You start by assuming the opposite: let √2 be a rational number, so √2 = p/q where p and q are co-prime integers. Then you square both sides and get 2q² = p², proving p² is even, so p is even. If p is even, then p=2k, so substituting gives q²=2k², meaning q is also even. But if both p and q are even, they are not co-prime, which contradicts our initial assumption.
I know this proof is tough to memorise, beta. I remember my son sitting right where you are, mumbling these steps like a prayer before his board exams, while I packed bhujia into little packets. His focus was on going abroad, and I knew his maths marks had to be perfect to get a scholarship. That’s why this proof matters—not for the marks, but for the future it can help build.
#exams#study
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