Prove that if x2 is an even integer, then x is an even integer by contraposition method.
Answers
Answer:
Let x be an integer,
To prove: If x² is even, then x is even.
Here we have to prove this by contra position method.
Let contra positive of the above statement is given by
"If x is not even, then x² is not even"
This statement can be proven as follows.
if x is not even then only possibility is x is odd.
We know that the product of two odd numbers is odd,
therefore x² = x * x is odd.
Therefore x² is not even.
Having proved the contra-positive, we infer the original statement
Answer:
The given is tha ∀x, x² is an even integer then x is an even integer
The contrapostive method is method in which we take the given statement and take the opposite of it and prove it to be true and by that result we deduct that the given statement to us is also right
Step 1. Form the contrapositive of the given statement. That is,
For all integers n, if n is not even, then n² is not even
But, we know that an integer is not even if, and only if, it is odd [by parity property]. So, the contrapositive becomes
For all integer n, if n is odd, then n² is odd
Step 2. Now prove the contrapositive using method of direct proof:
Suppose n is [particular but arbitrarily chosen] integer. [We must show that n² is also odd.] By definition of odd, we have
n = 2k + 1 for some integer k.
Then by substitution, we have
n . n = (2k + 1) . (2k + 1)
= 4k² + 2k + 2k + 1
= 2(2k² + 2k) + 1
Now ((2k2 + 2k) is an integer. [Because products and sums of integers are integers and 2 and k are both integers.] Hence, we have a form:
n . n = 2 . (some integer) + 1
or n² = 2 . (some integer) + 1
and so by definition of odd is n² odd.
Step 3. Therefore, the given statement is true by the logical equivalence between a statement and its contrapositve.