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 UA-cam who has done this way, I finally understand boolean algebra, thanks.
@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 :-)
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
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
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?
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
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"?
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
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...
(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
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.
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
You're so much clearer than my professor! Thank you.
Thanks Vlad for the tutorial finally you explained it in "dummie" terms so I could understand it finally! Subbed.
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 UA-cam who has done this way, I finally understand boolean algebra, thanks.
did you pass the exam BRO!?
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 :)
Have an exam tomorrow. My professor actually recommended over viewing this series. Fingers crossed haha
11 years later and I found u to be my savior
btw I'm majoring in computer science
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
This is the first time I've ever actually GOT this. Thanks so much for posting these!
fax this guy is the goat
The problem at 8:15. Did you use absorbtive law to get rid of that second A, or what that factoring? Or?
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!
@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.
At 2:35 he says he's taking A out of brackets, but there are no brackets.
the whole thing is a bracket
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
A'+BA'= A'(1+B)
So, the answer is: A'
Yes
Thanks a lot. Much appreciated
thank you vladimir, really pulling through 10 years later
In 06:37 can we say
A+AA' = A+0 = A? Am I right?
Yes
Yep.
Yep
Yeah that’s what I did
Why you didnt answer call from dimochka?
dimochka is dirsturbing бообще !!
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
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
I was looking for these rules on your website but, i didn't see them
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)
You explain boolean algebra well. Thank you a lot
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
i dint understand dis chapter in college...but u teached me lyk a game!!!
you could use 1and0 insted A AND B
Great Video, A lot of thanks for a simple way to understand this algebra.
Woahhh ,your video make it easier to understand... Thanks !
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!
Is the answer to the last question 'not A' ?
I had no problems throughout the semester but this is great revision for the exam, thanks heaps!
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?
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
Hi, first example A+AB = A+A=A then we have only one A and expression of B in brackets 1+B but why???
PLEASE HELP ME WITH A SOLUTION, WHAT IS THE RESOLUTION FOR (AB)' +AB THANKS
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
Thank You Very much Vladimir!Your explanation is clear,I wish you were my professor!
greatest video of all time thank you
Wooow...this totally Helped Me. Thanks A Lot.
Thank you for being so clear with the explanation!
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"?
can you help me with a problem?
So for the last problem I found A-negated, is that right?
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
but how did A get multiplied by 1?
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
Your explanation is really good.
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?
I learnt a lot from this honestly
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
lol i was like O_o wtf, i don't have skype installed on this PC
Thank you, I can now understand my class!!
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...
implement the following boolean expression are they are stated ( a ) = ABC + AB + AC ( salution thes question ) plzzzz help me
*+malik mushtaq*
ABC + AB + AC = AB(C + 1) + AC = AB + AC = A(B + C)
+Gootman Boats is this also correct bro? ABC+AB+AC= A(BC+B+C) =A(B(C+1)+C) = A(B+C)
JBluee Yes, that is another correct approach.
You saved my examn, thanks
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
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?
Thanks!! exam is next week XD so helpful
Your explanation is awesome!
Thank you SO MUCH for taking the time and making this videos, I understand everything now :D
A•A’ is 0 na how did you get 1?
A + A*A' should it not be A*A' = 0 using the complement law and A + 0 = A?
awesome..tryingdays to learn..but now with this im done under half hour
Wow. Why was it so hard to understand before? I'm such a stupid dummy
why is a always multiplied by one
i really dont get the first problem why A (1+B)
A+AB, A(1+B) ACCORDING TO THE RULES, IF U HAVE A+A IT EQAUALS A(OR 1) AND THAN 1+ ANYTHING IS JUST ITS SELF.
OMG!!! You are a lifesaver!!!
Factorizing A+AB gives you A(1+B).
In other words A+AB = A*1+A*B, right?
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.
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
How can we prove the following example
A.B + A'.B + A'.B' = A' + B
I understand that A*A or A+A= A, however what about A*B or A+B?
A*B is 1 only when both A=1 and B=1
A+B is 1 when either A=1 or B=1 (or both are 1)
There is no simplification for that
Thats a great video sir i didnt found it anywhere
Hey sir A negative multiplied by one is zero but you will show A negative
Yes if it was 0 it would be A because A is absorbed so its AB left 0*1 is 0 which is A
you just save me mate ! really thank you!
1:54
What what what do we have here?
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
Why do we know that A is multiplied by 1 ? why is it not 0?
It's ridiculous how helpful this is, it's like mathematical steroids or something. Thanks :D
It helped me a lot...thanks a lot man..gravity be with you 👍
8:02 -> "Someone calling me....eehhh by mistake" lololol!!!
great stuff... keep them coming
Thank you so much! This has been so helpful.
damn 8 years ago!!!
10 yo
You are the best sir. Thank you
Thanks for the great explanation, subscribed and liked
I’m watching this on the day of my exam
THANKS! YOU SAVED MY LIFE
Thanks for the easy explanations!
I am not too sure but maybe (A'+B'+C')*(A+B+C) might be C+B+A
very good videos you rly helping me a lot
Thank you this helped me a lot
cool stuff, thanks for the good explanations
thanks it makes me more understanding.
thanks for the lesson..really helpful (y)
Thank You!!