-0.000 282 005 915 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 915 37(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
-0.000 282 005 915 37(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:

|-0.000 282 005 915 37| = 0.000 282 005 915 37


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: 0.000 282 005 915 37.

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) 0.000 282 005 915 37 × 2 = 0 + 0.000 564 011 830 74;
  • 2) 0.000 564 011 830 74 × 2 = 0 + 0.001 128 023 661 48;
  • 3) 0.001 128 023 661 48 × 2 = 0 + 0.002 256 047 322 96;
  • 4) 0.002 256 047 322 96 × 2 = 0 + 0.004 512 094 645 92;
  • 5) 0.004 512 094 645 92 × 2 = 0 + 0.009 024 189 291 84;
  • 6) 0.009 024 189 291 84 × 2 = 0 + 0.018 048 378 583 68;
  • 7) 0.018 048 378 583 68 × 2 = 0 + 0.036 096 757 167 36;
  • 8) 0.036 096 757 167 36 × 2 = 0 + 0.072 193 514 334 72;
  • 9) 0.072 193 514 334 72 × 2 = 0 + 0.144 387 028 669 44;
  • 10) 0.144 387 028 669 44 × 2 = 0 + 0.288 774 057 338 88;
  • 11) 0.288 774 057 338 88 × 2 = 0 + 0.577 548 114 677 76;
  • 12) 0.577 548 114 677 76 × 2 = 1 + 0.155 096 229 355 52;
  • 13) 0.155 096 229 355 52 × 2 = 0 + 0.310 192 458 711 04;
  • 14) 0.310 192 458 711 04 × 2 = 0 + 0.620 384 917 422 08;
  • 15) 0.620 384 917 422 08 × 2 = 1 + 0.240 769 834 844 16;
  • 16) 0.240 769 834 844 16 × 2 = 0 + 0.481 539 669 688 32;
  • 17) 0.481 539 669 688 32 × 2 = 0 + 0.963 079 339 376 64;
  • 18) 0.963 079 339 376 64 × 2 = 1 + 0.926 158 678 753 28;
  • 19) 0.926 158 678 753 28 × 2 = 1 + 0.852 317 357 506 56;
  • 20) 0.852 317 357 506 56 × 2 = 1 + 0.704 634 715 013 12;
  • 21) 0.704 634 715 013 12 × 2 = 1 + 0.409 269 430 026 24;
  • 22) 0.409 269 430 026 24 × 2 = 0 + 0.818 538 860 052 48;
  • 23) 0.818 538 860 052 48 × 2 = 1 + 0.637 077 720 104 96;
  • 24) 0.637 077 720 104 96 × 2 = 1 + 0.274 155 440 209 92;
  • 25) 0.274 155 440 209 92 × 2 = 0 + 0.548 310 880 419 84;
  • 26) 0.548 310 880 419 84 × 2 = 1 + 0.096 621 760 839 68;
  • 27) 0.096 621 760 839 68 × 2 = 0 + 0.193 243 521 679 36;
  • 28) 0.193 243 521 679 36 × 2 = 0 + 0.386 487 043 358 72;
  • 29) 0.386 487 043 358 72 × 2 = 0 + 0.772 974 086 717 44;
  • 30) 0.772 974 086 717 44 × 2 = 1 + 0.545 948 173 434 88;
  • 31) 0.545 948 173 434 88 × 2 = 1 + 0.091 896 346 869 76;
  • 32) 0.091 896 346 869 76 × 2 = 0 + 0.183 792 693 739 52;
  • 33) 0.183 792 693 739 52 × 2 = 0 + 0.367 585 387 479 04;
  • 34) 0.367 585 387 479 04 × 2 = 0 + 0.735 170 774 958 08;
  • 35) 0.735 170 774 958 08 × 2 = 1 + 0.470 341 549 916 16;
  • 36) 0.470 341 549 916 16 × 2 = 0 + 0.940 683 099 832 32;
  • 37) 0.940 683 099 832 32 × 2 = 1 + 0.881 366 199 664 64;
  • 38) 0.881 366 199 664 64 × 2 = 1 + 0.762 732 399 329 28;
  • 39) 0.762 732 399 329 28 × 2 = 1 + 0.525 464 798 658 56;
  • 40) 0.525 464 798 658 56 × 2 = 1 + 0.050 929 597 317 12;
  • 41) 0.050 929 597 317 12 × 2 = 0 + 0.101 859 194 634 24;
  • 42) 0.101 859 194 634 24 × 2 = 0 + 0.203 718 389 268 48;
  • 43) 0.203 718 389 268 48 × 2 = 0 + 0.407 436 778 536 96;
  • 44) 0.407 436 778 536 96 × 2 = 0 + 0.814 873 557 073 92;
  • 45) 0.814 873 557 073 92 × 2 = 1 + 0.629 747 114 147 84;
  • 46) 0.629 747 114 147 84 × 2 = 1 + 0.259 494 228 295 68;
  • 47) 0.259 494 228 295 68 × 2 = 0 + 0.518 988 456 591 36;
  • 48) 0.518 988 456 591 36 × 2 = 1 + 0.037 976 913 182 72;
  • 49) 0.037 976 913 182 72 × 2 = 0 + 0.075 953 826 365 44;
  • 50) 0.075 953 826 365 44 × 2 = 0 + 0.151 907 652 730 88;
  • 51) 0.151 907 652 730 88 × 2 = 0 + 0.303 815 305 461 76;
  • 52) 0.303 815 305 461 76 × 2 = 0 + 0.607 630 610 923 52;
  • 53) 0.607 630 610 923 52 × 2 = 1 + 0.215 261 221 847 04;
  • 54) 0.215 261 221 847 04 × 2 = 0 + 0.430 522 443 694 08;
  • 55) 0.430 522 443 694 08 × 2 = 0 + 0.861 044 887 388 16;
  • 56) 0.861 044 887 388 16 × 2 = 1 + 0.722 089 774 776 32;
  • 57) 0.722 089 774 776 32 × 2 = 1 + 0.444 179 549 552 64;
  • 58) 0.444 179 549 552 64 × 2 = 0 + 0.888 359 099 105 28;
  • 59) 0.888 359 099 105 28 × 2 = 1 + 0.776 718 198 210 56;
  • 60) 0.776 718 198 210 56 × 2 = 1 + 0.553 436 396 421 12;
  • 61) 0.553 436 396 421 12 × 2 = 1 + 0.106 872 792 842 24;
  • 62) 0.106 872 792 842 24 × 2 = 0 + 0.213 745 585 684 48;
  • 63) 0.213 745 585 684 48 × 2 = 0 + 0.427 491 171 368 96;
  • 64) 0.427 491 171 368 96 × 2 = 0 + 0.854 982 342 737 92;

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:


0.000 282 005 915 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0000 1101 0000 1001 1011 1000(2)

6. Positive number before normalization:

0.000 282 005 915 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0000 1101 0000 1001 1011 1000(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:


0.000 282 005 915 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0000 1101 0000 1001 1011 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0000 1101 0000 1001 1011 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1111 0000 1101 0000 1001 1011 1000(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 1111 0000 1101 0000 1001 1011 1000


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 1111 0000 1101 0000 1001 1011 1000 =

0010 0111 1011 0100 0110 0010 1111 0000 1101 0000 1001 1011 1000


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 1111 0000 1101 0000 1001 1011 1000


Decimal number -0.000 282 005 915 37 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1111 0000 1101 0000 1001 1011 1000

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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