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https:/.../emt031080p.mp4

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    Welcome to the third video on
    binary numbers. In this video,
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    we're going to look at how to
    convert decimal numbers to
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    binary numbers, and in this
    particular video I'm going to
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    show you the place value table
    or addition method. In this
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    video. Our goal is to find out
    what place, why you combinations
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    build out the decimal number
    that needs to convert it into
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    binary. I can only use once for
    any of the place values.
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    As in the binary number system,
    I only have the digit one and
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    zero in a further video I'm
    going to show you a different
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    method called the division
    method, so it doesn't matter
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    which way you learn to convert
    binary numbers. Whatever method
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    you feel more comfortable with,
    just stick with that and then
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    your answer will be just as
    correct. If you are doing it
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    with the other method. So the
    table method, it's called the
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    place value table method because
    we are going to use the binary
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    place value. To convert decimal
    numbers for that again, let's
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    look at what the binary place
    value table looks like. So
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    remember that the place values
    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
    continue this as long as you
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    want it to, but these numbers
    also translate back down to
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    normal decimal numbers without
    the power form. So 2 to the
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    power of 0 would be 1. Two to
    the power of 1 would be 2. Two
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    to the power of two would be 4.
    To do a power of three is 8 to
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    two power of four is 16 and 2
    two power of five is 32. So if I
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    have got. Decimal number, such
    as the teen? How am I going to
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    use my knowledge of the place
    values to convert this number?
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    Into binary.
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    So what I'm going to do, I'm
    going to look at how many 30
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    twos would I need to use? Use to
    make the 13 while it's too big
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    because 13 is much much smaller
    than 32, so I'm not going to use
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    any of that. How many six teams
    am I going to need? I'm not
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    going to need any of the
    sixteens either, because 16 is 2
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    big 413, so the biggest place
    value I can use is 8 because
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    eight is the bigger out of the
    place values, which is smaller.
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    Done 13 so now if I'm used
    eight, what's the difference
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    between 13 and eight? What is
    the remainder that I still need
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    to build up from the rest of the
    place? Values so 13 -- 8 makes 5
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    and five can be built up from 4
    an one, which means that I'm not
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    going to use any of the tools
    and here is an important thing
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    that any of the place values
    that are in between the biggest
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    and smallest place value that I
    need to use.
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    All the way up to to the power
    of 0, which is one I need to
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    place zeros in between. Because
    these errors are very, very
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    important. These are the secret
    place Holder zeros. Remember
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    that in decimal numbers 502 and
    52 are very, very different. So
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    if I forgotten this placeholder
    0 here and I'm just writing them
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    52, I'm altering the value of
    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
    in here but so far it's not
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    important because I'm just
    telling by placing zeros in here
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    that I'm not using this place
    values. But these zeros are
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    don't really carry any essential
    information, so these kind of
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    zeros at the frontier the so
    called unnecessary 0 so I don't
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    have to write them out.
    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
    the place value table.
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    And the exit decimal
    numbers as well.
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    And then look at another
    decimal example.
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    42 in decimal, what would that
    be in binary? Now this is a much
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    much bigger number than the
    previous example, and looking at
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    42. Is 32 the biggest place
    value I can get out from 42 or
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    can I use the next place value
    up while the next place value up
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    would be 64? 'cause remember,
    these numbers are always double
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    up if I'm going from right to
    left, which means that 64 is too
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    big for my 42. So yes, 32 will
    be the biggest place value that
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    I can use now. What's the
    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
    at this place values, 10 can
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    easily build up from 8 and two
    again. Don't forget about the
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    placeholder zeros, so I need to
    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
    the 64 or not, it's usually your
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    choice, but in practical
    examples you don't really see
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    binary numbers starting with
    zeros, unless this is something
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    called like the fixed length
    place value table is like the 8
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    bit binaries, but at the moment
    we're just looking at general
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    binary numbers. Therefore, 42 in
    decimal is 101010 in binary, and
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    again 42 is an even number.
    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
    much, much bigger number.
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    75 in decimal, what would that
    look like in binary? Now I don't
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    know about you, but I'm getting
    a little bit tighter writing out
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    all the powers and they don't
    really give me that much extra
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    information. So what I'm going
    to do from now on just write out
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    the decimal place value
    equivalent, the one the two, the
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    four, the eight, the 16, the 32,
    the 64, and the next one is 120.
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    8 now as soon as you went over
    the number in question with
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    place values, then you know that
    that place where you will not be
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    used and the biggest place value
    that you can use to make 75 will
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    be 64. So again, what is the
    remainder? So what's the
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    difference between 75 and 64?
    Well easily calculated at 11.
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    That means 32 is too big. 16 is
    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
    difference between 8:00 and
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    11:00? Is 3 and three can be
    built up from two and one.
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    Again four, I haven't used so I
    need to place the zero down to
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    indicate that that place value
    is not used, so the final binary
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    answer is 1001011.
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    And I'm indicating that this
    is a binary number here.
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    Next example 122 in decimal.
    What does it look like in
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    binary? So again, start with the
    place values 124, eight, 1630,
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    two 64128, now 120. Just a
    slightly too big. But it means
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    that it's still not place value
    that I'm going to use.
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    So the biggest place for you I
    can use to build up 122 it's 64
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    again. So 64 will definitely be
    used. So what's the remainder?
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    How much more do I need to take
    out from this place values?
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    So 122 -- 64 four is too big
    for two to be taken away from.
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    So I need to borrow from here.
    Therefore one stays here and
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    then 12 appears. Here the
    difference between 12:00 and
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    4:00 is 8. Again, I will need
    to borrow because six is too
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    big for one, so zero will be
    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
    take 32 out from it.
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    The remainder in this
    case will be 26.
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    26 that can take
    a 16 out from it.
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    The remainder is 10 and I
    remember at 10 can be built up
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    from 8 and two there are two
    place values that I need to fill
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    up with zeros because these are
    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
    is
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    1111010.
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    The final example on
    converting decimal numbers
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    to binary numbers in this
    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
    256. The next one
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    would be 512.
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    But that is too big and so is
    256. I just needed to make sure
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    that this number is actually
    bigger than my place value,
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    because if I start with a place
    where you that is slightly too
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    small, then I will run out of
    digits that together. And don't
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    forget you can only put ones and
    zeros in here, so I can't say
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    that I'm using two of the 32
    because two is not part of my
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    number system. I only have got
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    one and 0. So the biggest place
    value I can take out is the 128,
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    so I'm going to use one of the
    128 and I need to see what's my
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    remainder. 9 -- 8 is one 4 -- 2
    is 2 and 2 -- 1 is 1. So I
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    still got quite a big number,
    but that number is smaller than
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    128 so I'm on the right track.
    Anytime when you got a remainder
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    here, your remainder should
    always be smaller than the last
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    place where you've used.
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    But 121 is less than 128, so I'm
    on the right track. So 64 I will
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    definitely use. But what's the
    remainder? Again? So four is too
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    big for one to be taken away
    from DEF running to borrow again
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    11 -- 4 makes it 7 again six I
    can take away 6 from 1 so need
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    to borrow again. 11 -- 6 gives
    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
    take away 32 gives me 7 -- 2
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    gifts, five, 5 -- 3 gifts two,
    so I'm using that after the 32
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    I'm going to use the 16 as well.
    So the difference here now 25 --
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    16. Is 9 an? I remember
    the nine can be easily build up
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    from 8:00 and 1:00, so I'm going
    to need two placeholder zeros in
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    here, so 249 in decimal is
    11111001. In binary. This was
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    our last example. I'm hoping
    that you understand how to
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    convert decimal numbers to
    binary numbers using the place
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    values. And that you probably
    find it easy in the next
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    minute, you will have some
    opportunities to practice
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    these questions yourself and
    you will have the answers
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    after.
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    So these are the practice
    questions.
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    And here are the answers.
Title:
https:/.../emt031080p.mp4
Video Language:
English
Duration:
13:57

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