You all must be familiar with the 'famous' proof by contradiction to prove the irrationality of 2. The following is a reminder of that proof. If you already know this, you may skip to the main question.
Problem:
Prove that 2 is irrational.
Solution:
Let us assume that 2 is rational. Then 2 can be expressed in the form of qp for coprime integers p and q with q=0.
Now
⟹⟹222q2===qpq2p2(squaring both sides)p2
Thus 2∣p2⟹2∣p.
So let p=2k for some integer k. Substituting p in 2q2=p2 gives
⟹⟹2q22q2q2===(2k)24k22k2
Thus 2∣q2⟹2∣q.
Hence 2 is a common factor of both p and q. But this contradicts the fact that p and q are coprime integers, hence our assumption is false.
∴2 is irrational.
(Q.E.D.)
Main Question
Let us do a similar procedure with 4.
Flawed problem:
Prove that 4 is irrational.
Flawed solution:
Step 1:
Let us assume that 4 is rational. Then 4 can be expressed in the form of qp for coprime integers p and q with q=0.
Step 2:
Now
⟹⟹444q2===qpq2p2(squaring both sides)p2
Step 3:
Thus 4∣p2⟹4∣p.
Step 4:
So let p=4k for some integer k. Substituting p in 4q2=p2 gives
⟹⟹4q24q2q2===(4k)216k24k2
Step 5:
Thus 4∣q2⟹4∣q.
Step 6:
Hence 4 is a common factor of both p and q. But this contradicts the fact that p and q are coprime integers, hence our assumption is false.
∴4 is irrational.
(Q.E.D.)
But we know that 4=2 is rational.
In which step in the flawed solution shown above has the error been committed for the first time?
Your answer seems reasonable.
Find out if you're right!