[Discrete Mathematics] Direct Proofs Examples
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- เผยแพร่เมื่อ 4 พ.ค. 2016
- In this video we tackle a divisbility proof and then prove that all integers are the difference of two squares.
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Writing out each step more clearly (as most professors require) would improve your discrete tutorials. I appreciate your effort
what program do you use to make these tutorials??
How do you know to go down a path showing 5j=2a and then j=2k ? Why not one of the other paths I'm sure exists? Is it just practice that makes the path to the final proof more easy to find?
Fun video. Well done!
Prove using direct proof :
a.) If n is even and m is odd, then the product of nm is always even.
b.) Prove that, for any odd integer n , the number 2n2+5n+4 must be odd
For proof that 7 divides a, can we say that j = 4k since 4 is a multiple of 4, and therefore still true? Or could we not do this because not every even number can be written as 4k since numbers like 2 or 6 would require k to not be an integer?
Also, do you have videos on the symbols you use? I'm a bit unclear on what they mean and how to use them?
I'm 2 years late, but for those who will be reading these, yes you can do it
Let's say 7|4a
7j=4a
Since 4a is divisible by 4, there 7j must also be divisible by 4
This means that j must be divisible by 4, and by that we should use j=4k
I am 3 years late but yeah, I've also been having problems with using 2k but then I realized 4k also serves the same purpose.
Assume 7|4a where every a is an integer
7j = 4a. Because 7 is odd then j needs to be even (j = 2k).
7(2k) = 4a. Simplify
7k = 2a. Because 7 is odd then j needs to be even (k = 2r).
7(2r) = 2a. Simplify
7r = a
7|a
I do not grasp why if 7r=a, then 7 is divisible by a.
It is a bit tricky, but true. The reason is because we know that r is a whole integer because we put j there to begin with to represent a whole integer and we eventually just transformed it to r. I am not sure how to explain it because I barely understand it myself.
I used j = 4k. So does that mean the original claim is valid?
Same here, ever figure to out?
That is the definition of integer division. If a*b=c then both a and b are dividers of c. Say a=7 and b=3, then c is 21. In that case 21/3 must be 7 and 21/7 must be 3.
Can you do it this way??
If 7|4a then there exists a number k (in the set of positive integers) such that 7k=4a;
If 7|a then there exists a number j (in the set of positive integers) such that 7j=a;
Then substituting a= 7j into 7k=4a,
7k=4(7j), then 7k=28j, which is k=4j.
If we substitute the value of k into 7k=4a, then 7(4j)=4a, which is 28j=4a
Thus 7j=a, proving 7|a.
Assume 7|4a.
Then 7j = 4a, and 4a/7 = j.
For 7|a, then it must be that a/7 = j/4.
For j to be an element of the positive integers, it must be that j is a multiple of 4.
That's an interesting way of proving it.
I guess you could also go from 7j=4a --> 7(j/4)=a,
which is of the form 7k=a, proving that 7|a as well.
I guess it's cool that there are so many ways of proving things.
Does anyone know what software he uses? It looks so clean I need it!
nah
Question: I enjoyed how you used a few examples to see a pattern and create a formula for such. My question is this: since I am not taking discrete mathematics currently (I am studying this independently), would I be allowed to do such thing in a formal proof? I feel like probably not, but how would you go about this if not? And how would I explain what I was doing? I am a physics major, and yet I'm required to still do proofs in physics, so I need to know for this semester.
You'd have to prove the formula holds using mathematical induction, generally.
Thank you!
Also a few examples of negative integers for the theorem "Every odd integer is a difference of two squares.":
2k + 1 = (k + 1)² - k²
-------------------------------------
1 = 1² - 0²
3 = 2² - 1²
5 = 3² - 2²
-1 = (0)² - (-1)² = 0² - 1²
-3 = (-1)² - (-2)² = 1² - 2²
-5 = (-2)² - (-3)² = 2² - 3²
✔
In the Divisibility Examples video you showed a counter example to disprove a question similar to this. If a=2 wouldn't that be a counter example to 5|4?
Yeah but the proposition is, if 5|2a than 5|a . 5|4 is not true, so you can't put in a=2
Shouldn't it be 5/j = 2a?
I don't get where the j come from , it suddenly appears !!
Watch the Divisibility video...
if you assign 5/2a to a number (j in this case) such that j = 5/2a (this reads j equals 5 divided by 2a), then you rearrange the both sides into 5j =2a
Prove 7|4a --> 7|a:
1. Assume 7|4a
2. 7j=4a=2(2a) [definition of divisibility: some j exists where 7j=4a; rewrite 4a as 2(2a) to demonstrate that it's even]
3. j=2k [j must be even to make 7j even because the 2 sides are equal, so j=2k for some k]
4. 7(2k)=2(2a) [substitute 2k for j]
5. 7k=2a [divide both sides by 2; now we can do steps 3-5 again essentially]
6. k=2m [LIKE STEP 3: k must be even to make 7k even, so declare m=2k for some m]
7. 7(2m)=2a [LIKE STEP 4: sub 2m for k]
8. 7m=a [LIKE STEP 5: divide both sides by 2]
9. 7|a [by definition of divisibility: some m exists where 7m=a, meaning 7|a]
W
@@itsnee lol i don't even remember this comment, the class was so long ago, but it looks right. Does w mean what? Does something seem off or unclear in my answer?
if 7/4a then7/a
Assume: 7j = 4a , for some intager j
j = 4k , for some intager k
so,
7(4k) = 4a
7k = a
=7/a
Why does j = 4k? I'm having a difficult time with these.. In the proof he does as an example, he uses the definition of odd/even as the next step and substitutes. So I'm curious where the 4k comes from! :P Thanks in advance
fuck math
@@drewrodrigues pretty much a number, lets say b is even if b = 2(a), a is an integer, but if thats true, then b = 4g, g is an integer too.
@@sakitoido If we have to describe 2 as 4(g), then g will be 0.5, which is not an integer.... i am confused
WHY ARE WE USING 4K
Ure voice💞
bruh
why do we assume that 5j=2a ?
That's what I'm confused about. Why does he need to use the 'j'.
How does 5j=2a?
that’s what I’m confused about too
Because 5 devides 2a which means 5 times some number is equal to 2a
the antecedent is clearly wrong since it's not the that case "all" even numbers are multiples of 5 (easiest counterexample being there's no integer j s.t. 5j=2)...but if you start with a wrong antecedent then you implication is always going to be true (F⇒F ∨ F⇒T) so I guess we can just move on...
Alguien más de la facultad de ciencias de la unam jajaja
why did you make j even?
ok..but why did he use?
Because if we assume 5j = 2a, and 5 is an odd number, we must assume j is an even number. This is because 5j must be even in order to equal 2a, which is an even number. Therefore, 5(2k) is an even number.
how do you get 5/a? i see 5/a=1/k
that's because K didn't exist from the begain he assumed it
I think j is a number which can be a fraction (1/2) or integer(1,2,3) so we denote that as J
It looks like 5/a, but it's actually 5|a which is: 5 divides into a, which means a is divisible by 5 (a/5 = no remainder)
@@boudyhesham5875 so the k only exists to prove that 5 can be divided by some integer? is there some basic definition that i'm missing?
we dont know if the squares are consecutive ! so its K+n not 1
I already lost when you said 5j
May you kindly help me with..... Prove that 7 is an odd number
But 5/a = 1/k... am lost
69 !!!
I am getting 2a/7
I think since that is 2(a/7) then that means 7|a, but idk
This fool said 69
You start off wrong. 5 does not 2a for EVERY a in Z. For example 5 does not divide 2a when a is 3. So the for all sign is wrong.
Lol. 69
Ughh
where the hell did the j come from ur imagination? .....
if a|b then then ac=b for some c. Since 5|2a that means that 5b=2a for some b. I just picked j since j,k,l,m,n are common variables in number theory.
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These examples are not helpful. You're skipping steps and not explaining your reasoning at all.
Shouldn't it be 5/j = 2a?
Thats what Im saying!