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**Unformatted text preview: **2/22/2020 Proof by Contradiction Proof by Contraposition Examples Example 1: Prove the following statement by contraposition:
If a product of two positive real numbers is greater than 100, then at least one of the number is
greater than 10. First, translate given statement from informal to formal language:
∀ positive real number r and s, if (r . s) > 100, then r > 10 or s > 10. Proof:
First, form the contrapositive of the given statement. That is,
∀ positive real number r and s, if r ≤ 10 and s ≤ 10, then (r. s) ≤ 100.
Now suppose r and s are positive real numbers and r ≤ 10 and s ≤ 10. Then, Since r ≤ 10 Multiply both side by s, we get r . s ≤ 10 . s ______________________(1) and since s ≤ 10 Multiply both sides by 10, we get 10 . s ≤ 10 . 10 i.e. 10 . s ≤ 100 ______________________(2)
Since, ' ≤ ' holds transitivity property, therefore from equation 1 and equation 2, we get
r . s ≤ 100
But this is what was to be shown.
This completes the proof. 1/3 2/22/2020 Proof by Contradiction Example 2: Prove the following statement by contraposition:
The negative of any irrational number is irrational.
First, translate given statement from informal to formal language:
∀ real numbers x, if x is irrational, then −x is irrational. Proof:
Form the contrapositive of the given statement. That is,
∀ real numbers x, if −x is not irrational, then x is not irrational.
Equivalently [because −(−x) = x],
∀ real numbers x, if x is rational, then −x is rational.
Now, we prove the contrapositive of the given statement using direct method of proof. Suppose x is
an [particular but arbitrarily chosen] rational number. [we must show that −x is also rational.] By
deﬁnition of rational, we have x = a/b for some integers a and b with b ≠ 0.
Then, x = −(a/b) = (−a)/b
Since both −a and b are integers with b ≠ 0, −x is rational [as was to be shown.]
This completes the proof. Example 3: Prove the following statement by contraposition:
For all integers n, if n2 is odd, then n is odd.
Proof:
Form the contrapositive of the given statement. That is,
For all integers n, if n is not odd, then n2 is not odd.
But, from the parity property, we know that an integer is not odd if, and only if, it is even. So, the
contrapositive statement becomes
2/3 2/22/2020 Proof by Contradiction For all integers n, if n is even, then n2 is even.
Now, we prove the contrapositive statement using the method of direct proof. Suppose n is
[particular but arbitrarily chosen] integer. [We must show that n2 is also even.] By deﬁnition of even,
we have n = 2k for some integer k So, by substitution we have n . n = (2k) . (2k)
= 2 (2.k.k)
Now (2.k.k) is an integer because products of integers are integer; and 2 and k are integers. Hence, n . n = 2 . (some integer)
or n2 = 2. (some integer)
and so by deﬁnition of even, n2 is even. Hence, the given statement is true by the logical
equivalence between a statement and its contrapositive. 3/3 ...

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- Spring '17
- Dr. David Tweedle