1 00:00:20,410 --> 00:00:24,656 Welcome to the third video on binary numbers. In this video, 2 00:00:24,656 --> 00:00:28,902 we're going to look at how to convert decimal numbers to 3 00:00:28,902 --> 00:00:32,762 binary numbers, and in this particular video I'm going to 4 00:00:32,762 --> 00:00:37,008 show you the place value table or addition method. In this 5 00:00:37,008 --> 00:00:41,640 video. Our goal is to find out what place, why you combinations 6 00:00:41,640 --> 00:00:45,886 build out the decimal number that needs to convert it into 7 00:00:45,886 --> 00:00:50,518 binary. I can only use once for any of the place values. 8 00:00:50,650 --> 00:00:54,966 As in the binary number system, I only have the digit one and 9 00:00:54,966 --> 00:00:58,950 zero in a further video I'm going to show you a different 10 00:00:58,950 --> 00:01:01,938 method called the division method, so it doesn't matter 11 00:01:01,938 --> 00:01:05,258 which way you learn to convert binary numbers. Whatever method 12 00:01:05,258 --> 00:01:08,910 you feel more comfortable with, just stick with that and then 13 00:01:08,910 --> 00:01:12,894 your answer will be just as correct. If you are doing it 14 00:01:12,894 --> 00:01:16,546 with the other method. So the table method, it's called the 15 00:01:16,546 --> 00:01:20,530 place value table method because we are going to use the binary 16 00:01:20,530 --> 00:01:24,504 place value. To convert decimal numbers for that again, let's 17 00:01:24,504 --> 00:01:28,827 look at what the binary place value table looks like. So 18 00:01:28,827 --> 00:01:32,757 remember that the place values were due to the 0. 19 00:01:33,540 --> 00:01:36,390 2 to one. Two to two. 20 00:01:37,190 --> 00:01:38,249 Two to three. 21 00:01:39,150 --> 00:01:44,205 2 to 4, two to five etc. You can continue this as long as you 22 00:01:44,205 --> 00:01:47,912 want it to, but these numbers also translate back down to 23 00:01:47,912 --> 00:01:51,619 normal decimal numbers without the power form. So 2 to the 24 00:01:51,619 --> 00:01:57,011 power of 0 would be 1. Two to the power of 1 would be 2. Two 25 00:01:57,011 --> 00:02:02,740 to the power of two would be 4. To do a power of three is 8 to 26 00:02:02,740 --> 00:02:08,469 two power of four is 16 and 2 two power of five is 32. So if I 27 00:02:08,469 --> 00:02:14,241 have got. Decimal number, such as the teen? How am I going to 28 00:02:14,241 --> 00:02:18,872 use my knowledge of the place values to convert this number? 29 00:02:19,630 --> 00:02:20,520 Into binary. 30 00:02:22,360 --> 00:02:26,826 So what I'm going to do, I'm going to look at how many 30 31 00:02:26,826 --> 00:02:31,611 twos would I need to use? Use to make the 13 while it's too big 32 00:02:31,611 --> 00:02:36,077 because 13 is much much smaller than 32, so I'm not going to use 33 00:02:36,077 --> 00:02:40,543 any of that. How many six teams am I going to need? I'm not 34 00:02:40,543 --> 00:02:44,371 going to need any of the sixteens either, because 16 is 2 35 00:02:44,371 --> 00:02:48,518 big 413, so the biggest place value I can use is 8 because 36 00:02:48,518 --> 00:02:52,346 eight is the bigger out of the place values, which is smaller. 37 00:02:52,430 --> 00:02:56,170 Done 13 so now if I'm used eight, what's the difference 38 00:02:56,170 --> 00:03:00,250 between 13 and eight? What is the remainder that I still need 39 00:03:00,250 --> 00:03:05,690 to build up from the rest of the place? Values so 13 -- 8 makes 5 40 00:03:05,690 --> 00:03:10,790 and five can be built up from 4 an one, which means that I'm not 41 00:03:10,790 --> 00:03:15,210 going to use any of the tools and here is an important thing 42 00:03:15,210 --> 00:03:19,290 that any of the place values that are in between the biggest 43 00:03:19,290 --> 00:03:22,350 and smallest place value that I need to use. 44 00:03:22,470 --> 00:03:28,758 All the way up to to the power of 0, which is one I need to 45 00:03:28,758 --> 00:03:32,688 place zeros in between. Because these errors are very, very 46 00:03:32,688 --> 00:03:36,225 important. These are the secret place Holder zeros. Remember 47 00:03:36,225 --> 00:03:40,941 that in decimal numbers 502 and 52 are very, very different. So 48 00:03:40,941 --> 00:03:45,657 if I forgotten this placeholder 0 here and I'm just writing them 49 00:03:45,657 --> 00:03:50,373 52, I'm altering the value of the number. So these places were 50 00:03:50,373 --> 00:03:51,945 placeholder, zeros are extremely 51 00:03:51,945 --> 00:03:58,241 important. Now. O's I could put in here but so far it's not 52 00:03:58,241 --> 00:04:02,251 important because I'm just telling by placing zeros in here 53 00:04:02,251 --> 00:04:06,662 that I'm not using this place values. But these zeros are 54 00:04:06,662 --> 00:04:10,672 don't really carry any essential information, so these kind of 55 00:04:10,672 --> 00:04:15,484 zeros at the frontier the so called unnecessary 0 so I don't 56 00:04:15,484 --> 00:04:19,895 have to write them out. Therefore I can conclude that 13 57 00:04:19,895 --> 00:04:21,499 in binary is 110. 58 00:04:22,360 --> 00:04:22,770 1. 59 00:04:27,000 --> 00:04:30,300 So again, just quickly right up the place value table. 60 00:04:32,310 --> 00:04:35,369 And the exit decimal numbers as well. 61 00:04:41,060 --> 00:04:44,427 And then look at another decimal example. 62 00:04:48,090 --> 00:04:54,978 42 in decimal, what would that be in binary? Now this is a much 63 00:04:54,978 --> 00:04:59,898 much bigger number than the previous example, and looking at 64 00:04:59,898 --> 00:05:06,005 42. Is 32 the biggest place value I can get out from 42 or 65 00:05:06,005 --> 00:05:11,255 can I use the next place value up while the next place value up 66 00:05:11,255 --> 00:05:15,005 would be 64? 'cause remember, these numbers are always double 67 00:05:15,005 --> 00:05:20,255 up if I'm going from right to left, which means that 64 is too 68 00:05:20,255 --> 00:05:25,505 big for my 42. So yes, 32 will be the biggest place value that 69 00:05:25,505 --> 00:05:30,380 I can use now. What's the remainder? 42 -- 32 gives me a 70 00:05:30,380 --> 00:05:31,880 nice and simple number. 71 00:05:32,020 --> 00:05:36,580 10 and again, when I'm looking at this place values, 10 can 72 00:05:36,580 --> 00:05:41,140 easily build up from 8 and two again. Don't forget about the 73 00:05:41,140 --> 00:05:46,460 placeholder zeros, so I need to place a 0 here under 16 here on 74 00:05:46,460 --> 00:05:49,120 the four and here on the one. 75 00:05:50,140 --> 00:05:55,348 Whether to put a base value on the 64 or not, it's usually your 76 00:05:55,348 --> 00:05:58,696 choice, but in practical examples you don't really see 77 00:05:58,696 --> 00:06:02,044 binary numbers starting with zeros, unless this is something 78 00:06:02,044 --> 00:06:06,508 called like the fixed length place value table is like the 8 79 00:06:06,508 --> 00:06:10,600 bit binaries, but at the moment we're just looking at general 80 00:06:10,600 --> 00:06:17,715 binary numbers. Therefore, 42 in decimal is 101010 in binary, and 81 00:06:17,715 --> 00:06:25,095 again 42 is an even number. So I'm expecting my last digit, 82 00:06:25,095 --> 00:06:28,170 the one to be 0. 83 00:06:31,740 --> 00:06:35,052 Let's up our game and look at a much, much bigger number. 84 00:06:36,230 --> 00:06:41,443 75 in decimal, what would that look like in binary? Now I don't 85 00:06:41,443 --> 00:06:46,255 know about you, but I'm getting a little bit tighter writing out 86 00:06:46,255 --> 00:06:51,067 all the powers and they don't really give me that much extra 87 00:06:51,067 --> 00:06:56,280 information. So what I'm going to do from now on just write out 88 00:06:56,280 --> 00:07:00,290 the decimal place value equivalent, the one the two, the 89 00:07:00,290 --> 00:07:06,305 four, the eight, the 16, the 32, the 64, and the next one is 120. 90 00:07:06,420 --> 00:07:11,321 8 now as soon as you went over the number in question with 91 00:07:11,321 --> 00:07:16,222 place values, then you know that that place where you will not be 92 00:07:16,222 --> 00:07:21,500 used and the biggest place value that you can use to make 75 will 93 00:07:21,500 --> 00:07:25,647 be 64. So again, what is the remainder? So what's the 94 00:07:25,647 --> 00:07:29,417 difference between 75 and 64? Well easily calculated at 11. 95 00:07:29,417 --> 00:07:34,695 That means 32 is too big. 16 is too big for eight will be 96 00:07:34,695 --> 00:07:36,957 sufficient. So if I'm using up 97 00:07:36,957 --> 00:07:41,550 an 8. What else do I need to difference between 8:00 and 98 00:07:41,550 --> 00:07:46,399 11:00? Is 3 and three can be built up from two and one. 99 00:07:47,100 --> 00:07:54,856 Again four, I haven't used so I need to place the zero down to 100 00:07:54,856 --> 00:08:01,504 indicate that that place value is not used, so the final binary 101 00:08:01,504 --> 00:08:03,166 answer is 1001011. 102 00:08:03,920 --> 00:08:06,110 And I'm indicating that this is a binary number here. 103 00:08:10,560 --> 00:08:17,347 Next example 122 in decimal. What does it look like in 104 00:08:17,347 --> 00:08:24,134 binary? So again, start with the place values 124, eight, 1630, 105 00:08:24,134 --> 00:08:31,538 two 64128, now 120. Just a slightly too big. But it means 106 00:08:31,538 --> 00:08:38,325 that it's still not place value that I'm going to use. 107 00:08:38,630 --> 00:08:44,120 So the biggest place for you I can use to build up 122 it's 64 108 00:08:44,120 --> 00:08:48,146 again. So 64 will definitely be used. So what's the remainder? 109 00:08:48,146 --> 00:08:52,904 How much more do I need to take out from this place values? 110 00:08:54,380 --> 00:08:59,855 So 122 -- 64 four is too big for two to be taken away from. 111 00:08:59,855 --> 00:09:04,235 So I need to borrow from here. Therefore one stays here and 112 00:09:04,235 --> 00:09:07,520 then 12 appears. Here the difference between 12:00 and 113 00:09:07,520 --> 00:09:12,265 4:00 is 8. Again, I will need to borrow because six is too 114 00:09:12,265 --> 00:09:17,375 big for one, so zero will be here. 11 will be here and 11 115 00:09:17,375 --> 00:09:18,835 -- 6 is 5. 116 00:09:20,690 --> 00:09:26,423 So 58 is the remainder 58. I can take 32 out from it. 117 00:09:29,690 --> 00:09:33,026 The remainder in this case will be 26. 118 00:09:35,090 --> 00:09:37,754 26 that can take a 16 out from it. 119 00:09:39,700 --> 00:09:44,796 The remainder is 10 and I remember at 10 can be built up 120 00:09:44,796 --> 00:09:50,284 from 8 and two there are two place values that I need to fill 121 00:09:50,284 --> 00:09:54,596 up with zeros because these are placeholders, the 0, four and 122 00:09:54,596 --> 00:09:56,556 one. So the binary equivalent. 123 00:09:57,510 --> 00:10:00,038 Of 122 in debt. 124 00:10:00,120 --> 00:10:04,312 Symbol is 125 00:10:04,312 --> 00:10:06,408 1111010. 126 00:10:16,840 --> 00:10:20,389 The final example on converting decimal numbers 127 00:10:20,389 --> 00:10:25,459 to binary numbers in this video is 249. Again, start 128 00:10:25,459 --> 00:10:27,487 with the place values. 129 00:10:29,320 --> 00:10:33,130 1248 130 00:10:34,140 --> 00:10:42,076 1630 Two, 64128 and 256. The next one 131 00:10:42,076 --> 00:10:45,052 would be 512. 132 00:10:45,680 --> 00:10:50,090 But that is too big and so is 256. I just needed to make sure 133 00:10:50,090 --> 00:10:53,030 that this number is actually bigger than my place value, 134 00:10:53,030 --> 00:10:56,852 because if I start with a place where you that is slightly too 135 00:10:56,852 --> 00:11:00,380 small, then I will run out of digits that together. And don't 136 00:11:00,380 --> 00:11:04,496 forget you can only put ones and zeros in here, so I can't say 137 00:11:04,496 --> 00:11:08,612 that I'm using two of the 32 because two is not part of my 138 00:11:08,612 --> 00:11:10,376 number system. I only have got 139 00:11:10,376 --> 00:11:16,488 one and 0. So the biggest place value I can take out is the 128, 140 00:11:16,488 --> 00:11:22,392 so I'm going to use one of the 128 and I need to see what's my 141 00:11:22,392 --> 00:11:29,403 remainder. 9 -- 8 is one 4 -- 2 is 2 and 2 -- 1 is 1. So I 142 00:11:29,403 --> 00:11:33,831 still got quite a big number, but that number is smaller than 143 00:11:33,831 --> 00:11:38,628 128 so I'm on the right track. Anytime when you got a remainder 144 00:11:38,628 --> 00:11:42,318 here, your remainder should always be smaller than the last 145 00:11:42,318 --> 00:11:43,794 place where you've used. 146 00:11:44,360 --> 00:11:50,552 But 121 is less than 128, so I'm on the right track. So 64 I will 147 00:11:50,552 --> 00:11:54,809 definitely use. But what's the remainder? Again? So four is too 148 00:11:54,809 --> 00:11:59,840 big for one to be taken away from DEF running to borrow again 149 00:11:59,840 --> 00:12:06,419 11 -- 4 makes it 7 again six I can take away 6 from 1 so need 150 00:12:06,419 --> 00:12:12,224 to borrow again. 11 -- 6 gives it 5 so I've got 57 as remainder 151 00:12:12,224 --> 00:12:14,159 after I've used the 64. 152 00:12:14,430 --> 00:12:20,820 The next place why is 32? So take away 32 gives me 7 -- 2 153 00:12:20,820 --> 00:12:26,784 gifts, five, 5 -- 3 gifts two, so I'm using that after the 32 154 00:12:26,784 --> 00:12:33,174 I'm going to use the 16 as well. So the difference here now 25 -- 155 00:12:33,174 --> 00:12:40,932 16. Is 9 an? I remember the nine can be easily build up 156 00:12:40,932 --> 00:12:48,030 from 8:00 and 1:00, so I'm going to need two placeholder zeros in 157 00:12:48,030 --> 00:12:54,036 here, so 249 in decimal is 11111001. In binary. This was 158 00:12:54,036 --> 00:12:59,496 our last example. I'm hoping that you understand how to 159 00:12:59,496 --> 00:13:04,410 convert decimal numbers to binary numbers using the place 160 00:13:04,410 --> 00:13:08,870 values. And that you probably find it easy in the next 161 00:13:08,870 --> 00:13:11,694 minute, you will have some opportunities to practice 162 00:13:11,694 --> 00:13:14,871 these questions yourself and you will have the answers 163 00:13:14,871 --> 00:13:15,224 after. 164 00:13:17,380 --> 00:13:19,456 So these are the practice questions. 165 00:13:25,220 --> 00:13:26,800 And here are the answers.