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The given statement is "If n is an even integer, then 2n is also an even integer".

First we tried to prove it with the direct proof. let's assume that n is an even integer that,

n = 2K, where K belongs to integers

Therefore, by substituting n = 2K in 2n,

2n = 2 x 2K

2n = 4K

2n = 2(2K)

2n = 2L (Since K is an integer, we can rewrite 2K to L by using the closure property)

Therefore, the given statement is correct by using the direct proof.

Now, it is the time to try with the contrapositive proof. According to the form of contrapositive, p -> q can be simply rewritten to

!q -> !p

So, in this case, let's say p = "n is an even integer" and q = "2n is an even integer". If we rewrite the statement, "if p then q", to !q -> !p, it is going to be "if not q then not p".

But, "not q" can be stated as "2n is not an even integer". Since 2n is already an even by the rules of Mathematics (an integer multiply by 2 is going to be even), I don't understand what is wrong with the contrapositive statement.

I'm sorry if my explanation is a bit long and contains mistakes.

gist076923
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Woshi
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1 Answers1

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There is nothing wrong with the contrapositive: if $2n$ is not even then $n$ is not even. According to the laws of Boolean logic, the result of implication is always true if the antecedent is false. In other words, if we start with a false premise, the conclusion can be anything (i.e. if pigs can fly so can I).

Vasili
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