Math Assignment Help | Math Homework Help
About Mathematics: Roger Bacon (1214-1294) has described Mathematics as the door and the key to the sciences. A neglect of mathematics hurts all the knowledge, because one is ignorant of it cannot know the other sciences or the things of the world. Mathematics or math assignment help as it is popularly called is a game of numbers. These numbers work under a set of rules that set the rule for the “number game i.e mathematics”. Maths is by far the easiest subject in the curriculum. The student needs to know the rules of the number game and he is a winner if he plays it well.
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Sample of math problems and solutions:
Problem 1. If in a group G, xy2 = y3x and xy2 = x3y, then show that x = y = e where e is the identity of G.
Solution:
xy2 = y3x
=>
x = y3xy-2
x2 = xy3xy-2 = xy2 yxy-2 = y3xyxy-2 x2y = y3xyxy-1 Now
yx 2 = x3y implies yx2 = xy3xyxy-1
=>
x2 = y-1xy3xyxy-1
=>
X2y = y-1xy3xyx
By (1) and (2) we obtain Y3xyxy-1 = y-1xy3xyx =>
y4xyx xy3xyxy
=>
y4xyx = xy2yxyx = y3xyxyxy
=>
(yx)2 = (yx)3
Interchange x and y in (3) to get
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(xy)2 = (yx)3 Now (3) and (4) imply (xy)2= (xy)3 (yx)2(yx)= (xy)3(yx) =>
e = xy2x => x-2 = y2
Further
xy2 = y3x => xx-2 = yx-2x
=>
x-1 = yx-1 => y = e. yx2 = x3y => ex2 = x3e => x = e.
Lastly Problem 2.
G is a group and there exist two relatively prime positive integers m and n such that ambm = bmam and anbn = bnan for all a, b ∊ G Prove that G is abelian.
Solution:
Since (m, n) = 1, we get mx + ny = 1 for some x, y ∊ Z. (ambn)mx = am(bnam)mx-1bn
Now
= am(bnam)mx(bnam)-1bn = (bnam)mxana-mb-nbn = (bnam)mx. Similarly it can be proved that (ambn)ny = (bnam)ny. From (1) and (2), we get ambn = (ambn)mx+ny = (bnam)mx+ny = bnam =>
Finally
amubnv = bnvamu
ab = a mx+ny bmx+ny
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= amx(anybmx)bny = amxbmx any bny
by (3)
= bmxamx bny any = bmx
+ ny
amx
+ ny
by hypothesis by (3)
= ba. Hence G is abelian.
Problem 3. Show that the equation x2ax = a-1 is solvable for x in a group G if and only if a is the cube of some element in G. Solution: Suppose x2ax = a-1 is solvable in G, there exists c ∊ G such that c2ac = a-1. This gives ca ca = c-1 => ca ca ca = ea = a => (ca)3 = a. Conversely let a = b3 for some b in G. Then x = b-2 is a solution of x2ax= a-1, since, x2ax =b4 3 -2 b b = b-3 while a-1 = b-3.
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