posted by Stephanie

Def: An interger "m" divides an integer "n" if there is an integer "q" such that n=mq.
?Suppose a, b, and c are integers such that a divides b and b divides c. Prove that a divides c.

1. drwls

If b is an integer factor of b, and b is an integer factor of x, then

b = ma and c = nb, where m and n are both integers.

It follows that c = m n a. Since m and n are both integers, so is mn. Therefore a is an integer factor of c, or, in other words, "a divides c".

## Similar Questions

1. ### math induction

prove the product of 4 consecutive integers is always divisible by 24 using the principles of math induction. Could anyone help me on this one?
2. ### discrete math

If a and b are positive integers, and m=lcm(a,b), explain why m divides any common multiple of a and b. The answer is in the definition of lcm:] the smallest multiple that is exactly divisible by every member of a set of numbers. So …
3. ### maths --plse help me..

Let a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)7 .
4. ### urgent -plsee

Let a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)7 .
5. ### maths --plse help me..

Let a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)^7 .
6. ### maths

Let a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)^7 .
7. ### maths

Let a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)^7 .
8. ### math analysis

Disprove: For every integer a, if 8 divides a, then 6 divides a. im still stuck. i wrote: Disproof. Suppose 8|a. by definition of divisibility, we know 8|a means there is an integer b with a=8b. There is not value of y.
9. ### DISCRETE MATHS

We need to show that 4 divides 1-n2 whenever n is an odd positive integer. If n is an odd positive integer then by definition n = 2k+1 for some non negative integer, k. Now 1 - n2 = 1 - (2k+1)2 = -4k2-4k = 4 (-k2-4k). k is a nonnegative …
10. ### math

Suppose n, a are integers. If n divides a, then n divides a^2.

More Similar Questions