Let n1 n2 n satisfy gcdn1 n2 1 prove that there is a set...

Transcript

00:01 In this question, we need to prove that there exist a bijection set with the given requirements.

00:06 And it is like n1 comma n2 and they belongs to natural number n.

00:13 And then here we have that gcd, greatest common divisor of n1 and n2 should be equals to 1.

00:22 That is basically they should be co -prime.

00:24 Now here we can say that n1, let n1 be equal to, p1 raised to the power alpha 1 multiplied with p2 raised to the power alpha 2 multiplied with and then it will go on up to pi raised to the power alpha i now here if we proceed further we can say that n2 value will be equals to q1 raised to the power b1 and then here it will be q2 raised to the power b2 and then this will go on up to to qj and it will be raised to the power bj.

01:06 Now here we must understand that pl and qm are prime numbers.

01:15 So it can be written over here as prime.

01:18 Also we can here write that l is having the range, let us write here that it belongs to 1 comma 2 and then it will go on a to i and then here it will be m that belongs to 1 comma 2 and this will go on up to j.

01:39 So now if we proceed further we can say that d belongs to natural number right.

01:47 So that means prime factorization of d must be a subset of prime factorization of n1 and n2 and hence we can write that d star will be equals to p1 raised to the power alpha 1, multiplied with p2 raised to the power alpha 2, multiplied with it will be pi raised to the power alpha i.

02:08 Now it will be multiplied with q1 raised to the power beta 1 multiplied with q2 raised to the power beta 2 and then this will go on up to q and then it will be j here and then raise to the power b of j where it must be noted that alpha l values are between 0 to 8.

02:30 A .l values.

02:32 And then here it must be noted that the value of beta m is also between bm and 0...

Question

Let n1, n2 ∈ N satisfy gcd(n1, n2) = 1. Prove that there is a set bijection (i.e. a one-to-one correspondence) {d ∈ N: d | n1n2} ↔ {d1 ∈ N: d1 | n1} × {d2 ∈ N: d2 | n2}

Question

Let n1, n2 ∈ N satisfy gcd(n1, n2) = 1. Prove that there is a set bijection (i.e. a one-to-one correspondence) {d ∈ N: d | n1n2} ↔ {d1 ∈ N: d1 | n1} × {d2 ∈ N: d2 | n2}

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