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Proof by contradiction question (How to negate)

The question says ‘Prove by contradiction that if n^2 is an even integer then n is also an even integer’, what would the assumption be that you make? (The negation) and how you would do the question,

Thanks!
Reply 1
Avatar for zxvs
zxvs
16
I think it'd be like "Assume there exists an even number n^2 such that n is odd."
is n^2 is an even integer, then n is an odd number
Reply 3
Avatar for fgcse2018
fgcse2018
OP
9
Original post by zxvs
I think it'd be like "Assume there exists an even number n^2 such that n is also odd."


So, do you not negate the ‘integer’ part to ‘fraction’? Or do you just leave it
Reply 4
Avatar for zxvs
zxvs
16
Original post by fgcse2018
So, do you not negate the ‘integer’ part to ‘fraction’? Or do you just leave it


I don't quite follow, sorry. I assume you're talking about proving rational numbers (which uses fractions)?
Reply 5
Avatar for fgcse2018
fgcse2018
OP
9
86289450-47D3-43BD-ABA7-E28BE0DD26B2.jpg.jpeg just part a
Original post by zxvs
I don't quite follow, sorry. I assume you're talking about proving rational numbers (which uses fractions)?


CCDB75C2-BD5F-4C4E-A207-00F0425E7070.jpg.jpeg
Just part a
Original post by fgcse2018
The question says ‘Prove by contradiction that if n^2 is an even integer then n is also an even integer’, what would the assumption be that you make? (The negation) and how you would do the question,

Thanks!


You literally just say:

"Assume, on the contrary, that if n2n^2 is even then nn is not even"

and work to a contradiction from there.
Reply 7
Avatar for zxvs
zxvs
16
I think my post got deleted lol whoops
Assume that there exists an even n2n^{2} where nn is odd

Spoiler

Original post by zxvs
I think it'd be like "Assume there exists an even number n^2 such that n is odd."

Original post by JackMac2904
Assume that there exists an even n2n^{2} where nn is odd
Both these forms are correct. Note that they only require you show the existence of one case where n is odd.

Statements of the form "if n^2 is even, n is odd" are incorrect negations; the way they are phrased, they require every case to be a counterexample, which is wrong.

To show why this distinction is important, suppose we had the statement "if n is prime, n is odd". This is false (n=2).

The correct negation says "there exists n prime with n even", which is true (n = 2 again).

The incorrect negation would be "if n is prime, n is even", which is false (take n = 3, 5, 7, ...).

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