0.001 087 172 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.001 087 172(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.001 087 172(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.001 087 172.

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) 0.001 087 172 × 2 = 0 + 0.002 174 344;
  • 2) 0.002 174 344 × 2 = 0 + 0.004 348 688;
  • 3) 0.004 348 688 × 2 = 0 + 0.008 697 376;
  • 4) 0.008 697 376 × 2 = 0 + 0.017 394 752;
  • 5) 0.017 394 752 × 2 = 0 + 0.034 789 504;
  • 6) 0.034 789 504 × 2 = 0 + 0.069 579 008;
  • 7) 0.069 579 008 × 2 = 0 + 0.139 158 016;
  • 8) 0.139 158 016 × 2 = 0 + 0.278 316 032;
  • 9) 0.278 316 032 × 2 = 0 + 0.556 632 064;
  • 10) 0.556 632 064 × 2 = 1 + 0.113 264 128;
  • 11) 0.113 264 128 × 2 = 0 + 0.226 528 256;
  • 12) 0.226 528 256 × 2 = 0 + 0.453 056 512;
  • 13) 0.453 056 512 × 2 = 0 + 0.906 113 024;
  • 14) 0.906 113 024 × 2 = 1 + 0.812 226 048;
  • 15) 0.812 226 048 × 2 = 1 + 0.624 452 096;
  • 16) 0.624 452 096 × 2 = 1 + 0.248 904 192;
  • 17) 0.248 904 192 × 2 = 0 + 0.497 808 384;
  • 18) 0.497 808 384 × 2 = 0 + 0.995 616 768;
  • 19) 0.995 616 768 × 2 = 1 + 0.991 233 536;
  • 20) 0.991 233 536 × 2 = 1 + 0.982 467 072;
  • 21) 0.982 467 072 × 2 = 1 + 0.964 934 144;
  • 22) 0.964 934 144 × 2 = 1 + 0.929 868 288;
  • 23) 0.929 868 288 × 2 = 1 + 0.859 736 576;
  • 24) 0.859 736 576 × 2 = 1 + 0.719 473 152;
  • 25) 0.719 473 152 × 2 = 1 + 0.438 946 304;
  • 26) 0.438 946 304 × 2 = 0 + 0.877 892 608;
  • 27) 0.877 892 608 × 2 = 1 + 0.755 785 216;
  • 28) 0.755 785 216 × 2 = 1 + 0.511 570 432;
  • 29) 0.511 570 432 × 2 = 1 + 0.023 140 864;
  • 30) 0.023 140 864 × 2 = 0 + 0.046 281 728;
  • 31) 0.046 281 728 × 2 = 0 + 0.092 563 456;
  • 32) 0.092 563 456 × 2 = 0 + 0.185 126 912;
  • 33) 0.185 126 912 × 2 = 0 + 0.370 253 824;
  • 34) 0.370 253 824 × 2 = 0 + 0.740 507 648;
  • 35) 0.740 507 648 × 2 = 1 + 0.481 015 296;
  • 36) 0.481 015 296 × 2 = 0 + 0.962 030 592;
  • 37) 0.962 030 592 × 2 = 1 + 0.924 061 184;
  • 38) 0.924 061 184 × 2 = 1 + 0.848 122 368;
  • 39) 0.848 122 368 × 2 = 1 + 0.696 244 736;
  • 40) 0.696 244 736 × 2 = 1 + 0.392 489 472;
  • 41) 0.392 489 472 × 2 = 0 + 0.784 978 944;
  • 42) 0.784 978 944 × 2 = 1 + 0.569 957 888;
  • 43) 0.569 957 888 × 2 = 1 + 0.139 915 776;
  • 44) 0.139 915 776 × 2 = 0 + 0.279 831 552;
  • 45) 0.279 831 552 × 2 = 0 + 0.559 663 104;
  • 46) 0.559 663 104 × 2 = 1 + 0.119 326 208;
  • 47) 0.119 326 208 × 2 = 0 + 0.238 652 416;
  • 48) 0.238 652 416 × 2 = 0 + 0.477 304 832;
  • 49) 0.477 304 832 × 2 = 0 + 0.954 609 664;
  • 50) 0.954 609 664 × 2 = 1 + 0.909 219 328;
  • 51) 0.909 219 328 × 2 = 1 + 0.818 438 656;
  • 52) 0.818 438 656 × 2 = 1 + 0.636 877 312;
  • 53) 0.636 877 312 × 2 = 1 + 0.273 754 624;
  • 54) 0.273 754 624 × 2 = 0 + 0.547 509 248;
  • 55) 0.547 509 248 × 2 = 1 + 0.095 018 496;
  • 56) 0.095 018 496 × 2 = 0 + 0.190 036 992;
  • 57) 0.190 036 992 × 2 = 0 + 0.380 073 984;
  • 58) 0.380 073 984 × 2 = 0 + 0.760 147 968;
  • 59) 0.760 147 968 × 2 = 1 + 0.520 295 936;
  • 60) 0.520 295 936 × 2 = 1 + 0.040 591 872;
  • 61) 0.040 591 872 × 2 = 0 + 0.081 183 744;
  • 62) 0.081 183 744 × 2 = 0 + 0.162 367 488;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.001 087 172(10) =

0.0000 0000 0100 0111 0011 1111 1011 1000 0010 1111 0110 0100 0111 1010 0011 00(2)

5. Positive number before normalization:

0.001 087 172(10) =

0.0000 0000 0100 0111 0011 1111 1011 1000 0010 1111 0110 0100 0111 1010 0011 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.001 087 172(10) =

0.0000 0000 0100 0111 0011 1111 1011 1000 0010 1111 0110 0100 0111 1010 0011 00(2) =

0.0000 0000 0100 0111 0011 1111 1011 1000 0010 1111 0110 0100 0111 1010 0011 00(2) × 20 =

1.0001 1100 1111 1110 1110 0000 1011 1101 1001 0001 1110 1000 1100(2) × 2-10


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): -10

Mantissa (not normalized):
1.0001 1100 1111 1110 1110 0000 1011 1101 1001 0001 1110 1000 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-10 + 2(11-1) - 1 =

(-10 + 1 023)(10) =

1 013(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;
  • 1 013 ÷ 2 = 506 + 1;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1013(10) =

011 1111 0101(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 1100 1111 1110 1110 0000 1011 1101 1001 0001 1110 1000 1100 =

0001 1100 1111 1110 1110 0000 1011 1101 1001 0001 1110 1000 1100


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) =
011 1111 0101

Mantissa (52 bits) =
0001 1100 1111 1110 1110 0000 1011 1101 1001 0001 1110 1000 1100


Decimal number 0.001 087 172 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0101 - 0001 1100 1111 1110 1110 0000 1011 1101 1001 0001 1110 1000 1100

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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