-0.000 282 005 956 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 956 6(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
-0.000 282 005 956 6(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:

|-0.000 282 005 956 6| = 0.000 282 005 956 6


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: 0.000 282 005 956 6.

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) 0.000 282 005 956 6 × 2 = 0 + 0.000 564 011 913 2;
  • 2) 0.000 564 011 913 2 × 2 = 0 + 0.001 128 023 826 4;
  • 3) 0.001 128 023 826 4 × 2 = 0 + 0.002 256 047 652 8;
  • 4) 0.002 256 047 652 8 × 2 = 0 + 0.004 512 095 305 6;
  • 5) 0.004 512 095 305 6 × 2 = 0 + 0.009 024 190 611 2;
  • 6) 0.009 024 190 611 2 × 2 = 0 + 0.018 048 381 222 4;
  • 7) 0.018 048 381 222 4 × 2 = 0 + 0.036 096 762 444 8;
  • 8) 0.036 096 762 444 8 × 2 = 0 + 0.072 193 524 889 6;
  • 9) 0.072 193 524 889 6 × 2 = 0 + 0.144 387 049 779 2;
  • 10) 0.144 387 049 779 2 × 2 = 0 + 0.288 774 099 558 4;
  • 11) 0.288 774 099 558 4 × 2 = 0 + 0.577 548 199 116 8;
  • 12) 0.577 548 199 116 8 × 2 = 1 + 0.155 096 398 233 6;
  • 13) 0.155 096 398 233 6 × 2 = 0 + 0.310 192 796 467 2;
  • 14) 0.310 192 796 467 2 × 2 = 0 + 0.620 385 592 934 4;
  • 15) 0.620 385 592 934 4 × 2 = 1 + 0.240 771 185 868 8;
  • 16) 0.240 771 185 868 8 × 2 = 0 + 0.481 542 371 737 6;
  • 17) 0.481 542 371 737 6 × 2 = 0 + 0.963 084 743 475 2;
  • 18) 0.963 084 743 475 2 × 2 = 1 + 0.926 169 486 950 4;
  • 19) 0.926 169 486 950 4 × 2 = 1 + 0.852 338 973 900 8;
  • 20) 0.852 338 973 900 8 × 2 = 1 + 0.704 677 947 801 6;
  • 21) 0.704 677 947 801 6 × 2 = 1 + 0.409 355 895 603 2;
  • 22) 0.409 355 895 603 2 × 2 = 0 + 0.818 711 791 206 4;
  • 23) 0.818 711 791 206 4 × 2 = 1 + 0.637 423 582 412 8;
  • 24) 0.637 423 582 412 8 × 2 = 1 + 0.274 847 164 825 6;
  • 25) 0.274 847 164 825 6 × 2 = 0 + 0.549 694 329 651 2;
  • 26) 0.549 694 329 651 2 × 2 = 1 + 0.099 388 659 302 4;
  • 27) 0.099 388 659 302 4 × 2 = 0 + 0.198 777 318 604 8;
  • 28) 0.198 777 318 604 8 × 2 = 0 + 0.397 554 637 209 6;
  • 29) 0.397 554 637 209 6 × 2 = 0 + 0.795 109 274 419 2;
  • 30) 0.795 109 274 419 2 × 2 = 1 + 0.590 218 548 838 4;
  • 31) 0.590 218 548 838 4 × 2 = 1 + 0.180 437 097 676 8;
  • 32) 0.180 437 097 676 8 × 2 = 0 + 0.360 874 195 353 6;
  • 33) 0.360 874 195 353 6 × 2 = 0 + 0.721 748 390 707 2;
  • 34) 0.721 748 390 707 2 × 2 = 1 + 0.443 496 781 414 4;
  • 35) 0.443 496 781 414 4 × 2 = 0 + 0.886 993 562 828 8;
  • 36) 0.886 993 562 828 8 × 2 = 1 + 0.773 987 125 657 6;
  • 37) 0.773 987 125 657 6 × 2 = 1 + 0.547 974 251 315 2;
  • 38) 0.547 974 251 315 2 × 2 = 1 + 0.095 948 502 630 4;
  • 39) 0.095 948 502 630 4 × 2 = 0 + 0.191 897 005 260 8;
  • 40) 0.191 897 005 260 8 × 2 = 0 + 0.383 794 010 521 6;
  • 41) 0.383 794 010 521 6 × 2 = 0 + 0.767 588 021 043 2;
  • 42) 0.767 588 021 043 2 × 2 = 1 + 0.535 176 042 086 4;
  • 43) 0.535 176 042 086 4 × 2 = 1 + 0.070 352 084 172 8;
  • 44) 0.070 352 084 172 8 × 2 = 0 + 0.140 704 168 345 6;
  • 45) 0.140 704 168 345 6 × 2 = 0 + 0.281 408 336 691 2;
  • 46) 0.281 408 336 691 2 × 2 = 0 + 0.562 816 673 382 4;
  • 47) 0.562 816 673 382 4 × 2 = 1 + 0.125 633 346 764 8;
  • 48) 0.125 633 346 764 8 × 2 = 0 + 0.251 266 693 529 6;
  • 49) 0.251 266 693 529 6 × 2 = 0 + 0.502 533 387 059 2;
  • 50) 0.502 533 387 059 2 × 2 = 1 + 0.005 066 774 118 4;
  • 51) 0.005 066 774 118 4 × 2 = 0 + 0.010 133 548 236 8;
  • 52) 0.010 133 548 236 8 × 2 = 0 + 0.020 267 096 473 6;
  • 53) 0.020 267 096 473 6 × 2 = 0 + 0.040 534 192 947 2;
  • 54) 0.040 534 192 947 2 × 2 = 0 + 0.081 068 385 894 4;
  • 55) 0.081 068 385 894 4 × 2 = 0 + 0.162 136 771 788 8;
  • 56) 0.162 136 771 788 8 × 2 = 0 + 0.324 273 543 577 6;
  • 57) 0.324 273 543 577 6 × 2 = 0 + 0.648 547 087 155 2;
  • 58) 0.648 547 087 155 2 × 2 = 1 + 0.297 094 174 310 4;
  • 59) 0.297 094 174 310 4 × 2 = 0 + 0.594 188 348 620 8;
  • 60) 0.594 188 348 620 8 × 2 = 1 + 0.188 376 697 241 6;
  • 61) 0.188 376 697 241 6 × 2 = 0 + 0.376 753 394 483 2;
  • 62) 0.376 753 394 483 2 × 2 = 0 + 0.753 506 788 966 4;
  • 63) 0.753 506 788 966 4 × 2 = 1 + 0.507 013 577 932 8;
  • 64) 0.507 013 577 932 8 × 2 = 1 + 0.014 027 155 865 6;

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:


0.000 282 005 956 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 0110 0010 0100 0000 0101 0011(2)

6. Positive number before normalization:

0.000 282 005 956 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 0110 0010 0100 0000 0101 0011(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:


0.000 282 005 956 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 0110 0010 0100 0000 0101 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 0110 0010 0100 0000 0101 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0101 1100 0110 0010 0100 0000 0101 0011(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 0101 1100 0110 0010 0100 0000 0101 0011


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 0101 1100 0110 0010 0100 0000 0101 0011 =

0010 0111 1011 0100 0110 0101 1100 0110 0010 0100 0000 0101 0011


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 0101 1100 0110 0010 0100 0000 0101 0011


Decimal number -0.000 282 005 956 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0101 1100 0110 0010 0100 0000 0101 0011

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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