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 July 5th, 2018, 12:28 AM #1 Newbie   Joined: Jul 2018 From: india Posts: 1 Thanks: 0 Immediate Help Required in discrete mathematics Hi Friends, I am new to this forum and my son is awfully struck in some discrete mathematics question. Your help is highly solicited. The questions are posted below: 1. Prove that if H, K are subgroups of a group G and H Ų K = G. Then either H=G or K=G 2. Let G be a group and a, b Є G. Then the equation x*a=b has a unique solution given by x= b* a 3. Linear sum W1+ W2 of two subspaces W1and W2 of a vector space V(F) is A subspace of V(F) 4. show that the function T: R2 → R2 such that T(0,1)=(3,4),T(3,1)=(2,2) And T(3,2)=(5,7) is not a L.T. 5. Let T:v →w be a linear transformation. Then T is onto iff p(T)= dim w. 6. show that the function T: R2 → R2 defined by T(x1,x2)=(x1-x2,x1+x2),for (x1,x2)Є R2 is bijective Please provide detailed steps as I do not know these mathematics, but when I give this to my son, he can understand. Thanks in advance friends.
 July 5th, 2018, 03:15 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,113 Thanks: 1909 2. For a, b ∈ G, where (G,*) is a group, let c ∈ G be the unique inverse of a, then x*a = b implies (x*a)*c = b*c, which simplifies to x = b*c. Note: the usual axioms of a group (G,*) specify that the group operation is associative and that each element of G has an inverse. It's an elementary theorem that no element has more than one inverse. 4. If T is a L.T., T(0,1) + T(3,1) = T(0+3,1+1) = T(3,2) = (5,7), but T(0,1) + T(3,1) = (3,4) + (2,2) = (5,6), not (5,7). 6. T is bijective as it has an inverse Q(x,y) = ((x+y)/2,(y-x)/2). Thanks from zylo

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