I dont know number 3(c), help me. Thank you This identity follows from the fact

Question

I dont know number 3(c), help me. Thank you

This identity follows from the fact that |ab| = p^alpha_1 + beta_1 _1 middot middot middot p^alpha_s + beta_s _s, together with the observation that for any two real numbers x and y, min(x, y) + max(x, y) = x + y Evaluate [198061, 231896] 2. Show that if g and m are positive integers, there are integers a and b such that (a, b) = g and [a, b] = m if and only if g | m. a) Show that for any three real numbers x, y, z, min(x, max(y, z)) = max(min x, y), min(x, z)). b) Show that (a, [b, c]) = [(a, b), (a, c)] c) It is not entirely accidental that if GCD (as an operation on two integers) is replaced by multiplication and LCM by addition, the relation in (b) becomes the ordinary distributive law, a(b + c) = ab + ac. On the other hand, show that the "dual" relation [a, (b, c)] = ([a, b], [a, c]) also holds, although the analogue a + bc = (a + b)(a + c) is generally false.

Explanation / Answer

3(c) Let (b,c) = r   

Since r is the GCD of b and c,

let b = rB and c = rC

Note that since r is the GCD, B and C are relatively prime.

So LCM of b and c is rBC.

Let (a,r) be s.

Let a = As and r = Rs (A and r are relatively prime)

=> [a,r] = ARs

Thus [a,(b,c)] = ARs (1)

[a,b] = [As,RsB] = Rs*(A,B)

[a,c] = [As,RsC] = Rs*(A,C)

So ([a,b],[A,c]) = (Rs*(A,B), Rs*(A,C)) = Rs ((A,B),(A,C))

Since we saw that B and C are relatively prime, the gcd of (A,C) and (A,C) is A

=> ([a,b],[a,c]) = ARs (2)

Comparing (1) and (2), the result is proved.

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