-0.000 282 005 914 93 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 93(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
-0.000 282 005 914 93(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. Start with the positive version of the number:

|-0.000 282 005 914 93| = 0.000 282 005 914 93


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 282 005 914 93.

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) 0.000 282 005 914 93 × 2 = 0 + 0.000 564 011 829 86;
  • 2) 0.000 564 011 829 86 × 2 = 0 + 0.001 128 023 659 72;
  • 3) 0.001 128 023 659 72 × 2 = 0 + 0.002 256 047 319 44;
  • 4) 0.002 256 047 319 44 × 2 = 0 + 0.004 512 094 638 88;
  • 5) 0.004 512 094 638 88 × 2 = 0 + 0.009 024 189 277 76;
  • 6) 0.009 024 189 277 76 × 2 = 0 + 0.018 048 378 555 52;
  • 7) 0.018 048 378 555 52 × 2 = 0 + 0.036 096 757 111 04;
  • 8) 0.036 096 757 111 04 × 2 = 0 + 0.072 193 514 222 08;
  • 9) 0.072 193 514 222 08 × 2 = 0 + 0.144 387 028 444 16;
  • 10) 0.144 387 028 444 16 × 2 = 0 + 0.288 774 056 888 32;
  • 11) 0.288 774 056 888 32 × 2 = 0 + 0.577 548 113 776 64;
  • 12) 0.577 548 113 776 64 × 2 = 1 + 0.155 096 227 553 28;
  • 13) 0.155 096 227 553 28 × 2 = 0 + 0.310 192 455 106 56;
  • 14) 0.310 192 455 106 56 × 2 = 0 + 0.620 384 910 213 12;
  • 15) 0.620 384 910 213 12 × 2 = 1 + 0.240 769 820 426 24;
  • 16) 0.240 769 820 426 24 × 2 = 0 + 0.481 539 640 852 48;
  • 17) 0.481 539 640 852 48 × 2 = 0 + 0.963 079 281 704 96;
  • 18) 0.963 079 281 704 96 × 2 = 1 + 0.926 158 563 409 92;
  • 19) 0.926 158 563 409 92 × 2 = 1 + 0.852 317 126 819 84;
  • 20) 0.852 317 126 819 84 × 2 = 1 + 0.704 634 253 639 68;
  • 21) 0.704 634 253 639 68 × 2 = 1 + 0.409 268 507 279 36;
  • 22) 0.409 268 507 279 36 × 2 = 0 + 0.818 537 014 558 72;
  • 23) 0.818 537 014 558 72 × 2 = 1 + 0.637 074 029 117 44;
  • 24) 0.637 074 029 117 44 × 2 = 1 + 0.274 148 058 234 88;
  • 25) 0.274 148 058 234 88 × 2 = 0 + 0.548 296 116 469 76;
  • 26) 0.548 296 116 469 76 × 2 = 1 + 0.096 592 232 939 52;
  • 27) 0.096 592 232 939 52 × 2 = 0 + 0.193 184 465 879 04;
  • 28) 0.193 184 465 879 04 × 2 = 0 + 0.386 368 931 758 08;
  • 29) 0.386 368 931 758 08 × 2 = 0 + 0.772 737 863 516 16;
  • 30) 0.772 737 863 516 16 × 2 = 1 + 0.545 475 727 032 32;
  • 31) 0.545 475 727 032 32 × 2 = 1 + 0.090 951 454 064 64;
  • 32) 0.090 951 454 064 64 × 2 = 0 + 0.181 902 908 129 28;
  • 33) 0.181 902 908 129 28 × 2 = 0 + 0.363 805 816 258 56;
  • 34) 0.363 805 816 258 56 × 2 = 0 + 0.727 611 632 517 12;
  • 35) 0.727 611 632 517 12 × 2 = 1 + 0.455 223 265 034 24;
  • 36) 0.455 223 265 034 24 × 2 = 0 + 0.910 446 530 068 48;
  • 37) 0.910 446 530 068 48 × 2 = 1 + 0.820 893 060 136 96;
  • 38) 0.820 893 060 136 96 × 2 = 1 + 0.641 786 120 273 92;
  • 39) 0.641 786 120 273 92 × 2 = 1 + 0.283 572 240 547 84;
  • 40) 0.283 572 240 547 84 × 2 = 0 + 0.567 144 481 095 68;
  • 41) 0.567 144 481 095 68 × 2 = 1 + 0.134 288 962 191 36;
  • 42) 0.134 288 962 191 36 × 2 = 0 + 0.268 577 924 382 72;
  • 43) 0.268 577 924 382 72 × 2 = 0 + 0.537 155 848 765 44;
  • 44) 0.537 155 848 765 44 × 2 = 1 + 0.074 311 697 530 88;
  • 45) 0.074 311 697 530 88 × 2 = 0 + 0.148 623 395 061 76;
  • 46) 0.148 623 395 061 76 × 2 = 0 + 0.297 246 790 123 52;
  • 47) 0.297 246 790 123 52 × 2 = 0 + 0.594 493 580 247 04;
  • 48) 0.594 493 580 247 04 × 2 = 1 + 0.188 987 160 494 08;
  • 49) 0.188 987 160 494 08 × 2 = 0 + 0.377 974 320 988 16;
  • 50) 0.377 974 320 988 16 × 2 = 0 + 0.755 948 641 976 32;
  • 51) 0.755 948 641 976 32 × 2 = 1 + 0.511 897 283 952 64;
  • 52) 0.511 897 283 952 64 × 2 = 1 + 0.023 794 567 905 28;
  • 53) 0.023 794 567 905 28 × 2 = 0 + 0.047 589 135 810 56;
  • 54) 0.047 589 135 810 56 × 2 = 0 + 0.095 178 271 621 12;
  • 55) 0.095 178 271 621 12 × 2 = 0 + 0.190 356 543 242 24;
  • 56) 0.190 356 543 242 24 × 2 = 0 + 0.380 713 086 484 48;
  • 57) 0.380 713 086 484 48 × 2 = 0 + 0.761 426 172 968 96;
  • 58) 0.761 426 172 968 96 × 2 = 1 + 0.522 852 345 937 92;
  • 59) 0.522 852 345 937 92 × 2 = 1 + 0.045 704 691 875 84;
  • 60) 0.045 704 691 875 84 × 2 = 0 + 0.091 409 383 751 68;
  • 61) 0.091 409 383 751 68 × 2 = 0 + 0.182 818 767 503 36;
  • 62) 0.182 818 767 503 36 × 2 = 0 + 0.365 637 535 006 72;
  • 63) 0.365 637 535 006 72 × 2 = 0 + 0.731 275 070 013 44;
  • 64) 0.731 275 070 013 44 × 2 = 1 + 0.462 550 140 026 88;

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).


5. 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:


0.000 282 005 914 93(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 1001 0001 0011 0000 0110 0001(2)

6. Positive number before normalization:

0.000 282 005 914 93(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 1001 0001 0011 0000 0110 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 914 93(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 1001 0001 0011 0000 0110 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 1001 0001 0011 0000 0110 0001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 1001 0001 0011 0000 0110 0001(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 0010 1110 1001 0001 0011 0000 0110 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(10)


10. 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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 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;

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

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


Exponent (adjusted) =

1011(10) =

011 1111 0011(2)


12. 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. 0010 0111 1011 0100 0110 0010 1110 1001 0001 0011 0000 0110 0001 =

0010 0111 1011 0100 0110 0010 1110 1001 0001 0011 0000 0110 0001


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

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

Exponent (11 bits) =
011 1111 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0010 1110 1001 0001 0011 0000 0110 0001


Decimal number -0.000 282 005 914 93 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 1001 0001 0011 0000 0110 0001

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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