MI5
				
				
			 
			
	
	
	
		
	
	
			
		
		
			
			
				
- 8
- 0
Theorem: If c|ab and (b, c) = 1 then c|a.
Proof: Consider (ab, ac) = a(b, c) = a. We have c|ab and clearly c|ac so c|a. It's not so clear to me why c|ac. Perhaps I'm missing something really obvious.
				
			Proof: Consider (ab, ac) = a(b, c) = a. We have c|ab and clearly c|ac so c|a. It's not so clear to me why c|ac. Perhaps I'm missing something really obvious.
 
 
		 
 
		 
 
		