The mistake in the “proof” is that the same symbol, k,
is used to represent two different quantities. By setting
m = 2k and n = 2k + 1, the proof implies that n = m + 1,
and thus it deduces the conclusion only for this one situation.
When m = 4 and n = 17, for instance, the computations
in the proof indicate that n − m = 1, but actually
n − m = 13. In other words, the proof does not deduce the
conclusion for an arbitrarily chosen even integer m and odd
integer n, and hence it is invalid.
The mistake in the “proof” is that the same symbol, k,is used to represent two different quantities. By settingm = 2k and n = 2k + 1, the proof implies that n = m + 1,and thus it deduces the conclusion only for this one situation.When m = 4 and n = 17, for instance, the computationsin the proof indicate that n − m = 1, but actuallyn − m = 13. In other words, the proof does not deduce theconclusion for an arbitrarily chosen even integer m and oddinteger n, and hence it is invalid.
การแปล กรุณารอสักครู่..
