Show that any integer of the form 6k+5 is also of the form 3k+2, but not conversely.
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Q. Show that any integer of the form 6k + 5 is also of the form 3j+2 but not conversely?
Ans. Let 6k + 5 be an integer. Then note that 6k + 5 = 6k + 3 + 2 = 3(2k + 1) + 2 so take j = 2k + 1 we see that 6k + 5 is of the form 3j + 2.
To show converse is false: Let 3j + 2 be an integer. Note that j is either even or odd. If j is even, then j = 2m for some integer m. Then 3j + 2 = 3(2m) + 2 = 6m + 2. In this case, 3j + 2 has remainder 2 when divided by 6. But any number of the form 6k + 5 has remainder 5 when divided by 6. Since a number cannot have two different remainders when divided by the same quiotient, we see that 3j+2 cannot be written as 6k+5 for even j.
Q. Show that any integer of the form 6k + 5 is also of the form 3j+2 but not conversely?
Ans. Let 6k + 5 be an integer. Then note that 6k + 5 = 6k + 3 + 2 = 3(2k + 1) + 2 so take j = 2k + 1 we see that 6k + 5 is of the form 3j + 2.
To show converse is false: Let 3j + 2 be an integer. Note that j is either even or odd. If j is even, then j = 2m for some integer m. Then 3j + 2 = 3(2m) + 2 = 6m + 2. In this case, 3j + 2 has remainder 2 when divided by 6. But any number of the form 6k + 5 has remainder 5 when divided by 6. Since a number cannot have two different remainders when divided by the same quiotient, we see that 3j+2 cannot be written as 6k+5 for even j.
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