Download √ 2 IS IRRATIONAL Recall the well ordering principle: Every non

√
2 IS IRRATIONAL
Recall the well ordering principle:
Every non-empty subset of N contains a least element.
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We may use this to prove that 2 is irrational.
Theorem.
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2 is irrational.
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• We argue by way of contradiction and suppose that 2 is rational.
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• We may then write 2 = a/b, where a and b are integers and b 6= 0. We may also
choose this expression so that a and b are both positive. (Why?)
Proof.
• Consider the set
√
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S = {k 2 | k and k 2 are positive integers}.
By its very construction, X is a subset of N.
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• We also know that a = b 2 lies in X, so X is non-empty.
• Since X is a non-empty subset of N, the
√ well ordering principle tells us that X
contains a least element. Call it s = t 2, where t is a positive integer.
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• Now consider the number s 2 − s. I claim that s 2 − s also lies in X.
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√
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• Note that s 2 − s = s 2 − t 2 = (s − t) 2.
• Because s lies in X, s is an integer. But so is t. It follows that s − t is also an
integer.
• We have
√
√
s − t = t 2 − t = ( 2 − 1)t > 0
√
2 > 1 and t > 0. Therefore s − t is a positive integer.
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• Moreover, we have s 2 − s = 2t − s, so s 2 − s is also an integer.
since
• What’s more,
√
√
s 2 − s = ( 2 − 1)s > 0,
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again since 2 > 1 and s > 0. It follows that s 2 − s is a positive integer.
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2 IS IRRATIONAL
2
√
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• In summary, we have shown that s 2 − s = (s − t) 2 is a positive √
integer with
the property that s − t is also a positive integer. This means that s 2 − s lies
in X.
• Now notice that
√
√
√
s − (s 2 − s) = 2s − s 2 = (2 − 2)s > 0
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since 2 > 2 and s > 0. It follows that s 2 − s < s.
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• Therefore s 2 − s is an element of X that is less than s. This
contradicts the fact that s is supposed to be a least element in X.
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• It follows that our original assumption must have been wrong, that is, 2 must
be a rational number. √
Remark. There is a more standard proof that 2 is irrational that uses the fact that
every positive integer can be expressed uniquely as the product of prime numbers. The
latter statement is called the fundamental theorem of arithmetic. We will prove the
FTA later in this course, but in the meantime, observe that the above proof only used
very elementary ideas.