Solution

194341

Problem
Dodgy proofs Back to problem
First name
Nishad Deulkar
Country
England
Age
0

M3. "All numbers are equal"
On the final step we have $a(a-b-c)=b(a-b-c)$ but recall that $a=b+c \Rightarrow a-b-c=0$ so in fact both the LHS and RSH are 0, giving us a division by 0 error.