Thank you so much for posting these videos. I am watching every single one of your series, it is extremely helpful and actually useful. Kicks my textbook's ass. And my teachers. Please keep exploring the world of electronics by posting videos. We need them.
Hey man, I've been watching your videos for about 2 weeks now, I have an exam this Friday and I'm happy to have found someone who could explain this thoroughly, I haven't found anyone on TH-cam who has done this way, I finally understand boolean algebra, thanks.
thanks for the video tutorial vladimiar. your so good about making it simple to explain the boolean algebra method... thanks a lot. you make the method easier to understand good job sir
Sarah It's because anything multiplied one still equals itself: 3*1=3 0*1=0 A*1=A So A+A*B can be written as A*1+A*B A can then be factored out leaving A(1+B) If that still doesn't make sense, distribute A to both terms in the parentheses
Braedon O thank you teachers feel the need to prove their intelligence so they over complicate things when they are talking to someone with the brain the size of a peanut. They could be talking to someone with down syndrome and pull out big words and numbers out their ass
@Supermanu15 say, if you multiply A by (1+B): first you multiply A by 1 then A by B and sum them up - you got A*1 + AB witch is A + AB that's why A(1 + B) = A + AB and vice versa :-) when I say "taking A out of brackets" I mean that A in A+AB is a common factor (we have A in both terms - in A which is same as A*1, and in AB). So if we it's in both terms separately (A*1 + AB) or together A(1+B) the result is the same. Hope it's clear. Please ask me more questions if something is not clear :-)
I think it should be 1. because A+(AA)negated will result in A+(A'+A') this gives you A+A'(1+A') then because 1+A' =1 (by law nullity) then you are left with A+A'(1) which is =A+A' and A+A'=1 by law of complement. thanks
Hello and thank you very much for this video its very easy to understand. I dont have a problem with A(1+B) for example but what if you have AB + A'B I saw that it works like A(B+'B) but is that a rule as a whole when you have another letter after the initial (in this case A) and only if there are repeating letters like in this case A and A and B or B. I guess what I am trying to ask is in which cases the expression can be simplified in this way? Thanks again
In Addition if one of the values equals 1 than the result always eqals 1 and in multiplication if both eqal 1 then the result is 1 other wise the result is 0...
Wait wait wait. The first equation A+AB, how did u assume the first A Is multiplied by 1 making the second part (1×B)?? Is this a common rule or is it the "something"?
HI, I WANT TO KNOW IF U LEAVE AN EXPRESSION LIKE THIS; NOT A+A NOT B NOT C, IS IT SIMPLIFIED? OR CAN IT BE SIMPLIFIED FURTHER BECAUSE THERE IS A and NOT A?
You have a mistake in your exercise , i guess.. at 6.12 you say -> A + (not)AA = A (1+(not)A) but in the previous minute you said , (not)A . A is 0 so wouldn't that be A + (not)A.A = A + 0 = A
How we know that A variable and B variable don't have the same value? Or better said. How we know that A != B ? Do we assume their values are different because they have different names?
Sarah A can have the value of 1 or 0 only, when you see A or B (any variable) it will have the value of 1. When the you see A' (or in the exam it might be a line on top of the variable) that means it is negated which is the opposite of A. This means the opposite would be 0. So A = 1, A' = 0.
(1+x') means 1 OR inverse x, and of course we know that when OR is being used, we need either one or both sides of the OR to be 1 for the output to be 1. If x = 1 then x' = 0 and 1 + 0 = 1. if x = 0 then x' = 1 and 1 + 1 = 1, hence 1 + x' will always equal 1 no matter the value of x or x' hope this was helpful
You're so much clearer than my professor! Thank you.
Thank you so much for posting these videos. I am watching every single one of your series, it is extremely helpful and actually useful. Kicks my textbook's ass. And my teachers.
Please keep exploring the world of electronics by posting videos. We need them.
Thanks Vlad for the tutorial finally you explained it in "dummie" terms so I could understand it finally! Subbed.
Hey man, I've been watching your videos for about 2 weeks now, I have an exam this Friday and I'm happy to have found someone who could explain this thoroughly, I haven't found anyone on TH-cam who has done this way, I finally understand boolean algebra, thanks.
did you pass the exam BRO!?
11 years later and I found u to be my savior
btw I'm majoring in computer science
Have an exam tomorrow. My professor actually recommended over viewing this series. Fingers crossed haha
thanks :)
I have exam in one hour :P
Haha
So do i lol
How did you do in yours ? Mine is in 4 hours.
he failed
How'd it go :)
This is the first time I've ever actually GOT this. Thanks so much for posting these!
fax this guy is the goat
I don’t know normal algebra very well, only really focused until 8th grade but I got the hang of these basic problems very easily due to your teaching
I REALLY UNDERSTOOD BOOLEAN WELL BECAUSE OF THESE VIDEOS.......
THANK YOU
I'm so happy you posted this, this saved my life for my test tomorrow :)
sameeee
You sound like Russian..well i have exam tomorrow and if i pass and if someday if we met then my friend the Vodka is on me
😀😀😆😝
Did u pass? Lol
Shit I wanna know too
How was it!? 😂
He left us on a cliffhanger
wow this was insanely helpful i had no idea what my teacher was talking about today and now I get it after this subbed and liked!
Awesome man! You're making so much more sense than my teacher!
Woahhh ,your video make it easier to understand... Thanks !
At 2:35 he says he's taking A out of brackets, but there are no brackets.
the whole thing is a bracket
You explain boolean algebra well. Thank you a lot
thanks for the video tutorial vladimiar. your so good about making it simple to explain the boolean algebra method... thanks a lot. you make the method easier to understand good job sir
greatest video of all time thank you
Great Video, A lot of thanks for a simple way to understand this algebra.
thank you vladimir, really pulling through 10 years later
The problem at 8:15. Did you use absorbtive law to get rid of that second A, or what that factoring? Or?
A'+BA'= A'(1+B)
So, the answer is: A'
Yes
Thanks a lot. Much appreciated
Wooow...this totally Helped Me. Thanks A Lot.
i dint understand dis chapter in college...but u teached me lyk a game!!!
How do I know that A is actually multiplied by one and I can take it out of brackets?
Sarah It's because anything multiplied one still equals itself: 3*1=3 0*1=0 A*1=A
So A+A*B can be written as A*1+A*B
A can then be factored out leaving A(1+B)
If that still doesn't make sense, distribute A to both terms in the parentheses
This whole time it wasn't making sense when my teacher was explaining it.. but the light just clicked on when Vladimir was explaining it.
A+0=A, A+A=A, A×A=A so why would I specifically choose A+1=A??
+Sheldon Cooper its multiplication (AND operator) not addition (OR operator). A*1 = A every time. if A=1 then 1*1=1 if A = 0 then 0*1=0
Braedon O thank you teachers feel the need to prove their intelligence so they over complicate things when they are talking to someone with the brain the size of a peanut. They could be talking to someone with down syndrome and pull out big words and numbers out their ass
@Supermanu15 say, if you multiply A by (1+B):
first you multiply A by 1
then A by B
and sum them up - you got A*1 + AB witch is A + AB
that's why A(1 + B) = A + AB and vice versa :-)
when I say "taking A out of brackets" I mean that A in A+AB is a common factor (we have A in both terms - in A which is same as A*1, and in AB). So if we it's in both terms separately (A*1 + AB) or together A(1+B) the result is the same.
Hope it's clear. Please ask me more questions if something is not clear :-)
Vladimir Keleshev
A'+B'= A'B'
This is equal or not.
In 06:37 can we say
A+AA' = A+0 = A? Am I right?
Yes
Yep.
Yep
Yeah that’s what I did
Thank you for being so clear with the explanation!
I learnt a lot from this honestly
Thank you so much! I have an exam on this tomorrow and up til now I didnt understand one bit of it but now it makes sense! Thank you!
Why you didnt answer call from dimochka?
dimochka is dirsturbing бообще !!
you could use 1and0 insted A AND B
At time code 0:28, why is it assumed A is 1? Thanks for the great videos!
A is actually A * 1...I think that is what the man was explaining
Thank You Very much Vladimir!Your explanation is clear,I wish you were my professor!
Your explanation is really good.
I had no problems throughout the semester but this is great revision for the exam, thanks heaps!
You saved my examn, thanks
Thank you, I can now understand my class!!
Your explanation is awesome!
You are the best sir. Thank you
8:02 -> "Someone calling me....eehhh by mistake" lololol!!!
Thank you SO MUCH for taking the time and making this videos, I understand everything now :D
awesome..tryingdays to learn..but now with this im done under half hour
OMG!!! You are a lifesaver!!!
Thats a great video sir i didnt found it anywhere
you just save me mate ! really thank you!
A+AA(negated) actually simplifies to A. 1+A(negated)=1 A*1=A
I think it should be 1. because A+(AA)negated will result in A+(A'+A')
this gives you A+A'(1+A') then because 1+A' =1 (by law nullity) then you are left with A+A'(1) which is =A+A' and A+A'=1 by law of complement. thanks
Thank you this helped me a lot
Thanks!! exam is next week XD so helpful
Hello and thank you very much for this video its very easy to understand. I dont have a problem with A(1+B) for example but what if you have AB + A'B I saw that it works like A(B+'B) but is that a rule as a whole when you have another letter after the initial (in this case A) and only if there are repeating letters like in this case A and A and B or B. I guess what I am trying to ask is in which cases the expression can be simplified in this way? Thanks again
thanks it makes me more understanding.
Thank you so much! This has been so helpful.
In Addition if one of the values equals 1 than the result always eqals 1 and in multiplication if both eqal 1 then the result is 1 other wise the result is 0...
Thanks for the easy explanations!
I dont understand. How do we know that A=1 in the first place?
watch the last video
@@themaydayman he doesnt explain it in the last video
@@deaconng I think I meant previous video (it was a year ago tho)
great stuff... keep them coming
At 2:14, why wasn't it AA + AB + AC + BC?
And the phrase "A is A multiplied by one" confuses me to no end.
A is the same thing as A times one
0:20 Why do you say it wouldn't be possible to simplify it further? A(1+B) works in regular algebra as far as I've learned =)
from my understanding it would be A+b=c so its not really a tangible answer I feel like that's what he was eluding to
great tutorials, thank you very much
It helped me a lot...thanks a lot man..gravity be with you 👍
Thanks for the great explanation, subscribed and liked
THANKS! YOU SAVED MY LIFE
Superb video, thank you very much
Thanks! Very Useful to me
Great work thanks
thanks for the lesson..really helpful (y)
Wait wait wait. The first equation A+AB, how did u assume the first A Is multiplied by 1 making the second part (1×B)?? Is this a common rule or is it the "something"?
very good videos you rly helping me a lot
@Supermanu15 You are welcome :-)
Hi, first example A+AB = A+A=A then we have only one A and expression of B in brackets 1+B but why???
Is the answer to the last question 'not A' ?
HI, I WANT TO KNOW IF U LEAVE AN EXPRESSION LIKE THIS; NOT A+A NOT B NOT C, IS IT SIMPLIFIED? OR CAN IT BE SIMPLIFIED FURTHER BECAUSE THERE IS A and NOT A?
I was looking for these rules on your website but, i didn't see them
lol i was like O_o wtf, i don't have skype installed on this PC
cool stuff, thanks for the good explanations
Thank you so much for this!!!!
Yes if it was 0 it would be A because A is absorbed so its AB left 0*1 is 0 which is A
thank you😍😍😍your amazing
I am not understanding your first step in how you are getting A + AB = A(1+B), can you please explain how you get the 1 in that function. Thanks
You have a mistake in your exercise , i guess..
at 6.12 you say -> A + (not)AA = A (1+(not)A)
but in the previous minute you said , (not)A . A is 0
so wouldn't that be
A + (not)A.A = A + 0 = A
Thank You Once Again Sir!!!
It's ridiculous how helpful this is, it's like mathematical steroids or something. Thanks :D
Thank You these video made it easy
How we know that A variable and B variable don't have the same value?
Or better said. How we know that A != B ?
Do we assume their values are different because they have different names?
Why is A automatically A multiplied by one?
Sarah A can have the value of 1 or 0 only, when you see A or B (any variable) it will have the value of 1. When the you see A' (or in the exam it might be a line on top of the variable) that means it is negated which is the opposite of A. This means the opposite would be 0. So A = 1, A' = 0.
I’m watching this on the day of my exam
I'm struggling to understand how he gets A(B+1) from A+AB.
me too honestly, didnt it say that A+A = A so shouldn't it be AB?
Thank You!!
Thanks, great video!
Factorizing A+AB gives you A(1+B).
In other words A+AB = A*1+A*B, right?
A + A*A' should it not be A*A' = 0 using the complement law and A + 0 = A?
So for the last problem I found A-negated, is that right?
Good job! Thank you so much
Thanks alot sir.
Great explanation
how did u declared (1+x') = 1. What rule is this? im sorry im confused.
(1+x') means 1 OR inverse x, and of course we know that when OR is being used, we need either one or both sides of the OR to be 1 for the output to be 1. If x = 1 then x' = 0 and 1 + 0 = 1. if x = 0 then x' = 1 and 1 + 1 = 1, hence 1 + x' will always equal 1 no matter the value of x or x'
hope this was helpful
damn 8 years ago!!!
10 yo
Is A+AA(bar)=A+0... [since AA(BAR)=0]
therefore,
A+AA(bar)=A...(Since A+0=A) a correct alt method?
YES its correct
Thank you....assignment question :) :P
PLEASE HELP ME WITH A SOLUTION, WHAT IS THE RESOLUTION FOR (AB)' +AB THANKS