$$\text { For all natural numbers a and } b , \operatorname { GCD } ( a , b ) \mid a b \text {. }$$

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The Answer of $$\text { For all natural numbers a and } b , \operatorname { GCD } ( a , b ) \mid a b \text {. }$$

For all natural numbers a and b,GCD⁡(a,b)∣ab. \text { For all natural numbers a and } b , \operatorname { GCD } ( a , b ) \mid a b \text {. } For all natural numbers a and b,GCD(a,b)∣ab.