Question

Can anyone provide some basic proof guidance? Any tips on determining when I should use direct, contraposition, or contradiction proofs?

Answer 1

Mind you, there is no simple way of determining which type of proof to use, and maybe some statements may be proved in multiple ways...

If you are given plenty of conditions, and conditionals, it probably calls for a direct proof.

If you are given something that seems so obvious, but uncannily difficult to prove directly, then you should probably use proof by contraposition.

And if you're given hardly anything to work with and just given something to prove, maybe contradiction is the way to go.

Answer 2

the basic idea i got so far was. "Start with direct. If it looks like it starts more complicated then it ends use contraposition. if that does not work then do contradiction."

Answer 3

Here's a nice example :)
Let k be an integer.

Show that if k² is odd, then k is odd.

Answer 4

see that i think would be contraposition because it starts with k^2 which seems more complex then k

Answer 5

Yeah... try proving that directly, though... it's a nightmare, and probably impossible.
Let k² be odd
then
k² = 2h + 1, where m is an integer
Then you have to show that
sqrt(2h + 1) is also odd
which is
well, there's no way to do that directly, that's the gist of it XD

Answer 6

ah ok i can see that
how about seeing that question and knowing to not to contradiction?

Answer 7

the only real example i have seen when it did not explicitly say "use contradiction" was proving the sqrt(2) is irrational

Answer 8

It can be proved by contradiction, but it's near identical to the contraposition proof, anyway... you want me to demonstrate it?

Answer 9

please

Answer 10

i wrote the contraposition one so you dont need to do that one

Answer 11

Let k be an even int. k=2n for some int n. k^2 = 4n^2 which by definition is even. therefore by proof of contraposition if k^2 is odd then k is odd.

Answer 12

To prove by contradiction, you assume the conditions then negate the consequence.
Suppose
k² is odd

Now suppose k is even.

Then k = 2h , where h is an integer.
Then k² = 4h²
but 4h² = 2(2h²)
Since 2 and h are integers, so is 2h²

Thus k² = (2h)² = 4h² is equal to twice an integer.
Thus k² is even, which contradicts our assumption.

Thus k² must have been odd.

Answer 13

See how identical it is to your contraposition proof?
Except the contraposition proof is shorter, so that's preferable.

Answer 14

ahhh ok i see. So is that common? That if I can do contraposition then i could do contradiction and visa versa?

Answer 15

Well, if you can do contraposition, surely you can do contradiction... Let's put it generally...
Say you can prove
p --> q
by contraposition. Then that means you can prove that
~q --> ~p

Then suppose you forget that and try to proceed by contradiction.
So you assume
p

And then you assume
~q

But you can show that if ~q, you can arrive at ~p, since as I said, you can prove it by contraposition.

So you'll arrive at ~p despite starting with p
And there's your contradiction
~p and p

LOL I PROVED THAT YOU CAN PROVE...
PROOFCEPTION!!!!

XD

Answer 16

lol alright nice. thank you i really appreciate your help. i think i getting it. Just a little stressful to do it on a 10 min quiz lol. Im worried if I start the wrong way I wont have time to fix it.

Answer 17

Takes practice... but as for the other way around, that is to say, if you can prove something by contradiction, then you can prove it by contraposition, no, because not everything you prove is in conditional form...
Like this rather stressful

Prove 1 > 0

Answer 18

lol what? that seems like a definition more then something you would prove

Answer 19

Yeah, but you can prove it on the assumption that there are infinitely many real numbers.

Answer 20

You want me to do it? XD

Answer 21

sure if you want lol. It couldn't hurt me any

Answer 22

Hehehe... I don't even think there's any other way to prove this BUT using contradiction...

So
Suppose 1 = 0, and let a be a real number.
1a = a, by identity property of multiplication
0a = 0, by zero property of multiplication

Since 1 = 0
1a = 0a , by multiplication property of equality.
a = 0, as assumed above.

Thus proving that all real numbers are equal to 0, which is a contradiction, as we assumed that there are infinitely many real numbers.

This it is not true that 1 = 0.

Answer 23

Now
Suppose 1 < 0
Then 1 is negative.
By the multiplication property of inequality, multiplying a negative number on both sides of the inequality would reverse the inequality, so
1(1) > 0(1)
1(1) = 1, by identity property of multiplication
0(1) = 0, by zero property of multiplication
Therefore
1 > 0

Which is a contradiction, as we assumed 1 < 0.

Thus, it is not true that 1 < 0

Answer 24

Now, we have shown that it is not true that
1 = 0
and
1 < 0

By the trichotomy property, it must be the case that
1 > 0

QED

Had fun with that? LOL

Answer 25

Believe me, it's the obvious things that are very very hard to prove.

Answer 26

ya i bet. alright man im gunna get back to studying. Thank you for your help with this. Im feeling a little better about it

Answer 27

No problem :)