Convert 52.234 377 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

How to convert the decimal number 52.234 377(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to the binary (base 2) the integer part: 52.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.

  • division = quotient + remainder;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

52(10) =


11 0100(2)


3. Convert to the binary (base 2) the fractional part: 0.234 377.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.234 377 × 2 = 0 + 0.468 754;
  • 2) 0.468 754 × 2 = 0 + 0.937 508;
  • 3) 0.937 508 × 2 = 1 + 0.875 016;
  • 4) 0.875 016 × 2 = 1 + 0.750 032;
  • 5) 0.750 032 × 2 = 1 + 0.500 064;
  • 6) 0.500 064 × 2 = 1 + 0.000 128;
  • 7) 0.000 128 × 2 = 0 + 0.000 256;
  • 8) 0.000 256 × 2 = 0 + 0.000 512;
  • 9) 0.000 512 × 2 = 0 + 0.001 024;
  • 10) 0.001 024 × 2 = 0 + 0.002 048;
  • 11) 0.002 048 × 2 = 0 + 0.004 096;
  • 12) 0.004 096 × 2 = 0 + 0.008 192;
  • 13) 0.008 192 × 2 = 0 + 0.016 384;
  • 14) 0.016 384 × 2 = 0 + 0.032 768;
  • 15) 0.032 768 × 2 = 0 + 0.065 536;
  • 16) 0.065 536 × 2 = 0 + 0.131 072;
  • 17) 0.131 072 × 2 = 0 + 0.262 144;
  • 18) 0.262 144 × 2 = 0 + 0.524 288;
  • 19) 0.524 288 × 2 = 1 + 0.048 576;
  • 20) 0.048 576 × 2 = 0 + 0.097 152;
  • 21) 0.097 152 × 2 = 0 + 0.194 304;
  • 22) 0.194 304 × 2 = 0 + 0.388 608;
  • 23) 0.388 608 × 2 = 0 + 0.777 216;
  • 24) 0.777 216 × 2 = 1 + 0.554 432;
  • 25) 0.554 432 × 2 = 1 + 0.108 864;
  • 26) 0.108 864 × 2 = 0 + 0.217 728;
  • 27) 0.217 728 × 2 = 0 + 0.435 456;
  • 28) 0.435 456 × 2 = 0 + 0.870 912;
  • 29) 0.870 912 × 2 = 1 + 0.741 824;
  • 30) 0.741 824 × 2 = 1 + 0.483 648;
  • 31) 0.483 648 × 2 = 0 + 0.967 296;
  • 32) 0.967 296 × 2 = 1 + 0.934 592;
  • 33) 0.934 592 × 2 = 1 + 0.869 184;
  • 34) 0.869 184 × 2 = 1 + 0.738 368;
  • 35) 0.738 368 × 2 = 1 + 0.476 736;
  • 36) 0.476 736 × 2 = 0 + 0.953 472;
  • 37) 0.953 472 × 2 = 1 + 0.906 944;
  • 38) 0.906 944 × 2 = 1 + 0.813 888;
  • 39) 0.813 888 × 2 = 1 + 0.627 776;
  • 40) 0.627 776 × 2 = 1 + 0.255 552;
  • 41) 0.255 552 × 2 = 0 + 0.511 104;
  • 42) 0.511 104 × 2 = 1 + 0.022 208;
  • 43) 0.022 208 × 2 = 0 + 0.044 416;
  • 44) 0.044 416 × 2 = 0 + 0.088 832;
  • 45) 0.088 832 × 2 = 0 + 0.177 664;
  • 46) 0.177 664 × 2 = 0 + 0.355 328;
  • 47) 0.355 328 × 2 = 0 + 0.710 656;
  • 48) 0.710 656 × 2 = 1 + 0.421 312;
  • 49) 0.421 312 × 2 = 0 + 0.842 624;
  • 50) 0.842 624 × 2 = 1 + 0.685 248;
  • 51) 0.685 248 × 2 = 1 + 0.370 496;
  • 52) 0.370 496 × 2 = 0 + 0.740 992;
  • 53) 0.740 992 × 2 = 1 + 0.481 984;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.234 377(10) =


0.0011 1100 0000 0000 0010 0001 1000 1101 1110 1111 0100 0001 0110 1(2)


5. Positive number before normalization:

52.234 377(10) =


11 0100.0011 1100 0000 0000 0010 0001 1000 1101 1110 1111 0100 0001 0110 1(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left so that only one non zero digit remains to the left of it:

52.234 377(10) =


11 0100.0011 1100 0000 0000 0010 0001 1000 1101 1110 1111 0100 0001 0110 1(2) =


11 0100.0011 1100 0000 0000 0010 0001 1000 1101 1110 1111 0100 0001 0110 1(2) × 20 =


1.1010 0001 1110 0000 0000 0001 0000 1100 0110 1111 0111 1010 0000 1011 01(2) × 25


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign: 0 (a positive number)


Exponent (unadjusted): 5


Mantissa (not normalized):
1.1010 0001 1110 0000 0000 0001 0000 1100 0110 1111 0111 1010 0000 1011 01


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


5 + 2(11-1) - 1 =


(5 + 1 023)(10) =


1 028(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:

  • division = quotient + remainder;
  • 1 028 ÷ 2 = 514 + 0;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above:

Exponent (adjusted) =


1028(10) =


100 0000 0100(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).

Mantissa (normalized) =


1. 1010 0001 1110 0000 0000 0001 0000 1100 0110 1111 0111 1010 0000 10 1101 =


1010 0001 1110 0000 0000 0001 0000 1100 0110 1111 0111 1010 0000


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (11 bits) =
100 0000 0100


Mantissa (52 bits) =
1010 0001 1110 0000 0000 0001 0000 1100 0110 1111 0111 1010 0000


Conclusion:

Number 52.234 377 converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0100 - 1010 0001 1110 0000 0000 0001 0000 1100 0110 1111 0111 1010 0000

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 1

      54
    • 0

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 1

      49
    • 0

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 1

      44
    • 1

      43
    • 1

      42
    • 1

      41
    • 0

      40
    • 0

      39
    • 0

      38
    • 0

      37
    • 0

      36
    • 0

      35
    • 0

      34
    • 0

      33
    • 0

      32
    • 0

      31
    • 0

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 1

      23
    • 1

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 1

      13
    • 1

      12
    • 0

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 1

      7
    • 0

      6
    • 1

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 0

      0

More operations of this kind:

52.234 376 = ? ... 52.234 378 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

52.234 377 to 64 bit double precision IEEE 754 binary floating point = ? Jan 26 18:23 UTC (GMT)
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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =


    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100