homework-6

Question

Answer 1

Using Direct Proof : Let’s assume 7|4a where ‘a’ is some integer.

So, 4a = 7x for some integer x. Here LHS is even and 7 in RHS is odd. So, x must be an even integer. Let x = 2y, for some integer y.

=> 4a = 7(2y) 

=> 2a = 7y. Now also LHS is even and 7 in RHS is odd. So, y must be an even integer. Let y = 2z, for some integer z.

=> 2a = 7(2z) 

=> a = 7z

So 7 is a factor of ‘a’. Therefore 7|a.

Comments

@KG
Kindly correct the sentence LHS is even and RHS is odd to LHS is even and 7 in RHS is odd. (Line no 2 & 4)
7x and 7y both are even.

---

Done.