-0.000 282 005 904 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 904 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
-0.000 282 005 904 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:

|-0.000 282 005 904 6| = 0.000 282 005 904 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: 0.000 282 005 904 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) 0.000 282 005 904 6 × 2 = 0 + 0.000 564 011 809 2;
  • 2) 0.000 564 011 809 2 × 2 = 0 + 0.001 128 023 618 4;
  • 3) 0.001 128 023 618 4 × 2 = 0 + 0.002 256 047 236 8;
  • 4) 0.002 256 047 236 8 × 2 = 0 + 0.004 512 094 473 6;
  • 5) 0.004 512 094 473 6 × 2 = 0 + 0.009 024 188 947 2;
  • 6) 0.009 024 188 947 2 × 2 = 0 + 0.018 048 377 894 4;
  • 7) 0.018 048 377 894 4 × 2 = 0 + 0.036 096 755 788 8;
  • 8) 0.036 096 755 788 8 × 2 = 0 + 0.072 193 511 577 6;
  • 9) 0.072 193 511 577 6 × 2 = 0 + 0.144 387 023 155 2;
  • 10) 0.144 387 023 155 2 × 2 = 0 + 0.288 774 046 310 4;
  • 11) 0.288 774 046 310 4 × 2 = 0 + 0.577 548 092 620 8;
  • 12) 0.577 548 092 620 8 × 2 = 1 + 0.155 096 185 241 6;
  • 13) 0.155 096 185 241 6 × 2 = 0 + 0.310 192 370 483 2;
  • 14) 0.310 192 370 483 2 × 2 = 0 + 0.620 384 740 966 4;
  • 15) 0.620 384 740 966 4 × 2 = 1 + 0.240 769 481 932 8;
  • 16) 0.240 769 481 932 8 × 2 = 0 + 0.481 538 963 865 6;
  • 17) 0.481 538 963 865 6 × 2 = 0 + 0.963 077 927 731 2;
  • 18) 0.963 077 927 731 2 × 2 = 1 + 0.926 155 855 462 4;
  • 19) 0.926 155 855 462 4 × 2 = 1 + 0.852 311 710 924 8;
  • 20) 0.852 311 710 924 8 × 2 = 1 + 0.704 623 421 849 6;
  • 21) 0.704 623 421 849 6 × 2 = 1 + 0.409 246 843 699 2;
  • 22) 0.409 246 843 699 2 × 2 = 0 + 0.818 493 687 398 4;
  • 23) 0.818 493 687 398 4 × 2 = 1 + 0.636 987 374 796 8;
  • 24) 0.636 987 374 796 8 × 2 = 1 + 0.273 974 749 593 6;
  • 25) 0.273 974 749 593 6 × 2 = 0 + 0.547 949 499 187 2;
  • 26) 0.547 949 499 187 2 × 2 = 1 + 0.095 898 998 374 4;
  • 27) 0.095 898 998 374 4 × 2 = 0 + 0.191 797 996 748 8;
  • 28) 0.191 797 996 748 8 × 2 = 0 + 0.383 595 993 497 6;
  • 29) 0.383 595 993 497 6 × 2 = 0 + 0.767 191 986 995 2;
  • 30) 0.767 191 986 995 2 × 2 = 1 + 0.534 383 973 990 4;
  • 31) 0.534 383 973 990 4 × 2 = 1 + 0.068 767 947 980 8;
  • 32) 0.068 767 947 980 8 × 2 = 0 + 0.137 535 895 961 6;
  • 33) 0.137 535 895 961 6 × 2 = 0 + 0.275 071 791 923 2;
  • 34) 0.275 071 791 923 2 × 2 = 0 + 0.550 143 583 846 4;
  • 35) 0.550 143 583 846 4 × 2 = 1 + 0.100 287 167 692 8;
  • 36) 0.100 287 167 692 8 × 2 = 0 + 0.200 574 335 385 6;
  • 37) 0.200 574 335 385 6 × 2 = 0 + 0.401 148 670 771 2;
  • 38) 0.401 148 670 771 2 × 2 = 0 + 0.802 297 341 542 4;
  • 39) 0.802 297 341 542 4 × 2 = 1 + 0.604 594 683 084 8;
  • 40) 0.604 594 683 084 8 × 2 = 1 + 0.209 189 366 169 6;
  • 41) 0.209 189 366 169 6 × 2 = 0 + 0.418 378 732 339 2;
  • 42) 0.418 378 732 339 2 × 2 = 0 + 0.836 757 464 678 4;
  • 43) 0.836 757 464 678 4 × 2 = 1 + 0.673 514 929 356 8;
  • 44) 0.673 514 929 356 8 × 2 = 1 + 0.347 029 858 713 6;
  • 45) 0.347 029 858 713 6 × 2 = 0 + 0.694 059 717 427 2;
  • 46) 0.694 059 717 427 2 × 2 = 1 + 0.388 119 434 854 4;
  • 47) 0.388 119 434 854 4 × 2 = 0 + 0.776 238 869 708 8;
  • 48) 0.776 238 869 708 8 × 2 = 1 + 0.552 477 739 417 6;
  • 49) 0.552 477 739 417 6 × 2 = 1 + 0.104 955 478 835 2;
  • 50) 0.104 955 478 835 2 × 2 = 0 + 0.209 910 957 670 4;
  • 51) 0.209 910 957 670 4 × 2 = 0 + 0.419 821 915 340 8;
  • 52) 0.419 821 915 340 8 × 2 = 0 + 0.839 643 830 681 6;
  • 53) 0.839 643 830 681 6 × 2 = 1 + 0.679 287 661 363 2;
  • 54) 0.679 287 661 363 2 × 2 = 1 + 0.358 575 322 726 4;
  • 55) 0.358 575 322 726 4 × 2 = 0 + 0.717 150 645 452 8;
  • 56) 0.717 150 645 452 8 × 2 = 1 + 0.434 301 290 905 6;
  • 57) 0.434 301 290 905 6 × 2 = 0 + 0.868 602 581 811 2;
  • 58) 0.868 602 581 811 2 × 2 = 1 + 0.737 205 163 622 4;
  • 59) 0.737 205 163 622 4 × 2 = 1 + 0.474 410 327 244 8;
  • 60) 0.474 410 327 244 8 × 2 = 0 + 0.948 820 654 489 6;
  • 61) 0.948 820 654 489 6 × 2 = 1 + 0.897 641 308 979 2;
  • 62) 0.897 641 308 979 2 × 2 = 1 + 0.795 282 617 958 4;
  • 63) 0.795 282 617 958 4 × 2 = 1 + 0.590 565 235 916 8;
  • 64) 0.590 565 235 916 8 × 2 = 1 + 0.181 130 471 833 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:


0.000 282 005 904 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0011 0011 0101 1000 1101 0110 1111(2)

6. Positive number before normalization:

0.000 282 005 904 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0011 0011 0101 1000 1101 0110 1111(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:


0.000 282 005 904 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0011 0011 0101 1000 1101 0110 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0011 0011 0101 1000 1101 0110 1111(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0011 0011 0101 1000 1101 0110 1111(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 0011 0011 0101 1000 1101 0110 1111


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 0011 0011 0101 1000 1101 0110 1111 =

0010 0111 1011 0100 0110 0010 0011 0011 0101 1000 1101 0110 1111


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 0011 0011 0101 1000 1101 0110 1111


Decimal number -0.000 282 005 904 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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