Question

Let
m and n be positive integers and let k be the least common multiple
of m and n. Show that mZ intersect nZ is equal to kZ. provide
justifications pleasw, thank you.

Answer #1

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Let
m and n be positive integers and let k be the least common multiple
of m and n. Show that mZ intersect nZ is equal to kZ. provide
justifications please, thank you.

Let m,n be any positive integers. Show that if m,n have no
common prime divisor (i.e. a divisor that is at the same time a
prime number), then m+n and m have no common prime divisor. (Hint:
try it indirectly)

Let
m,n be integers. show that the intersection of the ring generated
by n and the ring generated by m is the ring generated by their
least common multiple.

The least common multiple of nonzero integers a and b is the
smallest positive integer m such that a | m and b | m; m is usually
denoted [a,b]. Prove that
[a,b] = ab/(a,b) if a > 0 and b > 0.

Let X Geom(p). For positive integers n, k define
P(X = n + k | X > n) = P(X = n + k) / P(X > n) :
Show that P(X = n + k | X > n) = P(X = k) and then briefly
argue, in words, why this is true for geometric random
variables.

Prove that for fixed positive integers k and n, the number of
partitions of n is equal to the number of partitions of 2n + k into
n + k parts.
show by using bijection

Let a, b be positive integers and let a = k(a, b), b = h(a,
b).
Suppose that ab = n^2 show that k and h are perfect squares.

Let
m and n be positive integers. Exhibit an arrangement of the
integers between 1 and mn which has no increasing subsequence of
length m + 1, and no decreasing subsequence of length n + 1.

Let N* be the set of positive integers. The relation
∼ on N* is defined as follows: m ∼ n ⇐⇒ ∃k ∈
N* mn = k2
(a) Prove that ∼ is an equivalence relation.
(b) Find the equivalence classes of 2, 4, and 6.

Please include the argument in word, thanks
Let X ∼ Geom(p). For positive
integers n, k define
P(X = n + k | X > n) =
P(X = n + k) / P(X > n)
.
Show that P(X = n + k |
X > n) = P(X = k) and then briefly
argue, in words, why this is true
for geometric random variables.

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