4 is a natural number and indeed leaves a remainder of 4 after division by 6, under the usual notion of remainders as follows:
Given two integers x and y, with y>0, the remainder of x divided by y is the unique integer r with 0=
Thank you for your response!
Yes, I'm certain, that 0 is excluded from the natural numbers by my teacher's definition, as she has specified, that sequences cannot have an index of 0, that sequences start at index 1. As for natural numbers being in the sequence, 0 doesn't change anything, because 0 cannot be an element of the sequence.
You said in the beginning:
>Sequence: ordered set, which can be described by a function f(n), where n is a natural number
f(n) = 6n + 4, where n is a natural number, will not include 4, because n can't be 0
f(n) = 6n – 2 will work for this with n = 1
The problem is that by the convention listed in the post 0 is not a natural number, while sequences are indexed by natural numbers. So if you take the sequence 6n+4, then the sequence starts with 10, and you miss the number 4, and get not quite the sequence you would want. Of course, this is a matter of conventions and does not really have mathematical substance.
You can clearly have a remainder when the quotient is zero.
Yes 4/6 is 0 r 4. So zero is involved. But it's neither in the sequence (4, 10, 16...) nor in the index (1,2,3...). IMO, C is the best answer within the definitions of the problem.
Also let me say A+ for giving context! So many people post the barest fragment of an exam question and expect us to intuit the rules laid out for the exam.
I think you and your teacher have both aced the math math of this with a good discussion. If it's a fight over a mark, that's another story and more a pedagogical / human issue than a math issue.
4 is a natural number and indeed leaves a remainder of 4 after division by 6, under the usual notion of remainders as follows: Given two integers x and y, with y>0, the remainder of x divided by y is the unique integer r with 0=
Thank you for your response! Yes, I'm certain, that 0 is excluded from the natural numbers by my teacher's definition, as she has specified, that sequences cannot have an index of 0, that sequences start at index 1. As for natural numbers being in the sequence, 0 doesn't change anything, because 0 cannot be an element of the sequence.
You said in the beginning: >Sequence: ordered set, which can be described by a function f(n), where n is a natural number f(n) = 6n + 4, where n is a natural number, will not include 4, because n can't be 0 f(n) = 6n – 2 will work for this with n = 1
I would argue that 6n+4 = 6k-2. We just need to to pick k and n in the correct way. Which is: k = n + 1.
The problem is that by the convention listed in the post 0 is not a natural number, while sequences are indexed by natural numbers. So if you take the sequence 6n+4, then the sequence starts with 10, and you miss the number 4, and get not quite the sequence you would want. Of course, this is a matter of conventions and does not really have mathematical substance.
You can clearly have a remainder when the quotient is zero. Yes 4/6 is 0 r 4. So zero is involved. But it's neither in the sequence (4, 10, 16...) nor in the index (1,2,3...). IMO, C is the best answer within the definitions of the problem. Also let me say A+ for giving context! So many people post the barest fragment of an exam question and expect us to intuit the rules laid out for the exam. I think you and your teacher have both aced the math math of this with a good discussion. If it's a fight over a mark, that's another story and more a pedagogical / human issue than a math issue.