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Number Theory

· 2 min read

Based on some of the problems from Prof. Dragos Ghioca's problem set 1.

Question 1#

Let a,b∈Na, b \in \mathbb{N}. Show that if gcd⁡(a,b)=lcm(a,b)\gcd(a,b) = \text{lcm}(a,b), then a=ba = b.

Question Explanation#

a,ba, b here are natural numbers, i.e. the ones that are used to count things: 1,2,…1, 2, \dotso

Now, consider the set of all numbers that divide both aa and bb. The maximum of this set is defined to be gcd⁡(a,b)\gcd(a,b). Formally: gcd⁡(a,b):=max⁡{x:x∣a and x∣b}\gcd(a,b) := \max \{ x : x | a\ \text{and}\ x | b \}.

Similarly, consider the set of all numbers that are divisible by both aa and bb. The minimum of this set is defined to be lcm(a,b)\text{lcm}(a,b). Formally: lcm(a,b):=min⁡{x:a∣x and b∣x}\text{lcm}(a,b) := \min \{ x : a | x\ \text{and}\ b | x \}.

Solution 1
  1. a∣lcm(a,b)a | \text{lcm}(a,b) and b∣lcm(a,b)b | \text{lcm}(a,b), thus a≤lcm(a,b)a \leq \text{lcm}(a,b) and b≤lcm(a,b)b \leq \text{lcm}(a,b) (by definition).
  2. gcd⁡(a,b)∣a\gcd(a,b) | a and gcd⁡(a,b)∣b\gcd(a,b) | b, thus gcd⁡(a,b)≤a\gcd(a,b) \leq a and gcd⁡(a,b)≤b\gcd(a,b) \leq b (by definition).
  3. Combine both inequalities together:
  • gcd⁡(a,b)≤a≤lcm(a,b)\gcd(a,b) \leq a \leq \text{lcm}(a,b)
  • gcd⁡(a,b)≤b≤lcm(a,b)\gcd(a,b) \leq b \leq \text{lcm}(a,b)
  1. We are given that gcd⁡(a,b)=lcm(a,b)\gcd(a,b) = \text{lcm}(a,b), therefore inequalities are actually equalities:
  • gcd⁡(a,b)=a=lcm(a,b)\gcd(a,b) = a = \text{lcm}(a,b)
  • gcd⁡(a,b)=b=lcm(a,b)\gcd(a,b) = b = \text{lcm}(a,b)
  1. Hence, a=ba = b.