SirGMAT wrote:
If S and T are non-zero numbers and 1S+1T=S+T, which of the following must be true?
A. ST=1
B. S+T=1
C. 1/S=T
D. S/T=1
E. none of the above
Explanation provided:
1/S + 1/T = S+T;
T+S/ST = S+T→;
Cross-multiply: S+T=(S+T)∗ST;
(S+T)(ST−1)=0. Either S+T=0 or ST=1. Now, notice that if S+T=0 is true then none of the options must be true.
The correct answer is E
Question:
I understand the way of the equation, however, what I would have done is interfere at the following step:
S+T = (S+T)*ST
-> (S+T)/(S+T)=ST
-> ST = 1
Is there some rule which forbids me to take this step? Or is the only option to realize so, that you perform the above given equation as well and realise that "S+T = 0" negates all other options than E.... ??
Thanks in advance,
best regards
P.S. Sry if the format is terrible, this is the first question I am copying out of somewhere.
There was this fun derivation that my math teacher showed us in school. Just to demonstrate how cancelling of 0 could yield wrong results. He claimed that he could prove that 1=2 and hence all numbers are equal.
It goes as below:
Let,
\(a=b\)
Multiplying both sides by a, We get
\(a^2 = ab\)
Subtracting \(b^2\) from both sides
\(a^2 - b^2 = ab - b^2\)
\((a+b)(a-b) = b(a-b)\)
Cancelling \((a-b)\) on both sides,
\(a+b = b\)
Since \(a=b\)
\(a+a = a\)
\(2a = a\)
\(2=1\)