Just to clarify for anyone about to take the test and to not make a mistake like I did, zero is not considered as even or odd is wrong. Zero is considered as an even number.
yes! Even numbers are 2k where k⊆Z odd numbers are 2k + 1 where k⊆Z I can get 0 as a result if I fill in k=0 in 2k --> 2*0=0 and 0⊆Z I can't get 0 as a result from 2k + 1 where k⊆Z You can only get 0 if you fill in k=-0.5 since 2(-0.5) + 1 = 0 but -0.5⊆Z is wrong!!
Just putting a personal note here for future reference. The last example is a proof by contraposition and cases. So we can show the overall statement is true if the conclusion in that implication is true given we take into account the implication that not r is true i.e. both x and y are not even Also note that "both x and y are not even" could mean that "both are odd" and also "either x or y are odd". That's why we use the proof by cases to take all into account. Before using Without Loss of Generality to conclude that we can assume x to be odd. Again, we can do this because it doesn't matter in this context if "x is odd" or "y is odd", because the result will be the same.
I like "Discrete Mathematics and It's Applications" by Rosen and "Discrete and Combinatorial Mathematics" by Grimaldi. I use the first with my Discrete Math I course and the second with my Combinatorics course, which is essentially Discrete Math II. You can find my Combinatorics playlist at czcams.com/play/PLl-gb0E4MII1_QX6h6TzMW3rF_7Taapyd.html
Hello ma'am! May I ask if proof by cases is the same as the choose method? I’m confused about the two. The same goes for the Construction method and direct proofs. Are those two the same as well? Please enlighten me. Thanks.
At the end when you said this example made use of contraposition and proof by contradiction did you mean contrapositions and proof by cases? I don't see and proof by contradiction in that problem.
Sorry, could you clarify why you didn't need to do case 4 for the last question? (If x and y are integers and both xy and x+y are even, then both x and y are even")
Because I chose a proof by contraposition, I need to assume "not q". Based on the question, "q" would be that x and y are both even., so I have to assume that is not the case and then prove that "not q" implies "not p" so p implies q. If you are struggling, you might go back a few videos to the Proof by Contraposition video for some additional examples.
Hi!! Thanks a lot for this video it was very helpful, but i got a question, we negated R which is 'both X and Y are even', so the negation of this means 'both X and Y are odd' so why did we have multiple cases instead of just the case where both x and y are odd?
We know that the definition of an even integer is that it is an integer divisible by 2 evenly. So we represent an even integer by 2x where x is an integer in the same way we represent an odd by 2x+1 since an odd is an integer divisible by 2 with a remainder of 1.
Just to clarify for anyone about to take the test and to not make a mistake like I did, zero is not considered as even or odd is wrong. Zero is considered as an even number.
yes! Even numbers are 2k where k⊆Z
odd numbers are 2k + 1 where k⊆Z
I can get 0 as a result if
I fill in k=0 in 2k --> 2*0=0 and 0⊆Z
I can't get 0 as a result from 2k + 1 where k⊆Z
You can only get 0 if you fill in k=-0.5 since 2(-0.5) + 1 = 0
but -0.5⊆Z is wrong!!
Your lectures helped me greatly with my discrete maths preps. Thanks for existing ❤️
Thank you so much! I am self studying proofs and linear algebra, and these videos are incredible. The quality of instruction here is incredible!
5:40 zero is even, an even number is defined as any number divisible by 2 aka it can be formed as 2k for k an integer
which is the case for zero
Just putting a personal note here for future reference. The last example is a proof by contraposition and cases. So we can show the overall statement is true if the conclusion in that implication is true given we take into account the implication that not r is true i.e. both x and y are not even
Also note that "both x and y are not even" could mean that "both are odd" and also "either x or y are odd". That's why we use the proof by cases to take all into account. Before using Without Loss of Generality to conclude that we can assume x to be odd. Again, we can do this because it doesn't matter in this context if "x is odd" or "y is odd", because the result will be the same.
Hi! Just to confirm, in the end, you say it is a proof by contradiction and contraposition. Did you mean contraposition and cases?
Yes I did. Thanks for letting me know!
Like always, great content mom. Do you suggest any good book on the subject ?
I like "Discrete Mathematics and It's Applications" by Rosen and "Discrete and Combinatorial Mathematics" by Grimaldi. I use the first with my Discrete Math I course and the second with my Combinatorics course, which is essentially Discrete Math II. You can find my Combinatorics playlist at czcams.com/play/PLl-gb0E4MII1_QX6h6TzMW3rF_7Taapyd.html
Hello ma'am! May I ask if proof by cases is the same as the choose method? I’m confused about the two. The same goes for the Construction method and direct proofs. Are those two the same as well? Please enlighten me. Thanks.
At the end when you said this example made use of contraposition and proof by contradiction did you mean contrapositions and proof by cases? I don't see and proof by contradiction in that problem.
Correct. A proof by cases and contraposition.
3:55 why can you multiply both sides with n as a proof for case 3 but not for case 1? Case 1 would give n^2
hi, doing it by the direct method, how did the demonstration had been
Isn't 0 even?
yes it is even
no
It's ~0
Sorry, could you clarify why you didn't need to do case 4 for the last question? (If x and y are integers and both xy and x+y are even, then both x and y are even")
Because I chose a proof by contraposition, I need to assume "not q". Based on the question, "q" would be that x and y are both even., so I have to assume that is not the case and then prove that "not q" implies "not p" so p implies q. If you are struggling, you might go back a few videos to the Proof by Contraposition video for some additional examples.
Hi!! Thanks a lot for this video it was very helpful, but i got a question, we negated R which is 'both X and Y are even', so the negation of this means 'both X and Y are odd' so why did we have multiple cases instead of just the case where both x and y are odd?
The negation is NOT both X and Y are even. So it could be that they are both odd or that either X or Y are even.
is there is a proof by induction video?
Yes. I have two of them. Just go to my page and do a search
thank u
Welcome
how do we know that 2r and 2L is even based on the algebraic equation? 9:10
We know that the definition of an even integer is that it is an integer divisible by 2 evenly. So we represent an even integer by 2x where x is an integer in the same way we represent an odd by 2x+1 since an odd is an integer divisible by 2 with a remainder of 1.