-0.000 282 005 885 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 885 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
-0.000 282 005 885 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:

|-0.000 282 005 885 6| = 0.000 282 005 885 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: 0.000 282 005 885 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) 0.000 282 005 885 6 × 2 = 0 + 0.000 564 011 771 2;
  • 2) 0.000 564 011 771 2 × 2 = 0 + 0.001 128 023 542 4;
  • 3) 0.001 128 023 542 4 × 2 = 0 + 0.002 256 047 084 8;
  • 4) 0.002 256 047 084 8 × 2 = 0 + 0.004 512 094 169 6;
  • 5) 0.004 512 094 169 6 × 2 = 0 + 0.009 024 188 339 2;
  • 6) 0.009 024 188 339 2 × 2 = 0 + 0.018 048 376 678 4;
  • 7) 0.018 048 376 678 4 × 2 = 0 + 0.036 096 753 356 8;
  • 8) 0.036 096 753 356 8 × 2 = 0 + 0.072 193 506 713 6;
  • 9) 0.072 193 506 713 6 × 2 = 0 + 0.144 387 013 427 2;
  • 10) 0.144 387 013 427 2 × 2 = 0 + 0.288 774 026 854 4;
  • 11) 0.288 774 026 854 4 × 2 = 0 + 0.577 548 053 708 8;
  • 12) 0.577 548 053 708 8 × 2 = 1 + 0.155 096 107 417 6;
  • 13) 0.155 096 107 417 6 × 2 = 0 + 0.310 192 214 835 2;
  • 14) 0.310 192 214 835 2 × 2 = 0 + 0.620 384 429 670 4;
  • 15) 0.620 384 429 670 4 × 2 = 1 + 0.240 768 859 340 8;
  • 16) 0.240 768 859 340 8 × 2 = 0 + 0.481 537 718 681 6;
  • 17) 0.481 537 718 681 6 × 2 = 0 + 0.963 075 437 363 2;
  • 18) 0.963 075 437 363 2 × 2 = 1 + 0.926 150 874 726 4;
  • 19) 0.926 150 874 726 4 × 2 = 1 + 0.852 301 749 452 8;
  • 20) 0.852 301 749 452 8 × 2 = 1 + 0.704 603 498 905 6;
  • 21) 0.704 603 498 905 6 × 2 = 1 + 0.409 206 997 811 2;
  • 22) 0.409 206 997 811 2 × 2 = 0 + 0.818 413 995 622 4;
  • 23) 0.818 413 995 622 4 × 2 = 1 + 0.636 827 991 244 8;
  • 24) 0.636 827 991 244 8 × 2 = 1 + 0.273 655 982 489 6;
  • 25) 0.273 655 982 489 6 × 2 = 0 + 0.547 311 964 979 2;
  • 26) 0.547 311 964 979 2 × 2 = 1 + 0.094 623 929 958 4;
  • 27) 0.094 623 929 958 4 × 2 = 0 + 0.189 247 859 916 8;
  • 28) 0.189 247 859 916 8 × 2 = 0 + 0.378 495 719 833 6;
  • 29) 0.378 495 719 833 6 × 2 = 0 + 0.756 991 439 667 2;
  • 30) 0.756 991 439 667 2 × 2 = 1 + 0.513 982 879 334 4;
  • 31) 0.513 982 879 334 4 × 2 = 1 + 0.027 965 758 668 8;
  • 32) 0.027 965 758 668 8 × 2 = 0 + 0.055 931 517 337 6;
  • 33) 0.055 931 517 337 6 × 2 = 0 + 0.111 863 034 675 2;
  • 34) 0.111 863 034 675 2 × 2 = 0 + 0.223 726 069 350 4;
  • 35) 0.223 726 069 350 4 × 2 = 0 + 0.447 452 138 700 8;
  • 36) 0.447 452 138 700 8 × 2 = 0 + 0.894 904 277 401 6;
  • 37) 0.894 904 277 401 6 × 2 = 1 + 0.789 808 554 803 2;
  • 38) 0.789 808 554 803 2 × 2 = 1 + 0.579 617 109 606 4;
  • 39) 0.579 617 109 606 4 × 2 = 1 + 0.159 234 219 212 8;
  • 40) 0.159 234 219 212 8 × 2 = 0 + 0.318 468 438 425 6;
  • 41) 0.318 468 438 425 6 × 2 = 0 + 0.636 936 876 851 2;
  • 42) 0.636 936 876 851 2 × 2 = 1 + 0.273 873 753 702 4;
  • 43) 0.273 873 753 702 4 × 2 = 0 + 0.547 747 507 404 8;
  • 44) 0.547 747 507 404 8 × 2 = 1 + 0.095 495 014 809 6;
  • 45) 0.095 495 014 809 6 × 2 = 0 + 0.190 990 029 619 2;
  • 46) 0.190 990 029 619 2 × 2 = 0 + 0.381 980 059 238 4;
  • 47) 0.381 980 059 238 4 × 2 = 0 + 0.763 960 118 476 8;
  • 48) 0.763 960 118 476 8 × 2 = 1 + 0.527 920 236 953 6;
  • 49) 0.527 920 236 953 6 × 2 = 1 + 0.055 840 473 907 2;
  • 50) 0.055 840 473 907 2 × 2 = 0 + 0.111 680 947 814 4;
  • 51) 0.111 680 947 814 4 × 2 = 0 + 0.223 361 895 628 8;
  • 52) 0.223 361 895 628 8 × 2 = 0 + 0.446 723 791 257 6;
  • 53) 0.446 723 791 257 6 × 2 = 0 + 0.893 447 582 515 2;
  • 54) 0.893 447 582 515 2 × 2 = 1 + 0.786 895 165 030 4;
  • 55) 0.786 895 165 030 4 × 2 = 1 + 0.573 790 330 060 8;
  • 56) 0.573 790 330 060 8 × 2 = 1 + 0.147 580 660 121 6;
  • 57) 0.147 580 660 121 6 × 2 = 0 + 0.295 161 320 243 2;
  • 58) 0.295 161 320 243 2 × 2 = 0 + 0.590 322 640 486 4;
  • 59) 0.590 322 640 486 4 × 2 = 1 + 0.180 645 280 972 8;
  • 60) 0.180 645 280 972 8 × 2 = 0 + 0.361 290 561 945 6;
  • 61) 0.361 290 561 945 6 × 2 = 0 + 0.722 581 123 891 2;
  • 62) 0.722 581 123 891 2 × 2 = 1 + 0.445 162 247 782 4;
  • 63) 0.445 162 247 782 4 × 2 = 0 + 0.890 324 495 564 8;
  • 64) 0.890 324 495 564 8 × 2 = 1 + 0.780 648 991 129 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:


0.000 282 005 885 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1110 0101 0001 1000 0111 0010 0101(2)

6. Positive number before normalization:

0.000 282 005 885 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1110 0101 0001 1000 0111 0010 0101(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:


0.000 282 005 885 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1110 0101 0001 1000 0111 0010 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 1110 0101 0001 1000 0111 0010 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0000 1110 0101 0001 1000 0111 0010 0101(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 0000 1110 0101 0001 1000 0111 0010 0101


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 0000 1110 0101 0001 1000 0111 0010 0101 =

0010 0111 1011 0100 0110 0000 1110 0101 0001 1000 0111 0010 0101


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 0000 1110 0101 0001 1000 0111 0010 0101


Decimal number -0.000 282 005 885 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0000 1110 0101 0001 1000 0111 0010 0101

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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