-0.000 282 005 911 79 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 911 79(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
-0.000 282 005 911 79(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:

|-0.000 282 005 911 79| = 0.000 282 005 911 79


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: 0.000 282 005 911 79.

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) 0.000 282 005 911 79 × 2 = 0 + 0.000 564 011 823 58;
  • 2) 0.000 564 011 823 58 × 2 = 0 + 0.001 128 023 647 16;
  • 3) 0.001 128 023 647 16 × 2 = 0 + 0.002 256 047 294 32;
  • 4) 0.002 256 047 294 32 × 2 = 0 + 0.004 512 094 588 64;
  • 5) 0.004 512 094 588 64 × 2 = 0 + 0.009 024 189 177 28;
  • 6) 0.009 024 189 177 28 × 2 = 0 + 0.018 048 378 354 56;
  • 7) 0.018 048 378 354 56 × 2 = 0 + 0.036 096 756 709 12;
  • 8) 0.036 096 756 709 12 × 2 = 0 + 0.072 193 513 418 24;
  • 9) 0.072 193 513 418 24 × 2 = 0 + 0.144 387 026 836 48;
  • 10) 0.144 387 026 836 48 × 2 = 0 + 0.288 774 053 672 96;
  • 11) 0.288 774 053 672 96 × 2 = 0 + 0.577 548 107 345 92;
  • 12) 0.577 548 107 345 92 × 2 = 1 + 0.155 096 214 691 84;
  • 13) 0.155 096 214 691 84 × 2 = 0 + 0.310 192 429 383 68;
  • 14) 0.310 192 429 383 68 × 2 = 0 + 0.620 384 858 767 36;
  • 15) 0.620 384 858 767 36 × 2 = 1 + 0.240 769 717 534 72;
  • 16) 0.240 769 717 534 72 × 2 = 0 + 0.481 539 435 069 44;
  • 17) 0.481 539 435 069 44 × 2 = 0 + 0.963 078 870 138 88;
  • 18) 0.963 078 870 138 88 × 2 = 1 + 0.926 157 740 277 76;
  • 19) 0.926 157 740 277 76 × 2 = 1 + 0.852 315 480 555 52;
  • 20) 0.852 315 480 555 52 × 2 = 1 + 0.704 630 961 111 04;
  • 21) 0.704 630 961 111 04 × 2 = 1 + 0.409 261 922 222 08;
  • 22) 0.409 261 922 222 08 × 2 = 0 + 0.818 523 844 444 16;
  • 23) 0.818 523 844 444 16 × 2 = 1 + 0.637 047 688 888 32;
  • 24) 0.637 047 688 888 32 × 2 = 1 + 0.274 095 377 776 64;
  • 25) 0.274 095 377 776 64 × 2 = 0 + 0.548 190 755 553 28;
  • 26) 0.548 190 755 553 28 × 2 = 1 + 0.096 381 511 106 56;
  • 27) 0.096 381 511 106 56 × 2 = 0 + 0.192 763 022 213 12;
  • 28) 0.192 763 022 213 12 × 2 = 0 + 0.385 526 044 426 24;
  • 29) 0.385 526 044 426 24 × 2 = 0 + 0.771 052 088 852 48;
  • 30) 0.771 052 088 852 48 × 2 = 1 + 0.542 104 177 704 96;
  • 31) 0.542 104 177 704 96 × 2 = 1 + 0.084 208 355 409 92;
  • 32) 0.084 208 355 409 92 × 2 = 0 + 0.168 416 710 819 84;
  • 33) 0.168 416 710 819 84 × 2 = 0 + 0.336 833 421 639 68;
  • 34) 0.336 833 421 639 68 × 2 = 0 + 0.673 666 843 279 36;
  • 35) 0.673 666 843 279 36 × 2 = 1 + 0.347 333 686 558 72;
  • 36) 0.347 333 686 558 72 × 2 = 0 + 0.694 667 373 117 44;
  • 37) 0.694 667 373 117 44 × 2 = 1 + 0.389 334 746 234 88;
  • 38) 0.389 334 746 234 88 × 2 = 0 + 0.778 669 492 469 76;
  • 39) 0.778 669 492 469 76 × 2 = 1 + 0.557 338 984 939 52;
  • 40) 0.557 338 984 939 52 × 2 = 1 + 0.114 677 969 879 04;
  • 41) 0.114 677 969 879 04 × 2 = 0 + 0.229 355 939 758 08;
  • 42) 0.229 355 939 758 08 × 2 = 0 + 0.458 711 879 516 16;
  • 43) 0.458 711 879 516 16 × 2 = 0 + 0.917 423 759 032 32;
  • 44) 0.917 423 759 032 32 × 2 = 1 + 0.834 847 518 064 64;
  • 45) 0.834 847 518 064 64 × 2 = 1 + 0.669 695 036 129 28;
  • 46) 0.669 695 036 129 28 × 2 = 1 + 0.339 390 072 258 56;
  • 47) 0.339 390 072 258 56 × 2 = 0 + 0.678 780 144 517 12;
  • 48) 0.678 780 144 517 12 × 2 = 1 + 0.357 560 289 034 24;
  • 49) 0.357 560 289 034 24 × 2 = 0 + 0.715 120 578 068 48;
  • 50) 0.715 120 578 068 48 × 2 = 1 + 0.430 241 156 136 96;
  • 51) 0.430 241 156 136 96 × 2 = 0 + 0.860 482 312 273 92;
  • 52) 0.860 482 312 273 92 × 2 = 1 + 0.720 964 624 547 84;
  • 53) 0.720 964 624 547 84 × 2 = 1 + 0.441 929 249 095 68;
  • 54) 0.441 929 249 095 68 × 2 = 0 + 0.883 858 498 191 36;
  • 55) 0.883 858 498 191 36 × 2 = 1 + 0.767 716 996 382 72;
  • 56) 0.767 716 996 382 72 × 2 = 1 + 0.535 433 992 765 44;
  • 57) 0.535 433 992 765 44 × 2 = 1 + 0.070 867 985 530 88;
  • 58) 0.070 867 985 530 88 × 2 = 0 + 0.141 735 971 061 76;
  • 59) 0.141 735 971 061 76 × 2 = 0 + 0.283 471 942 123 52;
  • 60) 0.283 471 942 123 52 × 2 = 0 + 0.566 943 884 247 04;
  • 61) 0.566 943 884 247 04 × 2 = 1 + 0.133 887 768 494 08;
  • 62) 0.133 887 768 494 08 × 2 = 0 + 0.267 775 536 988 16;
  • 63) 0.267 775 536 988 16 × 2 = 0 + 0.535 551 073 976 32;
  • 64) 0.535 551 073 976 32 × 2 = 1 + 0.071 102 147 952 64;

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:


0.000 282 005 911 79(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1011 0001 1101 0101 1011 1000 1001(2)

6. Positive number before normalization:

0.000 282 005 911 79(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1011 0001 1101 0101 1011 1000 1001(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:


0.000 282 005 911 79(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1011 0001 1101 0101 1011 1000 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1011 0001 1101 0101 1011 1000 1001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1011 0001 1101 0101 1011 1000 1001(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 1011 0001 1101 0101 1011 1000 1001


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 1011 0001 1101 0101 1011 1000 1001 =

0010 0111 1011 0100 0110 0010 1011 0001 1101 0101 1011 1000 1001


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 1011 0001 1101 0101 1011 1000 1001


Decimal number -0.000 282 005 911 79 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1011 0001 1101 0101 1011 1000 1001

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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