THE ? ....PLEASE REPOST THE
QUESTION USING WORDS TO DESCRIBE ..IF YOU ARE NOT ABLE TO TYPE THE
SYMBOLS...
Let S and T be finite sets with n and m elements, respectively. .pdf
Let S and T be finite sets with n and m elements, respectively.
(a.) If m is greater than or equal to n, how many injective (one-to-one) functions F: S---> can be
defined?
(b.) If m=n, how many surjective (onto) functions f: S---> T can be defined?
Solution
n*m n^2.
Let S be a non empty subset of the real numbers. Assume that L is th.pdf
Let S be a non empty subset of the real numbers. Assume that L is the greatest lower bound of S
and L is not in S, prove that L is an accumulation point of S.
Solution
Let S be a bounded set and let GLB(S) be its greatest lower bound.
THAT IS THERE IS NUMBER L , SUCH THAT NO ELEMENT IN THE SET S IS
SMALLER THAN L..
FO EX ..IF S = [1,1/2 , 1/3,1/4...] ETC WE CAN SAY L= -1 , SINCE NO ELEMENT IN THE
SET IS LESS THAN L=-1 ..WE CAN EVEN SAY L=-0.5 AS A GLB , SINCE THAT IS ALSO
TRUE ..
Assume that GLB(S) is not an element of S...OK... IT CAN BE FROM THE DEFINITION OF
GLB , AS WE HAVE SHOWN ABOVE ...
Prove that GLB(S) is an acumulation point of S.
SO LET US TAKE ANY SEQUENCE [ S1,S2,...Sn] FROM SET S ....
SINCE S IS A BOUNDED SET AS GIVEN IN SOME SPACE SAY RN , THIS SEQUENCE
SHALL BE CONTAINED IN A BOUNDED CLOSED BALL & SO IS A COMPACT SUB
SET OF THE SPACE RN...
THE SEQUENCE THUS HAS A CONVERGING SUBSEQUENCE IN RN AND THE LIMIT
OF THAT CONVERGENCE IS THE ACCUMULATION POINT OF A ...
Let S be a bounded set and let GLB(S) be its greatest lower bound. A.pdf
Let S be a bounded set and let GLB(S) be its greatest lower bound. Assume that GLB(S) is not
an element of S.
Prove that GLB(S) is an acumulation point of S.
Solution
Let S be a bounded set and let GLB(S) be its greatest lower bound.
THAT IS THERE IS NUMBER L , SUCH THAT NO ELEMENT IN THE SET S IS
SMALLER THAN L..
FO EX ..IF S = [1,1/2 , 1/3,1/4...] ETC WE CAN SAY L= -1 , SINCE NO ELEMENT IN THE
SET IS LESS THAN L=-1 ..WE CAN EVEN SAY L=-0.5 AS A GLB , SINCE THAT IS ALSO
TRUE ..
Assume that GLB(S) is not an element of S...OK... IT CAN BE FROM THE DEFINITION OF
GLB , AS WE HAVE SHOWN ABOVE ...
Prove that GLB(S) is an acumulation point of S.
SO LET US TAKE ANY SEQUENCE [ S1,S2,...Sn] FROM SET S ....
SINCE S IS A BOUNDED SET AS GIVEN IN SOME SPACE SAY RN , THIS SEQUENCE
SHALL BE CONTAINED IN A BOUNDED CLOSED BALL & SO IS A COMPACT SUB
SET OF THE SPACE RN...
THE SEQUENCE THUS HAS A CONVERGING SUBSEQUENCE IN RN AND THE LIMIT
OF THAT CONVERGENCE IS THE ACCUMULATION POINT OF A ...
Let S be a nonempty bounded subset of R. Prove that sup S is unique..pdf
Let S be a nonempty bounded subset of R. Prove that sup S is unique.
Solution
The supremum is the least upper bound. That is, it is an upper bound (so everything in S is less
than or equal to it) AND it is the LEAST upper bound (so any other upper bound is larger).
We know that S is bounded, so there is some M such that every element of S is less than or equal
to M. So an upper bound exists. Now, the sup is the smallest of all the upper bounds.
Suppose we had two sups (by contradiction). That is suppose M and N were both the sup of S.
One of them is the smaller of the two (if they are the same number, then we don\'t actually have
two sups). Call the smaller one M and the larger one N. If both are supremums of S, then they
are both upper bounds of S. But now we have a problem: N is a sup, but there is an upper bound
(namely M) that is smaller than N. If N were a sup, it can\'t have any upper bounds that are
smaller. So N can\'t really be a sup of S, and we have a contradiction.
Thus, we must only have one supremum. So the supremum of S is unique..
Let S = { a,b,c,d } and T = { 1,2,3,4,5,6,7 }. How many functions a.pdf
Let S = { a,b,c,d } and T = { 1,2,3,4,5,6,7 }. How many functions are there from S into T?
explain your answer.
Solution
mappings from S to T since no of elements in S is 4 and no ofelements in T is 7 An
element in S can be mapped in 7 possible ways so 4 elements can be mapped in 7*7*7*7 ways
so in total 2401 ways.
Let R be the ring of 2 × 2 matrices of the form a b 0 0 wh.pdf
Let R be the ring of 2 × 2 matrices of the form
a b
0 0
where a, b ? R. Show that although R is a ring that has no identity, we can find a
subring S of R with an identity.
Solution
Take S to be the set of matrices of the form a 0 0 0 where a is in R. It is completely
obvious that S is closed under both multiplication and addition. [In fact S is isomorphic to R].
Now the (multiplicative) identity in S is: 1 0 0 0 again this is completely obvious (what happens
when you multiply something in S by the matrix on the left or right?). I hope this helps, if you
need more details write back..
Let R be a commutative ring with unity and let I,J be two ideals of .pdf
Let R be a commutative ring with unity and let I,J be two ideals of R such that I + J = R Show
that R/(I intersect J) is isomorphic to R/I
Solution
GIVEN R IS COMMUTATIVE RING WITH UNITY I & J ARE 2 IDEALS
OF R I+J=R TO SHOW THAT R/[I INTERSECT J ] IS ISOMORPHIC TO
[R/I] * [ R / J ] PROOF I IS AN IDEAL OF R ..SO I IS A SUB GROUP OF R
UNDER ADDITION I1*R1 & R1*I1 ARE ELEMENTS OF I FOR ANY ELEMENT I1 IN I
AND R1 IN R FURTHER SINCE R IS COMMUTATIVE.
Let p, q, and r be the following statements p Jamie is on the .pdf
Let p, q, and r be the following statements: p: Jamie is on the train. q: Sylvia is at the park. r:
Nigel is in the car. Translate the following statement into English: (p ^ ~q)-->r
Solution
answer: If jamie is on the train and sylvia is not at the park, then Nigel is in the car..
Let p, q, and r be the following statements p Jamie is on the .pdf
Let p, q, and r be the following statements:
p: Jamie is on the train.
q: Sylvia is at the park.
r: Nigel is in the car.
Translate the following statement into English: (~p q) r
Construct a truth table for ~q p
Solution
(~p V q) -> r ~p means negate V means or a->b means if a then b SO if Jamie is
on the train OR Jamie is on the train THEN Nigel is in the car..
Let P(x) be hte statement x=x2 if the domain consists of the int.pdf
Let P(x) be hte statement \"x=x2\" if the domain consists of the integers, What are the truth
values?
a) P(0)
b) p(1)
c) P(2)
d)P(-1)
Solution
a) T
b) T
c) T
d) F (-1 1)
e) T (there does an exist a (in fact, multiple x\'s) where x = x2. that\'s when the integers are 0 and
positive).
f) F ( negative integers do not equal their square).
Let H and K be finite abelian groups and let G be an infinite abelia.pdf
Let H and K be finite abelian groups and let G be an infinite abelian group. Let it be known that
G x H is isomorphic to G x K . Show by example that because G is infinite then H need not be
isomorphic to K.
Solution
Use the fundamental theorem for finitely generated abelian groups. First show that the finite
parts of G and H are equal, then show that the infinite parts are equal.
For example if G,H,K have infinite parts Z^n, Z^m and Z^ p, then
Z^(n+p)=Z^(m+p) implies n=m.
Let n be a non-negative integer and a and c be positive numbers. Use.pdf
Let n be a non-negative integer and a and c be positive numbers. Use probability to express
0a xne-cxdx in terms of a finite sum.
Solution
F(x) = P(X x) (also denoted FX(x)). F(x) is the cumulative probability to the left of
(and including) the point x. The survival function is the complement of the distribution function,
S(x) = 1 - F(x) = P(X > x). For a discrete random variable with probability function p(x), F(x) =
X wx p(w), and in this case F(x) is a step function (it has a jump at each point with non-zero
probability, while remaining constant until the next jump). If X has a continuous distribution
with density function f(x), then F(x) = R x -1 f(t)dt and F(x) is a continuous, differentiable, non-
decreasing function such that d dxF(x) = F0(x) = -S0(x) = f(x). If X has a mixed distribution,
then F(x) is continuous except at the points of non-zero probability mass, where F(x) will have a
jump. For any cdf P(a < X < b) = F(b) - F(a), lim x!1 F(x) = 1, lim x!-1 F(x) = 0..
Let G be a group. Recall that an automorphism of G is an isomorphi.pdf
Let G be a group. Recall that an automorphism of G is an
isomorphism f : G ? G.
(1) Show that the set of automorphisms of G forms a group; we denote
this group Aut(G).
(2) Show that Aut(Z/4Z) ~ Z/2Z.
=
(3) Calculate Aut(Z/2Z×Z/2Z), and show that this group is isomorphic
to a group we are already very familiar with.
Solution
(.
Let G be a group, G = n in N, and H is a subset of G. Then all tha.pdf
Let G be a group, |G| = n in N, and H is a subset of G. Then all that is necessary to show that
H<=G is that:
a. H is closed
b. H is closed and has the inverse property
c. H is closed, has the inverse and identity properties,
d. H is closed, has the inverse, identity, and associativity properties
Solution
d. H is closed, has the inverse, identity, and associativity properties.
Let I and J be ideals in a ring R.Solution H.pdf
Let I and J be ideals in a ring R.
Solution
Hi i have solved this question in my notebook, but i can\'t type it and post it here
because there\'s no time... please rate me lifesaver and i\'ll make sure the answer is in your
inbox! i don\'t do this generally but i am doing this due to lack of time....
Let L be a list of the 26 letters in the English alphabet which cons.pdf
Let L be a list of the 26 letters in the English alphabet which consists of 5 vowels and 21
consonants . Show that has a sublist consisting of four or more consecutive consonants.
Assuming L begins with a vowel, say A, show that L has a sublist consisting of five or more
consecutive consonants.
Solution
as there are 21 consonants so there will be 4 or more than 4 insublist.
Let Ho Age and Observance of traditions are independent. Let Ha be.pdf
Let Ho: Age and Observance of traditions are independent. Let Ha be: Age and Observance of
traditions are not independent. A type 1 error in this context would be
A)Saying age and observance are independent when they actually are independent.
B)Saying age and observerance are not independent when they actually are dependent.
C)Saying age and observance are dependent when they are not dependent.
D)Saying age and observance are independent when they actually are not independent.
Solution
C)Saying age and observance are dependent when they are not dependent..
Let L1 be the language that consists of strings that only contain ze.pdf
Let L1 be the language that consists of strings that only contain zeros (i.e. 0n ) let L2
be the language 0n 1n , let L3 be the language 0n 1n 0n , let L4 be the language
0n 1n 0n 1n , let L5 be the language 0n 1n 0n 1n 0n etc. The language I want my
Turing machine to recognize is the union 0 Li of all the languages L0, L1,
L2, L3 etc. Thus a string is in L i the string is empty (i = 0) or there is
some m > 0 such that the string is in Lm. The problem that needs to be solved is simply
to construct a Turing machine that would recongnize such languages. Any help (advice) would
be most appreciated.
Solution
Let L1, L2 be languages over the same alphabet S, then there are following operations on L1,
L2: L1 ? L2 = {w|w ? L1 or w ? L2} L1 ? L2 = {w = xy|x ? L1,y ? L2} L1 = {w ? S * |w ?/ L1}
L k 1 = (L1 ? L1) ... L1 | {z } , k-times, with L 0 1 = {e} L * 1 = {w ? S * |?k = 0,w ? L k 1 }.
This is called Kleene Star of language L1. L + 1 = {w ? S * |?k = 1,w ? L k 1 }.
Let L(x) be the statement x is a lion and C(x) be the statement .pdf
Let L(x) be the statement \"x is a lion\" and C(x) be the statement \"x drinks coffee.\" Let the
universe of discourse for x be the set of all creatures. Express the following predicate logic
statement in plain English.
Ax(L(x)>C(X))
a. Some lions drink coffee.
b. If lions are creatures, then they drink coffee.
c. All creatures drink coffee.
d. All lions drink coffee.
Solution
d. All lions drink coffee..
Let G be a commutative group with operation . Prove that the set of.pdf
Let G be a commutative group with operation *. Prove that the set of a in G such that a^3=e
(where e is the identity of G) is a subgroup of G.
Solution
I have written the answer on a paper and uploaded it at - http://i.imgur.com/yYepqIy.jpg.
Let Fn be the nth Fibonacci number. Deduce these properties WITHOUT .pdf
Let Fn be the nth Fibonacci number. Deduce these properties WITHOUT using the method of
mathematical induction.
Fn=2Fn-2+Fn-3
Solution
we know fibonacci numbers are related as
Fn = Fn-1+Fn-2
thus
Fn-1 = Fn-2 + Fn-3 thus
Fn = Fn-1 + Fn-2
= Fn-2 + Fn-3 +Fn-2
= 2 Fn-2 + Fn-3
thus ans.
Operation Blue Star - Saka Neela Tara
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Introduction to AI for Nonprofits with Tapp Network
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Full Book
for class 11th
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Let S = { a,b,c,d } and T = { 1,2,3,4,5,6,7 }. How many functions a.pdf
Let S = { a,b,c,d } and T = { 1,2,3,4,5,6,7 }. How many functions are there from S into T?
explain your answer.
Solution
mappings from S to T since no of elements in S is 4 and no ofelements in T is 7 An
element in S can be mapped in 7 possible ways so 4 elements can be mapped in 7*7*7*7 ways
so in total 2401 ways.
Let R be the ring of 2 × 2 matrices of the form a b 0 0 wh.pdf
Let R be the ring of 2 × 2 matrices of the form
a b
0 0
where a, b ? R. Show that although R is a ring that has no identity, we can find a
subring S of R with an identity.
Solution
Take S to be the set of matrices of the form a 0 0 0 where a is in R. It is completely
obvious that S is closed under both multiplication and addition. [In fact S is isomorphic to R].
Now the (multiplicative) identity in S is: 1 0 0 0 again this is completely obvious (what happens
when you multiply something in S by the matrix on the left or right?). I hope this helps, if you
need more details write back..
Let R be a commutative ring with unity and let I,J be two ideals of .pdf
Let R be a commutative ring with unity and let I,J be two ideals of R such that I + J = R Show
that R/(I intersect J) is isomorphic to R/I
Solution
GIVEN R IS COMMUTATIVE RING WITH UNITY I & J ARE 2 IDEALS
OF R I+J=R TO SHOW THAT R/[I INTERSECT J ] IS ISOMORPHIC TO
[R/I] * [ R / J ] PROOF I IS AN IDEAL OF R ..SO I IS A SUB GROUP OF R
UNDER ADDITION I1*R1 & R1*I1 ARE ELEMENTS OF I FOR ANY ELEMENT I1 IN I
AND R1 IN R FURTHER SINCE R IS COMMUTATIVE.
Let p, q, and r be the following statements p Jamie is on the .pdf
Let p, q, and r be the following statements: p: Jamie is on the train. q: Sylvia is at the park. r:
Nigel is in the car. Translate the following statement into English: (p ^ ~q)-->r
Solution
answer: If jamie is on the train and sylvia is not at the park, then Nigel is in the car..
Let p, q, and r be the following statements p Jamie is on the .pdf
Let p, q, and r be the following statements:
p: Jamie is on the train.
q: Sylvia is at the park.
r: Nigel is in the car.
Translate the following statement into English: (~p q) r
Construct a truth table for ~q p
Solution
(~p V q) -> r ~p means negate V means or a->b means if a then b SO if Jamie is
on the train OR Jamie is on the train THEN Nigel is in the car..
Let P(x) be hte statement x=x2 if the domain consists of the int.pdf
Let P(x) be hte statement \"x=x2\" if the domain consists of the integers, What are the truth
values?
a) P(0)
b) p(1)
c) P(2)
d)P(-1)
Solution
a) T
b) T
c) T
d) F (-1 1)
e) T (there does an exist a (in fact, multiple x\'s) where x = x2. that\'s when the integers are 0 and
positive).
f) F ( negative integers do not equal their square).
Let H and K be finite abelian groups and let G be an infinite abelia.pdf
Let H and K be finite abelian groups and let G be an infinite abelian group. Let it be known that
G x H is isomorphic to G x K . Show by example that because G is infinite then H need not be
isomorphic to K.
Solution
Use the fundamental theorem for finitely generated abelian groups. First show that the finite
parts of G and H are equal, then show that the infinite parts are equal.
For example if G,H,K have infinite parts Z^n, Z^m and Z^ p, then
Z^(n+p)=Z^(m+p) implies n=m.
Let n be a non-negative integer and a and c be positive numbers. Use.pdf
Let n be a non-negative integer and a and c be positive numbers. Use probability to express
0a xne-cxdx in terms of a finite sum.
Solution
F(x) = P(X x) (also denoted FX(x)). F(x) is the cumulative probability to the left of
(and including) the point x. The survival function is the complement of the distribution function,
S(x) = 1 - F(x) = P(X > x). For a discrete random variable with probability function p(x), F(x) =
X wx p(w), and in this case F(x) is a step function (it has a jump at each point with non-zero
probability, while remaining constant until the next jump). If X has a continuous distribution
with density function f(x), then F(x) = R x -1 f(t)dt and F(x) is a continuous, differentiable, non-
decreasing function such that d dxF(x) = F0(x) = -S0(x) = f(x). If X has a mixed distribution,
then F(x) is continuous except at the points of non-zero probability mass, where F(x) will have a
jump. For any cdf P(a < X < b) = F(b) - F(a), lim x!1 F(x) = 1, lim x!-1 F(x) = 0..
Let G be a group. Recall that an automorphism of G is an isomorphi.pdf
Let G be a group. Recall that an automorphism of G is an
isomorphism f : G ? G.
(1) Show that the set of automorphisms of G forms a group; we denote
this group Aut(G).
(2) Show that Aut(Z/4Z) ~ Z/2Z.
=
(3) Calculate Aut(Z/2Z×Z/2Z), and show that this group is isomorphic
to a group we are already very familiar with.
Solution
(.
Let G be a group, G = n in N, and H is a subset of G. Then all tha.pdf
Let G be a group, |G| = n in N, and H is a subset of G. Then all that is necessary to show that
H<=G is that:
a. H is closed
b. H is closed and has the inverse property
c. H is closed, has the inverse and identity properties,
d. H is closed, has the inverse, identity, and associativity properties
Solution
d. H is closed, has the inverse, identity, and associativity properties.
Let I and J be ideals in a ring R.Solution H.pdf
Let I and J be ideals in a ring R.
Solution
Hi i have solved this question in my notebook, but i can\'t type it and post it here
because there\'s no time... please rate me lifesaver and i\'ll make sure the answer is in your
inbox! i don\'t do this generally but i am doing this due to lack of time....
Let L be a list of the 26 letters in the English alphabet which cons.pdf
Let L be a list of the 26 letters in the English alphabet which consists of 5 vowels and 21
consonants . Show that has a sublist consisting of four or more consecutive consonants.
Assuming L begins with a vowel, say A, show that L has a sublist consisting of five or more
consecutive consonants.
Solution
as there are 21 consonants so there will be 4 or more than 4 insublist.
Let Ho Age and Observance of traditions are independent. Let Ha be.pdf
Let Ho: Age and Observance of traditions are independent. Let Ha be: Age and Observance of
traditions are not independent. A type 1 error in this context would be
A)Saying age and observance are independent when they actually are independent.
B)Saying age and observerance are not independent when they actually are dependent.
C)Saying age and observance are dependent when they are not dependent.
D)Saying age and observance are independent when they actually are not independent.
Solution
C)Saying age and observance are dependent when they are not dependent..
Let L1 be the language that consists of strings that only contain ze.pdf
Let L1 be the language that consists of strings that only contain zeros (i.e. 0n ) let L2
be the language 0n 1n , let L3 be the language 0n 1n 0n , let L4 be the language
0n 1n 0n 1n , let L5 be the language 0n 1n 0n 1n 0n etc. The language I want my
Turing machine to recognize is the union 0 Li of all the languages L0, L1,
L2, L3 etc. Thus a string is in L i the string is empty (i = 0) or there is
some m > 0 such that the string is in Lm. The problem that needs to be solved is simply
to construct a Turing machine that would recongnize such languages. Any help (advice) would
be most appreciated.
Solution
Let L1, L2 be languages over the same alphabet S, then there are following operations on L1,
L2: L1 ? L2 = {w|w ? L1 or w ? L2} L1 ? L2 = {w = xy|x ? L1,y ? L2} L1 = {w ? S * |w ?/ L1}
L k 1 = (L1 ? L1) ... L1 | {z } , k-times, with L 0 1 = {e} L * 1 = {w ? S * |?k = 0,w ? L k 1 }.
This is called Kleene Star of language L1. L + 1 = {w ? S * |?k = 1,w ? L k 1 }.
Let L(x) be the statement x is a lion and C(x) be the statement .pdf
Let L(x) be the statement \"x is a lion\" and C(x) be the statement \"x drinks coffee.\" Let the
universe of discourse for x be the set of all creatures. Express the following predicate logic
statement in plain English.
Ax(L(x)>C(X))
a. Some lions drink coffee.
b. If lions are creatures, then they drink coffee.
c. All creatures drink coffee.
d. All lions drink coffee.
Solution
d. All lions drink coffee..
Let G be a commutative group with operation . Prove that the set of.pdf
Let G be a commutative group with operation *. Prove that the set of a in G such that a^3=e
(where e is the identity of G) is a subgroup of G.
Solution
I have written the answer on a paper and uploaded it at - http://i.imgur.com/yYepqIy.jpg.
Let Fn be the nth Fibonacci number. Deduce these properties WITHOUT .pdf
Let Fn be the nth Fibonacci number. Deduce these properties WITHOUT using the method of
mathematical induction.
Fn=2Fn-2+Fn-3
Solution
we know fibonacci numbers are related as
Fn = Fn-1+Fn-2
thus
Fn-1 = Fn-2 + Fn-3 thus
Fn = Fn-1 + Fn-2
= Fn-2 + Fn-3 +Fn-2
= 2 Fn-2 + Fn-3
thus ans.
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Let S = { a,b,c,d } and T = { 1,2,3,4,5,6,7 }. How many functions a.pdf
Let S = { a,b,c,d } and T = { 1,2,3,4,5,6,7 }. How many functions a.pdf
Let R be the ring of 2 × 2 matrices of the form a b 0 0 wh.pdf
Let R be the ring of 2 × 2 matrices of the form a b 0 0 wh.pdf
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Let R be a commutative ring with unity and let I,J be two ideals of .pdf
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