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Welcome to the third video on[br]binary numbers. In this video,

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we're going to look at how to[br]convert decimal numbers to

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binary numbers, and in this[br]particular video I'm going to

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show you the place value table[br]or addition method. In this

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video. Our goal is to find out[br]what place, why you combinations

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build out the decimal number[br]that needs to convert it into

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binary. I can only use once for[br]any of the place values.

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As in the binary number system,[br]I only have the digit one and

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zero in a further video I'm[br]going to show you a different

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method called the division[br]method, so it doesn't matter

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which way you learn to convert[br]binary numbers. Whatever method

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you feel more comfortable with,[br]just stick with that and then

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your answer will be just as[br]correct. If you are doing it

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with the other method. So the[br]table method, it's called the

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place value table method because[br]we are going to use the binary

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place value. To convert decimal[br]numbers for that again, let's

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look at what the binary place[br]value table looks like. So

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remember that the place values[br]were due to the 0.

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2 to one. Two to two.

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Two to three.

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2 to 4, two to five etc. You can[br]continue this as long as you

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want it to, but these numbers[br]also translate back down to

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normal decimal numbers without[br]the power form. So 2 to the

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power of 0 would be 1. Two to[br]the power of 1 would be 2. Two

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to the power of two would be 4.[br]To do a power of three is 8 to

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two power of four is 16 and 2[br]two power of five is 32. So if I

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have got. Decimal number, such[br]as the teen? How am I going to

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use my knowledge of the place[br]values to convert this number?

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Into binary.

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So what I'm going to do, I'm[br]going to look at how many 30

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twos would I need to use? Use to[br]make the 13 while it's too big

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because 13 is much much smaller[br]than 32, so I'm not going to use

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any of that. How many six teams[br]am I going to need? I'm not

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going to need any of the[br]sixteens either, because 16 is 2

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big 413, so the biggest place[br]value I can use is 8 because

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eight is the bigger out of the[br]place values, which is smaller.

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Done 13 so now if I'm used[br]eight, what's the difference

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between 13 and eight? What is[br]the remainder that I still need

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to build up from the rest of the[br]place? Values so 13 -- 8 makes 5

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and five can be built up from 4[br]an one, which means that I'm not

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going to use any of the tools[br]and here is an important thing

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that any of the place values[br]that are in between the biggest

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and smallest place value that I[br]need to use.

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All the way up to to the power[br]of 0, which is one I need to

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place zeros in between. Because[br]these errors are very, very

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important. These are the secret[br]place Holder zeros. Remember

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that in decimal numbers 502 and[br]52 are very, very different. So

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if I forgotten this placeholder[br]0 here and I'm just writing them

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52, I'm altering the value of[br]the number. So these places were

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placeholder, zeros are extremely

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important. Now. O's I could put[br]in here but so far it's not

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important because I'm just[br]telling by placing zeros in here

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that I'm not using this place[br]values. But these zeros are

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don't really carry any essential[br]information, so these kind of

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zeros at the frontier the so[br]called unnecessary 0 so I don't

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have to write them out.[br]Therefore I can conclude that 13

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in binary is 110.

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1.

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So again, just quickly right up[br]the place value table.

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And the exit decimal[br]numbers as well.

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And then look at another[br]decimal example.

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42 in decimal, what would that[br]be in binary? Now this is a much

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much bigger number than the[br]previous example, and looking at

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42. Is 32 the biggest place[br]value I can get out from 42 or

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can I use the next place value[br]up while the next place value up

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would be 64? 'cause remember,[br]these numbers are always double

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up if I'm going from right to[br]left, which means that 64 is too

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big for my 42. So yes, 32 will[br]be the biggest place value that

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I can use now. What's the[br]remainder? 42 -- 32 gives me a

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nice and simple number.

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10 and again, when I'm looking[br]at this place values, 10 can

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easily build up from 8 and two[br]again. Don't forget about the

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placeholder zeros, so I need to[br]place a 0 here under 16 here on

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the four and here on the one.

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Whether to put a base value on[br]the 64 or not, it's usually your

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choice, but in practical[br]examples you don't really see

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binary numbers starting with[br]zeros, unless this is something

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called like the fixed length[br]place value table is like the 8

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bit binaries, but at the moment[br]we're just looking at general

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binary numbers. Therefore, 42 in[br]decimal is 101010 in binary, and

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again 42 is an even number.[br]So I'm expecting my last digit,

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the one to be 0.

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Let's up our game and look at a[br]much, much bigger number.

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75 in decimal, what would that[br]look like in binary? Now I don't

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know about you, but I'm getting[br]a little bit tighter writing out

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all the powers and they don't[br]really give me that much extra

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information. So what I'm going[br]to do from now on just write out

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the decimal place value[br]equivalent, the one the two, the

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four, the eight, the 16, the 32,[br]the 64, and the next one is 120.

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8 now as soon as you went over[br]the number in question with

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place values, then you know that[br]that place where you will not be

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used and the biggest place value[br]that you can use to make 75 will

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be 64. So again, what is the[br]remainder? So what's the

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difference between 75 and 64?[br]Well easily calculated at 11.

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That means 32 is too big. 16 is[br]too big for eight will be

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sufficient. So if I'm using up

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an 8. What else do I need to[br]difference between 8:00 and

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11:00? Is 3 and three can be[br]built up from two and one.

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Again four, I haven't used so I[br]need to place the zero down to

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indicate that that place value[br]is not used, so the final binary

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answer is 1001011.

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And I'm indicating that this[br]is a binary number here.

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Next example 122 in decimal.[br]What does it look like in

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binary? So again, start with the[br]place values 124, eight, 1630,

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two 64128, now 120. Just a[br]slightly too big. But it means

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that it's still not place value[br]that I'm going to use.

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So the biggest place for you I[br]can use to build up 122 it's 64

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again. So 64 will definitely be[br]used. So what's the remainder?

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How much more do I need to take[br]out from this place values?

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So 122 -- 64 four is too big[br]for two to be taken away from.

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So I need to borrow from here.[br]Therefore one stays here and

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then 12 appears. Here the[br]difference between 12:00 and

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4:00 is 8. Again, I will need[br]to borrow because six is too

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big for one, so zero will be[br]here. 11 will be here and 11

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-- 6 is 5.

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So 58 is the remainder 58. I can[br]take 32 out from it.

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The remainder in this[br]case will be 26.

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26 that can take[br]a 16 out from it.

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The remainder is 10 and I[br]remember at 10 can be built up

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from 8 and two there are two[br]place values that I need to fill

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up with zeros because these are[br]placeholders, the 0, four and

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one. So the binary equivalent.

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Of 122 in debt.

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Symbol[br]is

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1111010.

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The final example on[br]converting decimal numbers

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to binary numbers in this[br]video is 249. Again, start

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with the place values.

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1248

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1630 Two, 64128 and[br]256. The next one

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would be 512.

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But that is too big and so is[br]256. I just needed to make sure

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that this number is actually[br]bigger than my place value,

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because if I start with a place[br]where you that is slightly too

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small, then I will run out of[br]digits that together. And don't

0:11:00.380,0:11:04.496
forget you can only put ones and[br]zeros in here, so I can't say

0:11:04.496,0:11:08.612
that I'm using two of the 32[br]because two is not part of my

0:11:08.612,0:11:10.376
number system. I only have got

0:11:10.376,0:11:16.488
one and 0. So the biggest place[br]value I can take out is the 128,

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so I'm going to use one of the[br]128 and I need to see what's my

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remainder. 9 -- 8 is one 4 -- 2[br]is 2 and 2 -- 1 is 1. So I

0:11:29.403,0:11:33.831
still got quite a big number,[br]but that number is smaller than

0:11:33.831,0:11:38.628
128 so I'm on the right track.[br]Anytime when you got a remainder

0:11:38.628,0:11:42.318
here, your remainder should[br]always be smaller than the last

0:11:42.318,0:11:43.794
place where you've used.

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But 121 is less than 128, so I'm[br]on the right track. So 64 I will

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definitely use. But what's the[br]remainder? Again? So four is too

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big for one to be taken away[br]from DEF running to borrow again

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11 -- 4 makes it 7 again six I[br]can take away 6 from 1 so need

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to borrow again. 11 -- 6 gives[br]it 5 so I've got 57 as remainder

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after I've used the 64.

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The next place why is 32? So[br]take away 32 gives me 7 -- 2

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gifts, five, 5 -- 3 gifts two,[br]so I'm using that after the 32

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I'm going to use the 16 as well.[br]So the difference here now 25 --

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16. Is 9 an? I remember[br]the nine can be easily build up

0:12:40.932,0:12:48.030
from 8:00 and 1:00, so I'm going[br]to need two placeholder zeros in

0:12:48.030,0:12:54.036
here, so 249 in decimal is[br]11111001. In binary. This was

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our last example. I'm hoping[br]that you understand how to

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convert decimal numbers to[br]binary numbers using the place

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values. And that you probably[br]find it easy in the next

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minute, you will have some[br]opportunities to practice

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these questions yourself and[br]you will have the answers

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after.

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So these are the practice[br]questions.

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And here are the answers.