Martin Franke I noticed that, indeed acknowledging that might save a step. 13/2 = 6(r1) 6/2 = 3(r0) and 3/2 = 1(r1) Translating to 1101 For trinary it would be: 13/3 = 4(r1) 4/3 = 1(r1) Translating to 111 in trinary.
For fun in trinary: 244/3 = 81(r1) 81/3 = 27(r0) 27/3 = 9(r0) 9/3=3(r0) and 3/3=1(r0) Or 100001 in trinary 2014/3 = 671(r1) 671/3 = 223(r2) 223/3=74(r1) 74/3 = 24(r2) 24/3=8(r0) 8/3=2(r2) Or 2202121 in trinary
I believe this is just the usual way binary number is derived from decimal number. Step 1 - Keep dividing by 2 and note the remainder Step 2 - Reverse the remainders
zeN zeN Sometimes different bases are taught in a way that doesn't make this so obvious... E.g. If someone writes the binary 110110, having first leaned that the places represent powers of 2, I might have been inclined to count the digits, realize the leading one is 2^5, followed by 2^4, etc., and so do 2^5+2^4+2^2+2^1 = 32+16+4+2 = 54, but it's much quicker to do the reverse of what's demonstrated in this video and just start from 1 and say times 2 plus 1 times 2 times 2 plus 1 times 2 plus 1 times 2, which gives 54. I guess what I'm saying is, sometimes educators emphasize the theory and neglect the practical calculation techniques. Probably depends on where you went to school. I'm glad we have high quality math youtubers (for all different levels of math) helping fill these gaps.
This is a great trick, but I have to say you aren't actually "ignoring the remainder". You are just going back and placing the remainder beneath each number after the fact. This exact trick can be used to convert base 10 into any base. For instance, 445 to base 4 goes like this: 445/4=111(remainder 1); 111/4=27(remainder 3); 27/4=6(remainder 3); 6/4=1(remainder 2); 1/4=0(remainder 1) 1
@@fuseteam , right, you would want to keep track of remainders as you go in bases higher than 2 (technically you could do it after the fact, but it basically means doing the division a second time).
Thank you for pointing out it works for other bases too, even though it's a bit less efficient, it is still easier than what I have been using for 39 years.
This is the method I used when I was 13. Before that, around 9-10, if you had a 8 bit number, I checked if you could divide by 128, if you could you knew that was a “1”, subtract 128, and try 64 etc. That last method I invented myself for lack of having any books or other sources of information, just the English manual that came with my C64. (I didn’t know any English yet.)
Thank you much. You made the process unbelievably simple. I wish other videos are clear and concise like this. I have recommended this tutorial to my class fellows as well. Great job.
This is the best demonstration of how to use this method. If you think of divide and storing the remainder then you can use this method to convert to hexadecimal or any other base. You can understand how this works if you use the same method to convert a base ten to a base ten (which of course is the same number.) If you take 1326 and divide by 10 you end up with 132 with 6 remainder. You store the 6 in the ones place and 132 in the 10s place. Then divide the 132 by 10 you end up with 13 in the hundreds place with a remainder of 2 which you store in the tens place. And so no.
Chief, this is it. As silly as it seems I've always been messing up my denary to binary conversions SOMEHOW, but now I finally know an even easier method thanks to you. Thank you for helping us students out
There's one exception. When all numbers are even expect 1, then your on a power of two and only mark 1. For example, take 4,096. It goes down as the following. 4,096, 2,048, 1,024, 512, 256, 128, 64, 32, 16, 8, 4, 2, and 1. There's no remaining, so we tick 4,096 with an amount of 1
Actually, I don't think you are correct. Let's take a simpler power of 2, like 32. Our sequence written out is 1,2,4,8,16,32, so we would write 100000 for the binary representation, which is correct, and we followed all of Presh's rules.
Wish you went into a little bit of why it works. You can see why it works from the reverse operation. Given a a binary number, if you add a 0 to the end of it, you multiply the number by 2. If you add a 1 you multiple the number by 2 and add 1. The odd numbers correspond to cases where we multiplied by 2 and added 1 (because you can't multiply by 2 and get an odd number) and the even numbers correspond to cases where we just multiplied by 2.
Also note that this works for numbers of ANY base. Except you don't just count odd or even, you count what the remainder of the number is given division by the base. For example, the last digit of 52446 is 6 because 52446/10 has a remainder of 6. 5244/10 has a remainder of 4 and so on.
Not sure why there is so much criticism in the comments, but this is an elegant method to convert to Binary from Decimal. I get it that the remainders do not need to be "ignored", but the method of ignoring the remainders and replacing 1's for odd and 0's for evens also works just fine. Thank you to the OP.
Hmm. I always used to subtract the positional values. If X >= the positional value, then that bit is 1 and subtract the value, if it is not, then the bit is 0 and do not subtract. But I think it is easier by this halving method...
*I prefer this method because it goes horizontally, unlike the other method which is vertical. humans read horizontally.* And no need to write down the remainder like the other method; just sub each odd number with a 1. It's quicker.
Pretty basic trick. I've learned that in my first semester, but can you also proof correctness? :-P A proof sketch: Assume the binary encoding of the number. A division by 2 is identical to a the bit shift to the right. (This can by itself again be proven using sum formulas). The remainder by this division is the same as the least significant bit before this shift. So by shifting step by step until we reach 1 we shift all bits out of the binary encoding and wrote them down to the solution
You can reverse the process and go from Binary to Decimal by starting with 0 add the first digit, double the result, add the next digit, double that result, add the next digit, and repeat for all digits. For Example: 1010 0 +1 = 1, 2x1 +0 = 2, 2x2 +1 = 5, 2x5 +0 = 10 For Example: 0101 0 +0 = 0, 2x0 +1 = 1, 2x1 +0 = 2, 2x2 +1 = 5
I am taking a networking course for my college degree. We are at the point of converting decimal to binary. The way that the tutorials (and fellow) students explain it have made my head spin. After watching your video, I consider myself an "expert" (LOL). Thank you, very much, for this tutorial.
@@riteshbhartiya6155 When I was at school, I was taught that to write 13 in binary, you write out all the powers of 2 less than 13 ie 1,2,4 and 8 and see which set of these numbers sum to 13 and then write 0s and 1s as appropriate
@@kkgt6591 to find the binary equivalent of 13, write down all the powers of 2 below 13, ie 1,2,4 and 8 and see which of those numbers add to 13 There was nothing systematic about it and could take a long time
just playing around with this in excel, and discovered that reversing binary string with BIN2DEC function after 511 we are getting negative numbers. with 512 it's -512.. 513 goes as -511 and so on, up to 1023 which will be -1 and 1024 generates error.. damn
Coded as 8 bits. (0111 1111) is 511, the next "number" would be 512 (1000 0000), but the first digit (the one on the left) is the sign bit. So (1000 0000) is "translated" like ( - 0111 1111 -1) to get 1024 numbers, from -512 to 511. Conclusion ? Do your own function to find decimal and binary numbers \o/
A simple explanation for why this works is the fact you can check if a power of 2 goes into a number in its binary form. Let's say you wanted to check if 2 or 4 goes into 13. If you divide 13 by 2 you'll get 6 (discarding the remainder), which is even. If you divide 13 by 4 you get 3, which is odd. If the number is odd, then that power of 2 is part of the chosen numbers binary composition. So 2 is not a part of 13, while 4 is. The reason for this can be seen by how binary numbers are made up. For 13, it would take three 4's and a 1. But two 4's make an 8, so it would really be 8, 4, and 1. Every time the number 4 goes in an even number of times, that means a higher power of 2 can replace all the 4's. If you take the number 9, 4 goes in twice, which can be replaced by 8. So if you take a number and keep dividing it by 2, it is the same as finding the number if you divide it by 2, 4, 8, etc., and you can check if it's a part (where you put a 1) or not (where you put a 0).
Actually you’re keeping remainder information in memorized (wrote down) numbers. What does it mean that number is odd? It’s remainder by dividing by two is one. Same goes for even numbers.
Yes, some of you might very well have learnt this technique long ago, but that doesn't mean we all have. How many of you have gone through life oblivious of some piece of general knowledge that you suddenly discover everyone else around you seems to have known for years? That's the nature of general knowledge; it comprises those bits and pieces of knowledge you pick up along the way in life, much of it unconsciously. Personally this trick was entirely new to me.
Thank you so much. I just started my first semester and I think I might be at the back at the class in terms of performance. And that's all because I forget the concept of binary over the years now.
There is a more advanced technique You need to know the values of 2^x . And use the values as weight bars used to measure weights Like 13 Largest 2^x is 8 13-8=5 For 5 largest 2^x is 4 5-4= 1 So 13= 8+4+1 =(2^3)+(2^2)+0.(2^1) +(2^0) => 1101 Similarly 244=128+64+32+16+4=>11110100 I hope you get it
@@devgupta9781 Division of long numbers are difficult too. Here you need to remember the values of 2^x, only then it becomes easy. Of course I am taking about numbers greater than 100.
It may be easier write the remainder below the number, to get a better feel for it - for instance, 122 = Reverse (0 1 0 1 1 1 1) = 1 1 1 1 0 1 0 , another easy example 25 = reverse (1 0 0 1 1) = 1 1 0 0 1
The simplest method is still long division by 2.This sounds simple but is actually quite hard to do in your mind since you'll need to remember all the quotients to assign 1 or 0 later.Just keep dividing the number by 2 and store the remainder on your fingers (raised finger for 1) and read backwards.
The BEST and fastest way to convert decimals into binary. Do not use any other method and best for DCCL testing. Great job and thanks for the video. Excellent work.
The binary numbers are just the reminders of each division. It is better to directly write the remainder for each division, so you can keep track of each step.
It would be fantastic if you could make a video for this explanation i.e. conversion from binary to decimal: Under the rightmost digit write a 1, and then under each next digit to the left double that. Then cross out any columns with a 0 and add up the numbers. So for 1101, we would write 1101 -----1 Then we double so we have 2, 4, and 8. 1101 8421 Now cross out the column with a 0. 111 841 Add up the numbers 8 + 4 + 1 = 13.
Thank you for your tutorial. Please make sure that you are doing this type of calc in the same direction (Right to left) since it does effect the answer. I was trying to figure out 93 and kept getting 29.
This is... the method they teach you in school? I find it more easy to use the “doubling method” as I call it: Just remember that *1 = 1* Adding a 0 in a Binary number equals to do *x • 2* where x is the binary number to which you are adding a zero. Adding a 1 instead equals to do *(x • 2) + 1* So, for an example: 1110010 = I know from memory that 111 is 7 Let’s add two 0s 7 • 2 = 14 14 • 2 = 28 Now let’s add a 1 (28 • 2) + 1 = 57 Finally, let’s add the final 0 57 • 2 = 114 Seems hard to do on paper, but it’s pretty quick
Oh my god man I really thought I wasn't cut out for math but you just made this brilliant explanation about binary numbers which just blew my mind of how clear and well-explained that was. Thanks for the great video!!!
It is a great method. The trouble is you don't know why it works. If you use the ... 2048 1024 512 256 128 64 32 16 8 4 2 1 you know why but more cumbersome. Great vid.
Very nice... I was taught a grid system at school the equivalent to the decimal units, tens, hundreds etc but one, two, four, eight, sixteen etc and it was horrible!... this seems much better (and yes I am older than 101111 yrs old) "There are 10 types of people in the world; those who understand binary and those that don't!".
Ditto I have been coding since 1980 age 11 I was also taught to start from the left (MSB) keep moving right with a running remainder. Which was fine back in the 8 bit days. I never tried to flip the calculation on its head and let the odds and evens do the work. It seems obvious how it works and I'm a bit miffed that I only just discovered it. It seems so easy it feels like cheating.
It’s so simple how and why it works once you see it. Binary places go up doubling each time, so dividing by half is the inverse. DIV by 2 is either going to be remainder 1 or 0. 0 is even, 1 is odd. Obvious right. But treating each column as a simple odd even uses these facts conjoined and QED!
Это работает с любой системой счисления(там где пишем код в другой системе счисления, надо писать остаток от деления на СС) Например: 153(10) = ?(6) 4 - 25 - 153 4 mod 6 - 25 mod 6 - 153 mod 6 ? = 413(6)
This works for any base. For base n, divide the decimal number repeatedly by the base (n). The remainder for each step is the digit for that place. So to convert 254 to base 16 (hexadecimal): 254÷16 = 15 remainder 14 (E) 15÷16 = 0 remainder 15 (F) So 255 in hexadecimal is FE.
Thanks this really helps though it only helps in answering objective questions Please can you do a method that can be permitted in the theory aspect of an exam? I stay in Nigeria😧
Matthew Smith There are a few explanations given on stackexchange: math.stackexchange.com/questions/86207/converting-decimalbase-10-numbers-to-binary-by-repeatedly-dividing-by-2 I also give a proof in this post on my blog (skip down to the section with the proof and the discussion of dividing by 2 and ignoring the remainder) mindyourdecisions.com/blog/2014/08/27/the-egyptian-method-russian-peasant-multiplication-video-and-a-proof/#.VSLhBPnF98E
? It is exactly the same thing. When you have to decide "even/odd" you do exactly that: figuring out if the reminder needs to be 0 or 1. And instead of dividing and taking the reminders in the oppoisite direction, you just start writing on the right side an work yourself to the left, which when reading the digits "reverses" the reminders.
Broooo bless you! My computer teacher at VoTech was trying to explain this to me and i had no freking clue what he was talking about. ITS SO SIMPLE xD Thanks brother!
People claiming it's just the usual way of converting to binary: yes, it is that. But remember how much extra variables and scribbling you do for that. While now you only need to keep one number in mind, and directly start writing the binary form from right to left. For smaller numbers, it will look like an extra ordinary feat to the onlookers and it saves a lot of time Thanks for sharing this @mindyourdecisions it is very useful
You are not actually ignoring the remainders. The ones and zeroes you write below are, in fact, the remainders.
Martin Franke I noticed that, indeed acknowledging that might save a step.
13/2 = 6(r1) 6/2 = 3(r0) and 3/2 = 1(r1)
Translating to 1101
For trinary it would be:
13/3 = 4(r1) 4/3 = 1(r1)
Translating to 111 in trinary.
For fun in trinary:
244/3 = 81(r1) 81/3 = 27(r0) 27/3 = 9(r0) 9/3=3(r0) and 3/3=1(r0)
Or 100001 in trinary
2014/3 = 671(r1) 671/3 = 223(r2) 223/3=74(r1) 74/3 = 24(r2) 24/3=8(r0) 8/3=2(r2)
Or 2202121 in trinary
kek
shut up
Yes, the algorithm works for any base.
I believe this is just the usual way binary number is derived from decimal number.
Step 1 - Keep dividing by 2 and note the remainder
Step 2 - Reverse the remainders
exactly... sane thing came to my mind... nothing much new
So true
Ya right
Yup the usual method
yeah, you can think of it as extracting the digits from a binary number:
1101 mod 2 = 1
110 mod 2 = 0
11 mod 2 = 1
1 mod 2 = 1
2014 TH-cam: nah
2019 TH-cam: Certainly, everybody wants to hear about this
I’m actually doing binary tho wth
well i did miss tgat one so its nice YT reminded it ;)
I need it for uni
This is just the same thing as normal conversion....
zeN zeN Sometimes different bases are taught in a way that doesn't make this so obvious... E.g. If someone writes the binary 110110, having first leaned that the places represent powers of 2, I might have been inclined to count the digits, realize the leading one is 2^5, followed by 2^4, etc., and so do 2^5+2^4+2^2+2^1 = 32+16+4+2 = 54, but it's much quicker to do the reverse of what's demonstrated in this video and just start from 1 and say times 2 plus 1 times 2 times 2 plus 1 times 2 plus 1 times 2, which gives 54.
I guess what I'm saying is, sometimes educators emphasize the theory and neglect the practical calculation techniques. Probably depends on where you went to school. I'm glad we have high quality math youtubers (for all different levels of math) helping fill these gaps.
Right bro
He is just taking LCM without drawing rows and columns
This is a great trick, but I have to say you aren't actually "ignoring the remainder". You are just going back and placing the remainder beneath each number after the fact. This exact trick can be used to convert base 10 into any base. For instance, 445 to base 4 goes like this:
445/4=111(remainder 1); 111/4=27(remainder 3); 27/4=6(remainder 3); 6/4=1(remainder 2); 1/4=0(remainder 1)
1
yes i agree tho in base 4 you can't reproduce the remainders after the fact
@@fuseteam , right, you would want to keep track of remainders as you go in bases higher than 2 (technically you could do it after the fact, but it basically means doing the division a second time).
@@SgtSupaman yep while in binary you just have to look at the last digit of each result to reproduce the remainders
Thank you for pointing out it works for other bases too, even though it's a bit less efficient, it is still easier than what I have been using for 39 years.
TH-cam is a better teacher than your average college professor. Why can't teachers make it this simple? Quick, straight to the point, and simple.
Is this what you call a trick ?? This is the very original procedure to convert a decimal number into its binary equivalent.
I never learned it today, but some people might have already knew that so...¯\_(ツ)_/¯
This is the method I used when I was 13. Before that, around 9-10, if you had a 8 bit number, I checked if you could divide by 128, if you could you knew that was a “1”, subtract 128, and try 64 etc. That last method I invented myself for lack of having any books or other sources of information, just the English manual that came with my C64. (I didn’t know any English yet.)
I think he meant a trick for beginners rather than going online and finding some sort of converter
if this is too basic for you you can just leave
Гальванизированный Труп don’t forget: That this video is several years old, and TH-cam dragged this out from the void.
Thank you much. You made the process unbelievably simple. I wish other videos are clear and concise like this. I have recommended this tutorial to my class fellows as well. Great job.
I haven't understood this the entire semester and I got it in 3 minutes!!
Exactly buddy😂😂😂
our teacher explained this to us in a different way but this one is so much easier ... thank you
This is the best demonstration of how to use this method. If you think of divide and storing the remainder then you can use this method to convert to hexadecimal or any other base.
You can understand how this works if you use the same method to convert a base ten to a base ten (which of course is the same number.) If you take 1326 and divide by 10 you end up with 132 with 6 remainder. You store the 6 in the ones place and 132 in the 10s place. Then divide the 132 by 10 you end up with 13 in the hundreds place with a remainder of 2 which you store in the tens place. And so no.
Very useful! Faster than the method my teacher taught me.
I have an exam on Tuesday and I'll have to use this, wish me luck!
Best method ever. Handy in exams. Thanks a lot
Anuj Kumar this is the normal methode
This is the exact same method normally used to calculate this sort of thing
@@sangramjitchakraborty7845 not everywhere, we were never taught this method
@@sangramjitchakraborty7845 I wasn't taught this method...
This is possible in any base. Just write sown the remainder after dividing by the base you want to get.
Finally someone explains this in a way that a fifth grader can understand, and not just assuming that everybody knows where to start! Thank you?
Chief, this is it. As silly as it seems I've always been messing up my denary to binary conversions SOMEHOW, but now I finally know an even easier method thanks to you. Thank you for helping us students out
There's one exception. When all numbers are even expect 1, then your on a power of two and only mark 1. For example, take 4,096. It goes down as the following. 4,096, 2,048, 1,024, 512, 256, 128, 64, 32, 16, 8, 4, 2, and 1. There's no remaining, so we tick 4,096 with an amount of 1
Actually, I don't think you are correct. Let's take a simpler power of 2, like 32. Our sequence written out is 1,2,4,8,16,32, so we would write 100000 for the binary representation, which is correct, and we followed all of Presh's rules.
Wish you went into a little bit of why it works. You can see why it works from the reverse operation. Given a a binary number, if you add a 0 to the end of it, you multiply the number by 2. If you add a 1 you multiple the number by 2 and add 1. The odd numbers correspond to cases where we multiplied by 2 and added 1 (because you can't multiply by 2 and get an odd number) and the even numbers correspond to cases where we just multiplied by 2.
Also note that this works for numbers of ANY base. Except you don't just count odd or even, you count what the remainder of the number is given division by the base. For example, the last digit of 52446 is 6 because 52446/10 has a remainder of 6. 5244/10 has a remainder of 4 and so on.
Not sure why there is so much criticism in the comments, but this is an elegant method to convert to Binary from Decimal. I get it that the remainders do not need to be "ignored", but the method of ignoring the remainders and replacing 1's for odd and 0's for evens also works just fine. Thank you to the OP.
Never my teachers explain this to me. Because they ignore it!!!
👏🏼👏🏼👏🏼
I just learned more in 2 minutes 50 seconds and the short time I spent reading the comments, than I have in my hour long lecture. Thanks @everyone
Ummm... How else did we use to solve for binary ... That was the method from beginning
Hmm. I always used to subtract the positional values. If X >= the positional value, then that bit is 1 and subtract the value, if it is not, then the bit is 0 and do not subtract. But I think it is easier by this halving method...
One way I've seen is to calculate the digits in big endian order, instead of this alg, which does it in little endian order.
Awesome,the method you thought was easier and accurate than my teacher thought me.
Studying computer science and was so stuck on this , thank your video helped a lot.
Me tooo , I'm a wannabe tho
Why ignore the remainder when it's what you are going to write down at the end anyway?
because you can reproduce it either way
@@fuseteam wait repeat
But slowly
@@dhwaneelkapadia3265 the remainders are derivable from the oddness of quotients
@@dhwaneelkapadia3265 I think he means that you'll always get remainder 0 on even numbers and remainder 1 in odd numbers so you needn't note them
*I prefer this method because it goes horizontally, unlike the other method which is vertical. humans read horizontally.*
And no need to write down the remainder like the other method; just sub each odd number with a 1. It's quicker.
This really really helped me! The remainder part tripped me up, until you came along. Everyone wants to be so deep, but simplicity is key!
I love watching videos of people teaching my native language
Pretty basic trick. I've learned that in my first semester, but can you also proof correctness? :-P
A proof sketch:
Assume the binary encoding of the number. A division by 2 is identical to a the bit shift to the right. (This can by itself again be proven using sum formulas). The remainder by this division is the same as the least significant bit before this shift. So by shifting step by step until we reach 1 we shift all bits out of the binary encoding and wrote them down to the solution
daaaaaang that's wild.
You can reverse the process and go from Binary to Decimal by starting with 0 add the first digit, double the result, add the next digit, double that result, add the next digit, and repeat for all digits.
For Example: 1010
0 +1 = 1, 2x1 +0 = 2, 2x2 +1 = 5, 2x5 +0 = 10
For Example: 0101
0 +0 = 0, 2x0 +1 = 1, 2x1 +0 = 2, 2x2 +1 = 5
I am taking a networking course for my college degree. We are at the point of converting decimal to binary. The way that the tutorials (and fellow) students explain it have made my head spin. After watching your video, I consider myself an "expert" (LOL). Thank you, very much, for this tutorial.
it's basically the same way as the normal way of converting a decimal number to binary.
So you knew the trick already. Big fluffu deal. Some of us didn't and had to use a longer method
Is there really a longer method or you call dividing by 2 and writing remainders in reverse order a longer method.
@@andywright8803 which longer method were you using?
@@riteshbhartiya6155 When I was at school, I was taught that to write 13 in binary, you write out all the powers of 2 less than 13 ie 1,2,4 and 8 and see which set of these numbers sum to 13 and then write 0s and 1s as appropriate
@@kkgt6591 to find the binary equivalent of 13, write down all the powers of 2 below 13, ie 1,2,4 and 8 and see which of those numbers add to 13
There was nothing systematic about it and could take a long time
just playing around with this in excel, and discovered that reversing binary string with BIN2DEC function after 511 we are getting negative numbers. with 512 it's -512.. 513 goes as -511 and so on, up to 1023 which will be -1 and 1024 generates error.. damn
Coded as 8 bits.
(0111 1111) is 511, the next "number" would be 512 (1000 0000), but the first digit (the one on the left) is the sign bit. So (1000 0000) is "translated" like ( - 0111 1111 -1) to get 1024 numbers, from -512 to 511.
Conclusion ? Do your own function to find decimal and binary numbers \o/
A simple explanation for why this works is the fact you can check if a power of 2 goes into a number in its binary form. Let's say you wanted to check if 2 or 4 goes into 13. If you divide 13 by 2 you'll get 6 (discarding the remainder), which is even. If you divide 13 by 4 you get 3, which is odd. If the number is odd, then that power of 2 is part of the chosen numbers binary composition. So 2 is not a part of 13, while 4 is. The reason for this can be seen by how binary numbers are made up. For 13, it would take three 4's and a 1. But two 4's make an 8, so it would really be 8, 4, and 1. Every time the number 4 goes in an even number of times, that means a higher power of 2 can replace all the 4's. If you take the number 9, 4 goes in twice, which can be replaced by 8. So if you take a number and keep dividing it by 2, it is the same as finding the number if you divide it by 2, 4, 8, etc., and you can check if it's a part (where you put a 1) or not (where you put a 0).
I’m here in 2018. Why in the hell wasn’t my textbook this easy? I was so lost but found this video and now I get it. Thanks!
A really helpful video on binary numbers. Easy for exams.
why only for exam???
@@pradeep422 i think what he really means is that this is good for exams too
Actually you’re keeping remainder information in memorized (wrote down) numbers. What does it mean that number is odd? It’s remainder by dividing by two is one. Same goes for even numbers.
Can you explain this again?
@@kyliestaraway2492 hmm
This was very clever. Is there a shortcut for hexadecimal?
Thanks
This method also works with trinary, quaternary, quinary, sexternary, septenary, octanary, nonary, etc
Yes, some of you might very well have learnt this technique long ago, but that doesn't mean we all have. How many of you have gone through life oblivious of some piece of general knowledge that you suddenly discover everyone else around you seems to have known for years? That's the nature of general knowledge; it comprises those bits and pieces of knowledge you pick up along the way in life, much of it unconsciously. Personally this trick was entirely new to me.
I have a quiz on Decimal, Binary, and Hexadecimal in two hours. Live saver. Truly. Thank you, sir!
That’s the most clean way to calculate binary number I have ever seen !
Thank you so much. I just started my first semester and I think I might be at the back at the class in terms of performance. And that's all because I forget the concept of binary over the years now.
In alternative, you can just annotate the remainders and then read them backwards to get the same answer :P
That is the method my teacher taught us.
both work similarly
There is a more advanced technique
You need to know the values of 2^x . And use the values as weight bars used to measure weights
Like 13
Largest 2^x is 8
13-8=5
For 5 largest 2^x is 4
5-4= 1
So 13= 8+4+1 =(2^3)+(2^2)+0.(2^1) +(2^0) => 1101
Similarly
244=128+64+32+16+4=>11110100
I hope you get it
Your technique is even more time consuming and difficult too.
@@devgupta9781 Division of long numbers are difficult too. Here you need to remember the values of 2^x, only then it becomes easy.
Of course I am taking about numbers greater than 100.
It may be easier write the remainder below the number, to get a better feel for it - for instance,
122 = Reverse (0 1 0 1 1 1 1) = 1 1 1 1 0 1 0 , another easy example 25 = reverse (1 0 0 1 1) = 1 1 0 0 1
The simplest method is still long division by 2.This sounds simple but is actually quite hard to do in your mind since you'll need to remember all the quotients to assign 1 or 0 later.Just keep dividing the number by 2 and store the remainder on your fingers (raised finger for 1) and read backwards.
u suck at math
Teaching is a skill, he is an excellent skilled teacher.
The BEST and fastest way to convert decimals into binary. Do not use any other method and best for DCCL testing. Great job and thanks for the video. Excellent work.
Completely understood after watching this, a simple and easy way to do it. Thank you
why you never tell me this 😡
The binary numbers are just the reminders of each division. It is better to directly write the remainder for each division, so you can keep track of each step.
Did anyone see that Presh's last answer is 2014=11111011110, which is actually the year this vid is published? Isn't this a bonus?
It would be fantastic if you could make a video for this explanation i.e. conversion from binary to decimal:
Under the rightmost digit write a 1, and then under each next digit to the left double that. Then cross out any columns with a 0 and add up the numbers.
So for 1101, we would write
1101
-----1
Then we double so we have 2, 4, and 8.
1101
8421
Now cross out the column with a 0.
111
841
Add up the numbers 8 + 4 + 1 = 13.
Wouldn't noting remainders be faster?
Yes, that would take 1 pass. He takes 2 passes. And he has to store all the intermediate results for the 2nd pass.
Wow. Thank you so much. Am a beginner student in computer science. This formula of converting decimals to binary is mine. Am well taught
Wow! Thanks a lot man. Never knew its even possible to calculate binary digits
This by far is the best method I have been shown so far. Thank you!!
GENIUS, my teacher just literally told us to do binary by adding.
Thank you for your tutorial. Please make sure that you are doing this type of calc in the same direction (Right to left) since it does effect the answer. I was trying to figure out 93 and kept getting 29.
Its so much easier to understand on video then on paper. Thanks!
Simple and to the point, love that its not complicated :)
You can write down instantly the binary numbers from right to left. In this way you don't need to remember which number was even or odd.
Fast and simple. Why they don't teach it the easy way from the start? The world may never know.
Thanks for posting this mathematical challenges, they are fantastic!. They really have improved my thinking skills.
This is... the method they teach you in school?
I find it more easy to use the “doubling method” as I call it:
Just remember that *1 = 1*
Adding a 0 in a Binary number equals to do *x • 2* where x is the binary number to which you are adding a zero.
Adding a 1 instead equals to do *(x • 2) + 1*
So, for an example:
1110010 =
I know from memory that 111 is 7
Let’s add two 0s
7 • 2 = 14
14 • 2 = 28
Now let’s add a 1
(28 • 2) + 1 = 57
Finally, let’s add the final 0
57 • 2 = 114
Seems hard to do on paper, but it’s pretty quick
Sir your teaching technique is so helpful.
What is genius in this? This is the tradition method I learnt in school.
Oh my god man I really thought I wasn't cut out for math but you just made this brilliant explanation about binary numbers which just blew my mind of how clear and well-explained that was.
Thanks for the great video!!!
It is a great method. The trouble is you don't know why it works.
If you use the ... 2048 1024 512 256 128 64 32 16 8 4 2 1 you know why but more cumbersome.
Great vid.
Love the simplicity and thinking behind this
it acts like tutorial during my exams👍👍👍
This was so helpful to a newbie! Thank you!
Very nice... I was taught a grid system at school the equivalent to the decimal units, tens, hundreds etc but one, two, four, eight, sixteen etc and it was horrible!... this seems much better (and yes I am older than 101111 yrs old)
"There are 10 types of people in the world; those who understand binary and those that don't!".
Ditto
I have been coding since 1980 age 11
I was also taught to start from the left (MSB) keep moving right with a running remainder. Which was fine back in the 8 bit days.
I never tried to flip the calculation on its head and let the odds and evens do the work. It seems obvious how it works and I'm a bit miffed that I only just discovered it.
It seems so easy it feels like cheating.
holy shit,this is so easy,the long one was boring af but this will probably save me tomorrow
And did it?
You do realize how much this is gonna help students, right?
This is awesome, easy to do. Will use this method this coming exams. Thank you very much 😊
It’s so simple how and why it works once you see it.
Binary places go up doubling each time, so dividing by half is the inverse.
DIV by 2 is either going to be remainder 1 or 0.
0 is even, 1 is odd. Obvious right.
But treating each column as a simple odd even uses these facts conjoined and QED!
Это работает с любой системой счисления(там где пишем код в другой системе счисления, надо писать остаток от деления на СС)
Например:
153(10) = ?(6)
4 - 25 - 153
4 mod 6 - 25 mod 6 - 153 mod 6
? = 413(6)
this was the best and the easiest method i found on the net thanks a lot sir
This works for any base. For base n, divide the decimal number repeatedly by the base (n). The remainder for each step is the digit for that place.
So to convert 254 to base 16 (hexadecimal):
254÷16 = 15 remainder 14 (E)
15÷16 = 0 remainder 15 (F)
So 255 in hexadecimal is FE.
Thanks this really helps though it only helps in answering objective questions
Please can you do a method that can be permitted in the theory aspect of an exam? I stay in Nigeria😧
this will be my cheat sheet for my test tomorrow thanks!
HOLY SHIT! this is so much easier than the technique that is taught by the professor damn
great video, teacher explained it a bit weird but it clicked when you explained it, thank you
Best trick ever on decimal to binary.
your method is the easiest that I have ever learned.. thaks a lot
Well explained..can you also do the reverse? how to use the binary to convert back to the original number.
i wasn't well when my teacher thought this but now i know thanks man
Really,i understand better here than my teacher😊
Can you explain why this works?
Matthew Smith There are a few explanations given on stackexchange: math.stackexchange.com/questions/86207/converting-decimalbase-10-numbers-to-binary-by-repeatedly-dividing-by-2
I also give a proof in this post on my blog (skip down to the section with the proof and the discussion of dividing by 2 and ignoring the remainder) mindyourdecisions.com/blog/2014/08/27/the-egyptian-method-russian-peasant-multiplication-video-and-a-proof/#.VSLhBPnF98E
Thanks for your reply, looks like there's no easy answer! The russian peasant multiplication thing is awesome! A video everyone can use!
My teacher explained it similar, but much more confusing. This clicked perfectly for me.
Hey! The last example you showed was the year this video was posted!
Me: *Graduates*
TH-cam: It’s time you learn this
The fact that this isn't the default way taught everywhere is a crime.
Awesome method easier than remainder method
?
It is exactly the same thing.
When you have to decide "even/odd" you do exactly that: figuring out if the reminder needs to be 0 or 1.
And instead of dividing and taking the reminders in the oppoisite direction, you just start writing on the right side an work yourself to the left, which when reading the digits "reverses" the reminders.
Broooo bless you! My computer teacher at VoTech was trying to explain this to me and i had no freking clue what he was talking about. ITS SO SIMPLE xD Thanks brother!
AMAZING! much easier than the trick which my computer science teacher taught me.
THANK YOU ITS VERY HELPFUL 👍 For my exam
People claiming it's just the usual way of converting to binary: yes, it is that. But remember how much extra variables and scribbling you do for that.
While now you only need to keep one number in mind, and directly start writing the binary form from right to left.
For smaller numbers, it will look like an extra ordinary feat to the onlookers and it saves a lot of time
Thanks for sharing this @mindyourdecisions it is very useful
How would you do it if the number you start with had a decimal?
I think u would ignore the decimal?
@@nami2075 I did that and then did the number after the decimal separate and got the correct answer.
Drew Schultz Nice!
Drew Schultz What number was it exactly?