0:00:20.340,0:00:24.685
In this video, we're going to[br]look at the binary fractions

0:00:24.685,0:00:29.030
again, but from a slightly[br]different angle. I will show you

0:00:29.030,0:00:32.585
an alternative method to convert[br]decimal fractions into binary

0:00:32.585,0:00:37.325
fractions, which will work in[br]most cases. I will show you some

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examples, and again I will draw[br]your attention to the

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limitations of this method.

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So let's look at an example. So[br]what happens if you would need

0:00:49.768,0:00:55.480
to convert 13.3125 into binary?[br]From here on, I'm going to split

0:00:55.480,0:01:00.716
the number I'm going to convert[br]the whole number and the

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fraction part separately. The[br]whole number 413. We're going to

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use the normal way of converting[br]decimal numbers into binary's,

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so use the place where you

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table. 124 eight 16 So what[br]combination of these makes up 13

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while 8 + 4 makes 12[br]+ 1 makes 13, so the

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whole part of this decimal[br]fraction is 1101.

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How about the decimal part? What[br]I'm going to do now? I'm going

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to just separately right down[br]the decimal part.

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And the trick is here to keep[br]doubling the number. So what's

0:01:48.846,0:01:51.111
the double of three 125?

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Double of five is 10, carried[br]A1 double of two is 4 + 1

0:01:57.524,0:02:03.614
makes 5 double of two is 2 and[br]double of three is 6. Now what

0:02:03.614,0:02:08.892
happened in here is that I do[br]not have any overflow into the

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whole number part of this part[br]of the structure and therefore

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I'm going to record 0 after[br]the radix point.

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Then I'm going to keep doubling.[br]Obviously double of 00, so I

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don't even need to think about[br]that. 2 * 5 is 10, carried A one

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2 * 2 is 4 + 1 makes it five.[br]2 * 6 is 12. So record the two

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and the one is an overflow. So[br]this digit. Now I'm going to

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pick up and record as the 2nd[br]digit of the.

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Binary fraction.

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Imagine like if you were picking[br]this one up from here recorded

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here. So from now it disappears.[br]The next step that I'm going to

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do, I'm going to double again,[br]but without the whole part. So

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I'm just going to 55.

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Which may extend double of two[br]is 4 + 1 is 5 and there again I

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have got no overflow into the[br]whole parts, so I'm going to

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record a 0 here.

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Double again 2 * 5 is 10.

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There is an overflow. This is my[br]last digit here because from

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here on I've got no more[br]fractional parts, so you stop

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when you end up with a zero in[br]the fractional part. Now pulling

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the two together.

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13.3125, in decimal[br]is the same

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as 1101 Radix[br].0101 in binary.

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The second example is[br]9.1875. Again separates the

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number into whole and[br]fractional part. The whole

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part is 9.

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Which is 8 + 1, so eight[br]no four, no two and one.

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The decimal part now.

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0.1875 Let's keep doubling it. 2[br]* 5 is 10 carry one. 2 *

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7 is 14 + 1 makes it 15[br]carried A one 2 * 8016 + 1

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is 17, carried A one. 2 * 1[br]is 2 + 1 is 3 again. I

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did not have any overflow into[br]the whole number part, so the

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1st digit behind the radix point[br]that I'm going to record.

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Is a 0.

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Double again.

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2 * 5 is 10, carried a warm 2[br]* 7 is 14 + 1 makes it 15.

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Carry the one 2 * 3 is 6 +[br]1, seven again no overflow into

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the whole part. So I'm going to[br]record 0 here.

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Double again 2 * 5 is 10.[br]Carried a wamp. 2 * 7 is 14

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+ 1 makes it 15. Now I have[br]gotten overflowing here, so I'm

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going to record this.

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As the next digit after the[br]radix point and then double

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again, don't forget that this[br]one is not here anymore because

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I picked up and recorded it in

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here. So 2 * 5 is 10.

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So the next digit is 1 again. So[br]for the two things together.

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9.1875 in decimal[br]is the same

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as 1001 radix[br].0011 in binary.

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The next number is 0.6875.[br]Luckily, this number doesn't

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have any whole parts, so we just[br]need to concentrate on the

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decimal fraction part so I can[br]just simply keep doubling this

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number. 2 * 5 is 10. Carry[br]one 2 * 7 is 14 +

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1 is 15, carried A one 2[br]* 8016 + 1 is 17.

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Either one 2 * 6 is.

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12 + 1 is 13, so I've got an

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overflow. So the 1st digit[br]I'm going to record behind

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the radix point is 1.

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Double 2 * 5 is 10, carried A[br]one 2 * 7 is 14 + 1 makes it

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15. Carry the one 2 * 3 is 6 +[br]1. Seven this case I did not

0:07:07.575,0:07:12.346
have an overflow, so the next[br]day did after the radix point is

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0 double again.

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2 * 510 carried A one 2 *[br]7 is 14 + 1 makes it 15

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overflow, so the next digit is[br]1. Remember that's gone now and

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O makes no difference there. 2 *[br]5 is 10, so it's 1.0. So we

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have got one more digit here,[br]which is a one so 0.6875 in

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decimal. As the same is O radix

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.1011. In binary.

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Now let's look at a nice and[br]easy decimal number. The one

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that we didn't quite know how to[br]deal with at the end of the last

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video. So let's look at 3.4. So[br]let's separate the number again

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in two whole and fractional[br]parts, so three is 2 + 1, which

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is 1 one and the fractional[br]part. Let's just double 2 * 4 is

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0.8, so after the radix point,[br]the 1st digit will be 0.

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2 * 8 is 16.

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So 1.6 the next digit is 1.

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2 * 6 is 12, so 2.

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.1 again carry the one.

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Double of two is 4 no[br]carry, 004 is 8 again, no

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overflow. Double F8 is 16 so[br]I've got one here now.

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WF6 is 12. I've got another one[br]here. Now double F2 is 4.

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Put down a zero and hold on.[br]I'm repeating myself look.

0:09:06.750,0:09:10.646
Point 4.8. 6248624862

0:09:10.646,0:09:18.035
so this.[br]Simple decimal fraction 3.4 is

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an infinitely recurring binary[br]fraction, so that's again shows

0:09:23.102,0:09:28.732
you some difficulties when it[br]comes to converting that simple

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fractions to binary fractions.[br]So this would be.

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3.4 in decimal[br]would be 1

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one radix .01100110.

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212345678 places.

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As I mentioned it in the last[br]video, this is something that's

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fundamentally inherent property[br]of the binary number system. We

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can't really do anything about[br]it, but by using more binary

0:10:03.802,0:10:07.812
digits to represent the decimal[br]numbers, we can minimize this

0:10:07.812,0:10:15.535
problem. Let's look[br]at another simple

0:10:15.535,0:10:21.416
example, 4.715. Separate[br]it again to whole and

0:10:21.416,0:10:27.019
fractional. Part 4 is just 100.[br]Remember this is 1, two and four

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and the fractional part will be[br]0.715. Now let's keep doubling

0:10:31.760,0:10:38.656
it. 2 * 5 is then carried A1[br]three times, one is 2 + 1 is

0:10:38.656,0:10:45.552
three 2 * 7 is 14, so I've got[br]one of the overflow, so the 1st

0:10:45.552,0:10:49.431
digit in after the radix point[br]will be one.

0:10:49.490,0:10:53.901
Now those two unnecessary[br]anymore. So double again 2 * 3

0:10:53.901,0:10:59.916
is Six 2 * 4 is 8, so there[br]is no overflow. This digit will

0:10:59.916,0:11:01.119
be at 0.

0:11:02.090,0:11:08.516
Double it again. 2 * 6 is 12,[br]carried A one 2 * 8016 + 1 makes

0:11:08.516,0:11:12.674
17. I've got an overflow here[br]now, so that's number one.

0:11:13.470,0:11:20.529
Double it again. 2 * 2 is[br]four 2 * 7 is 14.

0:11:21.170,0:11:23.340
So next digit is 1.

0:11:24.230,0:11:30.080
Double against that digits gone[br]2 * 4 is eight 2 * 4 is 8

0:11:30.080,0:11:35.150
no overflow, so this digit[br]phobia, 0 double 2 * 8 is 16,

0:11:35.150,0:11:42.170
carried A one 2 * 8 is 16 + 1[br]is 17. So there is one as an

0:11:42.170,0:11:47.630
overflow that's gone. Now double[br]again 2 * 6 is 12, carried A one

0:11:47.630,0:11:54.260
2 * 7 is 14 + 1 makes it 15,[br]so I've got one as an overflow.

0:11:54.300,0:12:01.440
Double again 2 * 2 is[br]four 2 * 5 is 10.

0:12:01.970,0:12:04.190
Overflow, so that's another one

0:12:04.190,0:12:10.380
in there. Times 2 is 8000 is the[br]next digit? Well, I don't know

0:12:10.380,0:12:14.780
about you, but I'm getting[br]exhausted in here and look there

0:12:14.780,0:12:19.580
is not even a sign anywhere for[br]a repetition, so this function

0:12:19.580,0:12:23.580
looks even worse than the[br]previous one. And again just

0:12:23.580,0:12:28.780
look at it how simple this is in[br]decimal. So yes, the binary

0:12:28.780,0:12:32.780
number system indeed have got[br]quite a few limitations which

0:12:32.780,0:12:35.180
can get quite a bit annoying

0:12:35.180,0:12:39.500
well. What kind of things have[br]been discovered about the binary

0:12:39.500,0:12:43.380
number system? Well, basically[br]we know that not all decimal

0:12:43.380,0:12:46.872
fractions can be expressed as a[br]finite binary fraction.

0:12:46.872,0:12:50.752
Unfortunately, this cannot be[br]avoided, but can be minimized by

0:12:50.752,0:12:55.408
using more bids. Also, if you[br]look at the examples through the

0:12:55.408,0:13:00.064
video again, you can see that[br]the radix point is different for

0:13:00.064,0:13:03.944
different numbers, so the[br]position of the radix point is

0:13:03.944,0:13:05.884
changing from number to number.

0:13:05.930,0:13:09.857
That can get quite confusing for[br]the computer, but Luckily for

0:13:09.857,0:13:13.784
this problem we do have a[br]solution and that is the

0:13:13.784,0:13:17.711
floating point notation which we[br]will talk about in more details

0:13:17.711,0:13:21.638
in one of the following videos.[br]For now I've prepared some

0:13:21.638,0:13:26.279
examples for you, so please look[br]at them. Try them and you will

0:13:26.279,0:13:29.849
find the answers later, so these[br]are the practice questions.

0:13:35.580,0:13:37.560
And here are the answers.