-0.000 282 005 914 386 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 386 7(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
-0.000 282 005 914 386 7(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:

|-0.000 282 005 914 386 7| = 0.000 282 005 914 386 7


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: 0.000 282 005 914 386 7.

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) 0.000 282 005 914 386 7 × 2 = 0 + 0.000 564 011 828 773 4;
  • 2) 0.000 564 011 828 773 4 × 2 = 0 + 0.001 128 023 657 546 8;
  • 3) 0.001 128 023 657 546 8 × 2 = 0 + 0.002 256 047 315 093 6;
  • 4) 0.002 256 047 315 093 6 × 2 = 0 + 0.004 512 094 630 187 2;
  • 5) 0.004 512 094 630 187 2 × 2 = 0 + 0.009 024 189 260 374 4;
  • 6) 0.009 024 189 260 374 4 × 2 = 0 + 0.018 048 378 520 748 8;
  • 7) 0.018 048 378 520 748 8 × 2 = 0 + 0.036 096 757 041 497 6;
  • 8) 0.036 096 757 041 497 6 × 2 = 0 + 0.072 193 514 082 995 2;
  • 9) 0.072 193 514 082 995 2 × 2 = 0 + 0.144 387 028 165 990 4;
  • 10) 0.144 387 028 165 990 4 × 2 = 0 + 0.288 774 056 331 980 8;
  • 11) 0.288 774 056 331 980 8 × 2 = 0 + 0.577 548 112 663 961 6;
  • 12) 0.577 548 112 663 961 6 × 2 = 1 + 0.155 096 225 327 923 2;
  • 13) 0.155 096 225 327 923 2 × 2 = 0 + 0.310 192 450 655 846 4;
  • 14) 0.310 192 450 655 846 4 × 2 = 0 + 0.620 384 901 311 692 8;
  • 15) 0.620 384 901 311 692 8 × 2 = 1 + 0.240 769 802 623 385 6;
  • 16) 0.240 769 802 623 385 6 × 2 = 0 + 0.481 539 605 246 771 2;
  • 17) 0.481 539 605 246 771 2 × 2 = 0 + 0.963 079 210 493 542 4;
  • 18) 0.963 079 210 493 542 4 × 2 = 1 + 0.926 158 420 987 084 8;
  • 19) 0.926 158 420 987 084 8 × 2 = 1 + 0.852 316 841 974 169 6;
  • 20) 0.852 316 841 974 169 6 × 2 = 1 + 0.704 633 683 948 339 2;
  • 21) 0.704 633 683 948 339 2 × 2 = 1 + 0.409 267 367 896 678 4;
  • 22) 0.409 267 367 896 678 4 × 2 = 0 + 0.818 534 735 793 356 8;
  • 23) 0.818 534 735 793 356 8 × 2 = 1 + 0.637 069 471 586 713 6;
  • 24) 0.637 069 471 586 713 6 × 2 = 1 + 0.274 138 943 173 427 2;
  • 25) 0.274 138 943 173 427 2 × 2 = 0 + 0.548 277 886 346 854 4;
  • 26) 0.548 277 886 346 854 4 × 2 = 1 + 0.096 555 772 693 708 8;
  • 27) 0.096 555 772 693 708 8 × 2 = 0 + 0.193 111 545 387 417 6;
  • 28) 0.193 111 545 387 417 6 × 2 = 0 + 0.386 223 090 774 835 2;
  • 29) 0.386 223 090 774 835 2 × 2 = 0 + 0.772 446 181 549 670 4;
  • 30) 0.772 446 181 549 670 4 × 2 = 1 + 0.544 892 363 099 340 8;
  • 31) 0.544 892 363 099 340 8 × 2 = 1 + 0.089 784 726 198 681 6;
  • 32) 0.089 784 726 198 681 6 × 2 = 0 + 0.179 569 452 397 363 2;
  • 33) 0.179 569 452 397 363 2 × 2 = 0 + 0.359 138 904 794 726 4;
  • 34) 0.359 138 904 794 726 4 × 2 = 0 + 0.718 277 809 589 452 8;
  • 35) 0.718 277 809 589 452 8 × 2 = 1 + 0.436 555 619 178 905 6;
  • 36) 0.436 555 619 178 905 6 × 2 = 0 + 0.873 111 238 357 811 2;
  • 37) 0.873 111 238 357 811 2 × 2 = 1 + 0.746 222 476 715 622 4;
  • 38) 0.746 222 476 715 622 4 × 2 = 1 + 0.492 444 953 431 244 8;
  • 39) 0.492 444 953 431 244 8 × 2 = 0 + 0.984 889 906 862 489 6;
  • 40) 0.984 889 906 862 489 6 × 2 = 1 + 0.969 779 813 724 979 2;
  • 41) 0.969 779 813 724 979 2 × 2 = 1 + 0.939 559 627 449 958 4;
  • 42) 0.939 559 627 449 958 4 × 2 = 1 + 0.879 119 254 899 916 8;
  • 43) 0.879 119 254 899 916 8 × 2 = 1 + 0.758 238 509 799 833 6;
  • 44) 0.758 238 509 799 833 6 × 2 = 1 + 0.516 477 019 599 667 2;
  • 45) 0.516 477 019 599 667 2 × 2 = 1 + 0.032 954 039 199 334 4;
  • 46) 0.032 954 039 199 334 4 × 2 = 0 + 0.065 908 078 398 668 8;
  • 47) 0.065 908 078 398 668 8 × 2 = 0 + 0.131 816 156 797 337 6;
  • 48) 0.131 816 156 797 337 6 × 2 = 0 + 0.263 632 313 594 675 2;
  • 49) 0.263 632 313 594 675 2 × 2 = 0 + 0.527 264 627 189 350 4;
  • 50) 0.527 264 627 189 350 4 × 2 = 1 + 0.054 529 254 378 700 8;
  • 51) 0.054 529 254 378 700 8 × 2 = 0 + 0.109 058 508 757 401 6;
  • 52) 0.109 058 508 757 401 6 × 2 = 0 + 0.218 117 017 514 803 2;
  • 53) 0.218 117 017 514 803 2 × 2 = 0 + 0.436 234 035 029 606 4;
  • 54) 0.436 234 035 029 606 4 × 2 = 0 + 0.872 468 070 059 212 8;
  • 55) 0.872 468 070 059 212 8 × 2 = 1 + 0.744 936 140 118 425 6;
  • 56) 0.744 936 140 118 425 6 × 2 = 1 + 0.489 872 280 236 851 2;
  • 57) 0.489 872 280 236 851 2 × 2 = 0 + 0.979 744 560 473 702 4;
  • 58) 0.979 744 560 473 702 4 × 2 = 1 + 0.959 489 120 947 404 8;
  • 59) 0.959 489 120 947 404 8 × 2 = 1 + 0.918 978 241 894 809 6;
  • 60) 0.918 978 241 894 809 6 × 2 = 1 + 0.837 956 483 789 619 2;
  • 61) 0.837 956 483 789 619 2 × 2 = 1 + 0.675 912 967 579 238 4;
  • 62) 0.675 912 967 579 238 4 × 2 = 1 + 0.351 825 935 158 476 8;
  • 63) 0.351 825 935 158 476 8 × 2 = 0 + 0.703 651 870 316 953 6;
  • 64) 0.703 651 870 316 953 6 × 2 = 1 + 0.407 303 740 633 907 2;

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:


0.000 282 005 914 386 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 0100 0011 0111 1101(2)

6. Positive number before normalization:

0.000 282 005 914 386 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 0100 0011 0111 1101(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:


0.000 282 005 914 386 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 0100 0011 0111 1101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 0100 0011 0111 1101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 1000 0100 0011 0111 1101(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 1101 1111 1000 0100 0011 0111 1101


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 1101 1111 1000 0100 0011 0111 1101 =

0010 0111 1011 0100 0110 0010 1101 1111 1000 0100 0011 0111 1101


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 1101 1111 1000 0100 0011 0111 1101


Decimal number -0.000 282 005 914 386 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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