Is m and n an integer?

The numbers m and n are integers.

Is the binomial coefficient always an integer?

See my post here for a simple purely arithmetical proof that every binomial coefficient is an integer. The proof shows how to rewrite any binomial coefficient fraction as a product of fractions whose denominators are all coprime to any given prime p.

Can M n be rational?

Note that you can use an argument similar to this one to show that any repeating decimal is a rational number. i. Yes, since m and n are integers, so are m + n and mn (because sums and products of integers are integers).

Are integers closed under division?

b) The set of integers is not closed under the operation of division because when you divide one integer by another, you don’t always get another integer as the answer. For example, 4 and 9 are both integers, but 4 ÷ 9 = 4/9.

How do you prove n is an integer?

How can we proof it? Theorem: If n is an odd integer, then n2 is an odd integer. Proof: Since n is an odd integer, there exists an integer k such that n=2k+1. Therefore, n2 = (2k+1)2 = 4k2+4k+1 = 2(2k2+2k)+1.

Is n choose k an integer?

Then (nk) is an integer.

Is n choose k always divisible by n?

Clearly n divides into n! If n is prime, then by Euclid’s Lemma, n can’t divide k! or (n-k)! Thus since C(n,k) = n!/[k!( n-k)!] is an integer and so n must divide into it.

Which number is rational but not integer?

In other words, any integer a can be written as a = a/1, which is a rational number. Thus, every integer is a rational number. Clearly, 3/2,-5/3, etc. are rational numbers but they are not integers.

Which of the following does not represent an integer?

(d)(-12) / 5 is the correct answer It is a fractional no. which represents -2.4 . And -2.4 is not an integers PLZ MARK AS A BRAINLIEST.

Is (N m) an integer?

It follows that ( N M) is an integer. Another way to think of it is combinatorially, which is of course the motivation. Let 1 ≤ k ≤ n. Consider the set S of all sequences of k distinct numbers among { 1, …, n }. The size of S is n ⋅ ( n − 1) ⋯ ( n − k + 1) = n! ( n − k)!.

Is (n – 1 m – 1) a natural number?

We saw that ( n − 1 m) and ( n − 1 m − 1) are natural number, therefore ( n m) is also a natural number, by pascal’s rule. Therefore k + 1 ∈ S which implies, by strong induction, that S = N. Hence ( n m) ∈ N, ∀ 0 ≤ m ≤ n.

Is there a non-combinatorial way to find the natural number N-1?

Therefore Well, one noncombinatorial way is to induct on n using Pascal’s triangle; that is, using the fact that ( n k) = ( n − 1 k − 1) + ( n − 1 k) (easy to verify directly) and that each ( n − 1 0) is just 1. As Jonas mentioned, it counts something so it has to be a natural number.

What is the non-combinatorial way to find the product of integers?

Thus n! divides every product of n consecutive integers, since it has a smaller power of every prime divisor. Therefore Well, one noncombinatorial way is to induct on n using Pascal’s triangle; that is, using the fact that ( n k) = ( n − 1 k − 1) + ( n − 1 k) (easy to verify directly) and that each ( n − 1 0) is just 1.