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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.

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