Discrete Math 1.7.3 Proof by Contradiction
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- เผยแพร่เมื่อ 28 ส.ค. 2024
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The full playlist for Discrete Math I (Rosen, Discrete Mathematics and Its Applications, 7e) can be found at • Discrete Math I (Entir...
For the second example, why did you assume n was odd? If n being even was the ‘q’, I thought we were assuming ‘not p’ instead of ‘not q,’ which is what you did.
I think my initial slide is unclear. If we have just one statement, like radical 2 is irrational, we assume it is false and prove by contradiction that it must be true. If we have a conditional statement like example 2, then we assume if p then not q and prove by contradiction that q must be true.
For 3n+2 case of proof you could use contrapositive or contradiction, both look similar as in you assumed n is not what is expected so assumption is getting violated. My question is how to know which type of proof to apply for a case and what are differences between the above 2 ways . Thanks , appreciate your teaching very much !!
cshivani There is no ‘right’ method for proof, so just go with what you are comfortable with and what requires the least amount of work!
For the second example... Would it still be considered correct if I used this working...
assume n is odd such that n = 2c + 1 where c is any integer. So 3n+2 = 6c+5 and is even. We know even = odd + odd, so 6c must be odd, which is a contradiction...
A reply would be much appreciated!
excellent
Tank you kim..... ❤️❤️❤️❤️❤️
Hi what software diid you use ? did you use surface pro ?
Tnt 1111 I use Power Point and the new ‘Recording’ tab on my Surface Pro. I then export the video.
thank you BTW nice channel ;)
Hi Kim, how do you solve a proof problem with log format.
sam nwabuaso The trick to proof is knowing the theorems and definitions surrounding what you are trying to prove. So to do a proof involving logarithms, you’d want to know any definitions and theorems regarding logarithms.
Yes exactly because to me its complicated, but I was asked to prove if the logarithm was irrational. Example log 6 base 4
can you please tell me why a must be even
a^2 is even because it is equal to 2b^2. 2b^2 complies with the definition of an even number (2k, k for any number). Also because a square of an even number is even and the square of an odd number is odd, a = even.
Thanks!
3n+2 is even then n is even
assume n is odd
n=2n-1
3(2n-1)=-2
6n-3=-2
6n=1
This is false, therefore by proof of contradiction, n has to be even
wouldnt this work better and is simpler?
We can't say 3n+2 is even. We use 2n to be even because whether a value is even or odd, if you multiply it by 2 it is even. We can't say that about 3n+2. If n is odd, then 3n+2 is odd, but if n is even, then 3n+2 is even.
the contradiction is not in n is even or odd
the contradiction is n=2k-1, k element of Z, substituting on 3n+2=3(2k-1)+2=6k-3+2=6k-1 so 6k-1 is odd you proved that 3n+2=6k-1 in other words an odd number, but the contradiction is that you asummed that is even tho
you got a contradiction too, for k= even 2k-1 it gives an odd number is better to use 2k+1 for odd and 2k for evven