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Cen fah a bhfuil mise agus thusa ag caint francais agus spanish?
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∙ 14y ago
"Cén fáth a bhfuil mise agus tusa ag caint Fraincis agus Spáinnis" means "Why are we talking French and Spanish?"
Wiki User
∙ 14y ago
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What does 'Cé thusa' mean in Irish?
Irish = Cé thusa? English Translation = Who are you?
How do you pronounce Fuair mé thusa?
"Fuair mé thusa" is pronounced foo-ir may huh-suh.
How do you say 'who are you' in Gaelic?
In (Scottish) Gaelic it is Cò thusa?In Irish (Gaelic) it is Cé thusa?
I found you in Irish Gaelic?
Fuair mé thusa
What diesCha toigh leam meach ach thusa mean?
"meach" should be "neach" "Cha toigh leam neach ach thusa" means "I don't like anyone but you"
How do you say you are everything in Gaelic?
's thusa gach nì
What is the Gaelic for 'and you'?
IRISH (Gaelic): agus tusa (one person); agus sibhse (more than one) (SCOTS) Gaelic: agus thusa; agus sibhse
What are the Gaelic words for 'you'?
IRISH: tú or tusa (more emphatic) one person sibh or sibhse more than one person.SCOTTISH GAELIC: thu or thusa (one person, familiar) sibh or sibhse (more than one, or polite form for one) as in French.
What is the Gaelic for 'know thyself'?
In Irish, fios (information, knowledge, science) or eolas (knowledge, guidance, experience).In Scottish Gaelic, fios.Just so people know (yes I am from Ireland)The proper spelling of Irish in Irish is Gaeilge not Gaelic. Gaelic is acculy an Irish sport a bit like soccer.
How do you say 'sod off' in Scottish Gaelic?
If 'sod off' means 'get lost' there are numerous expressions:Thalla!Càirich ort às an-seo! get lost, shift yourself out of here!Rach air seachran! = get lost, go astray!Thalla is tarraing! = get lost and get stuffed!Thoir an t-sitig ort! = get lost! scram!Truis a-mach! = get lost, go away!Thoir do chasan leat! = 1 take off, run away! 2 get lost!Thalla 's bheir ort! = oh get lost!Tarr às! = 1 get lost! 2 escape!Thalla is tarraing thusa! = buzz off, go and get lost!
How many people lost their jobs during 2008-2010 in America?
First, there's a difference between being "without a job" and being "unemployed". The elderly, the young, and others not looking for work are without jobs. To be "unemployed" you have to be in the labor force and to be in the labor force you have to be working or looking for work. One catch is: as unemployment rises, people give up looking for work thus they aren't in the labor force, so they aren't unemployed. So unemployment drops without any bone losing a job. And when the economy improves and people can get jobs people who weren't in the loabor force start to look for jobs thus putting themselves in the labor force and unemployed both. Thusa unemployment rises even though unemployment might be rising too. FYI there is "frictional" unemployment like when a nurse quits a job and takes a few days to get another and "structural unemployment" where the people don't have the skills to match the jobs that need filling or the potential employees are too far away to take the job so they stay unemployed.
State and prove Lagrange's theorem?
THEOREM:The order of a subgroup H of group G divides the order of G.First we need to define the order of a group or subgroupDefinition:If G is a finite group (or subgroup) then the order of G is the number of elements of G.Lagrange's Theorem simply states that the number of elements in any subgroup of a finite group must divide evenly into the number of elements in the group. Note that the {A, B} subgroup of the Atayun-HOOT! group has 2 elements while the Atayun-HOOT! group has 4 members. Also we can recall that the subgroups of S3, the permutation group on 3 objects, that we found cosets of in the previous chapter had either 2 or 3 elements -- 2 and 3 divide evenly into 6.A consequence of Lagrange's Theorem would be, for example, that a group with 45 elements couldn't have a subgroup of 8 elements since 8 does not divide 45. It could have subgroups with 3, 5, 9, or 15 elements since these numbers are all divisors of 45.Now that we know what Lagrange's Theorem says let's prove it. We'll prove it by extablishing that the cosets of a subgroup aredisjoint -- different cosets have no member in common, andeach have the same number of members as the subgroup.This leads to the conclusion that a subset with n elements has k cosets each with n elements. These cosets do not overlap and together they contain every element in the group. This means that the group must have kn elements -- a multiple of n. We'll accomplish our proof with two lemmas. Lemma:If H is a finite subgroup of a group G and H contains n elements then any right coset of H contains n elements.Proof:For any element x of G, Hx = {h • x | h is in H} defines a right coset of H. By the cancellation law each h in H will give a different product when multiplied on the left onto x. Thus each element of H will create a corresponding unique element of Hx. Thus Hx will have the same number of elements as H.Lemma:Two right cosets of a subgroup H of a group G are either identical or disjoint.Proof:Suppose Hx and Hy have an element in common. Then for some elements h1 and h2 of Hh1 • x = h2 • yThis implies that x = h1-1 • h2 • y. Since H is closed this means there is some element h3 (which equals h1-1 • h2) of H such that x = h3 • y. This means that every element of Hx can be written as an element of Hyby the correspondenceh • x = (h • h3) • yfor every h in H. We have shown that if Hx and Hy have a single element in common then every element of Hx is in Hy. By a symmetrical argument it follows that every element of Hy is in Hx and therefore the "two" cosets must be the same coset.Since every element g of G is in some coset (namely it's in Hg since e, the identity element is in H) the elements of G can be distributed among H and its right cosets without duplication. If k is the number of right cosets and n is the number of elements in each coset then |G| = kn.Alternate Proof:In the last chapter we showed that a • b-1 being an element of H was equivalent to a and b being in the same right coset of H. We can use this Idea establish Lagrange's Theorem.Define a relation on G with a ~ b if and only if a • b-1 is in H. Lemma: The relation a ~ b is an equivalence relation.Proof:We need to establish the three properties of an equivalence relation -- reflexive, symmetrical and transitive.(1) Reflexive:Since a • a-1 = e and e is in H it follows that for any a in Ga ~ a(2) Symmetrical:If a ~ b then a • b-1 is in H. Then the inverse of a • b-1 is in H. But the inverse of a • b-1 is b • a-1 sob ~ a(3) Transitive:If a ~ b and b ~ c then both a • b-1 and b • c-1 are in H. Therefore their product (a • b-1) • (b • c-1) is in H. But the product is simply a • c-1. Thusa ~ cAnd we have shown that the relation is an equivalence relation.It remains to show that the (disjoint) equivalence classes each have as many elements as H.Lemma:The number of elements in each equivalence class is the same as the number of elements in H.Proof:For any a in G the elements of the equivalence class containing a are exactly the solutions of the equationa • x-1 = hWhere h is any element of H. By the cancellation law each member h of H will give a different solution. Thus the equivalence classes have the same number of elements as H.One of the imediate results of Lagrange's Theorem is that a group with a prime number of members has no nontrivial subgroups. (why?)Definition:if H is a subgroup of G then the number of left cosets of H is called the index of H in G and is symbolized by (G:H). From our development of Lagrange's theorem we know that|G| = |H| (G:H)
