The procedure of proof should follow the following format. The answer should lik
ID: 3701852 • Letter: T
Question
The procedure of proof should follow the following format. The answer should like that.
Example
Prove that f :R×R?R×Rsuch that f(x,y) = (2y,-x) is a bijection.
Proof:
(1)Let (a,b) and (c,d) be arbitrary elements ofR×Rsuch that a?c or b?d.
(2)By the definition of f, f(a,b) = (2b,-a) and f(c,d) = (2d, -c)
(3)Case 1: if a?c, then -a?-c, so (2b,-a)?(2d, -c)
(4)Case 2: if b?d, then 2b?2d, so (2b,-a)?(2d, -c)
(5)By (3) and (4), in all cases we have f(a,b)?f(c,d). This proves that f is one-to-one.
(6)Let (u,v) be an arbitrary element of R×R. By definition of f, we have f(-v, u/2) = (u,v)
(7)By (6), (-v,u/2)?f-1(u,v), which shows that f-1(u,v)??. This proves f is onto.
(8)By (5) and (7), f is one-to-one and onto, so f is a bijection
6. Consider the following attempt at defining a function f 6,6) for each r ? {1, . . . ,6), define f(z) to be a number y ? {1, . . . .6) such that ry Is f well-defined? Prove that your answer is correct 1 (mod 7).Explanation / Answer
1) Let (a,b) be arbitary elements of f such that f:{a} ----> {b}, such that (ab) = 1(mod 7) and a belongs to {1,2,3,4,5,6} and b belongs to {1,2,3,4,5,6}.
2)By the definition of f, f(a) ----> f(b), such that (ab) = 1(mod 7).
3) Case 1: When a = 1 and b1 = 1, (ab1) = 1 === 1 (mod 7) and thus f(a) ----> f(b1)
Case 2: When a = 1 and b2 = 1, (ab2) = 1 === 1(mod 7) and thus f(a) -----> f( 2)
Thus, we see that ab1 = ab2 = 1(mod 7), which tell us that b1 = b2 for all a
4) Thus, we have f(a,b1) = f(a,b2) where b1 = b2 for all a, which clearly states that f is well defined.
Please let me know in case of any clarifications required. Thanks!
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