-0.000 282 005 961 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 961(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
-0.000 282 005 961(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:

|-0.000 282 005 961| = 0.000 282 005 961


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: 0.000 282 005 961.

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) 0.000 282 005 961 × 2 = 0 + 0.000 564 011 922;
  • 2) 0.000 564 011 922 × 2 = 0 + 0.001 128 023 844;
  • 3) 0.001 128 023 844 × 2 = 0 + 0.002 256 047 688;
  • 4) 0.002 256 047 688 × 2 = 0 + 0.004 512 095 376;
  • 5) 0.004 512 095 376 × 2 = 0 + 0.009 024 190 752;
  • 6) 0.009 024 190 752 × 2 = 0 + 0.018 048 381 504;
  • 7) 0.018 048 381 504 × 2 = 0 + 0.036 096 763 008;
  • 8) 0.036 096 763 008 × 2 = 0 + 0.072 193 526 016;
  • 9) 0.072 193 526 016 × 2 = 0 + 0.144 387 052 032;
  • 10) 0.144 387 052 032 × 2 = 0 + 0.288 774 104 064;
  • 11) 0.288 774 104 064 × 2 = 0 + 0.577 548 208 128;
  • 12) 0.577 548 208 128 × 2 = 1 + 0.155 096 416 256;
  • 13) 0.155 096 416 256 × 2 = 0 + 0.310 192 832 512;
  • 14) 0.310 192 832 512 × 2 = 0 + 0.620 385 665 024;
  • 15) 0.620 385 665 024 × 2 = 1 + 0.240 771 330 048;
  • 16) 0.240 771 330 048 × 2 = 0 + 0.481 542 660 096;
  • 17) 0.481 542 660 096 × 2 = 0 + 0.963 085 320 192;
  • 18) 0.963 085 320 192 × 2 = 1 + 0.926 170 640 384;
  • 19) 0.926 170 640 384 × 2 = 1 + 0.852 341 280 768;
  • 20) 0.852 341 280 768 × 2 = 1 + 0.704 682 561 536;
  • 21) 0.704 682 561 536 × 2 = 1 + 0.409 365 123 072;
  • 22) 0.409 365 123 072 × 2 = 0 + 0.818 730 246 144;
  • 23) 0.818 730 246 144 × 2 = 1 + 0.637 460 492 288;
  • 24) 0.637 460 492 288 × 2 = 1 + 0.274 920 984 576;
  • 25) 0.274 920 984 576 × 2 = 0 + 0.549 841 969 152;
  • 26) 0.549 841 969 152 × 2 = 1 + 0.099 683 938 304;
  • 27) 0.099 683 938 304 × 2 = 0 + 0.199 367 876 608;
  • 28) 0.199 367 876 608 × 2 = 0 + 0.398 735 753 216;
  • 29) 0.398 735 753 216 × 2 = 0 + 0.797 471 506 432;
  • 30) 0.797 471 506 432 × 2 = 1 + 0.594 943 012 864;
  • 31) 0.594 943 012 864 × 2 = 1 + 0.189 886 025 728;
  • 32) 0.189 886 025 728 × 2 = 0 + 0.379 772 051 456;
  • 33) 0.379 772 051 456 × 2 = 0 + 0.759 544 102 912;
  • 34) 0.759 544 102 912 × 2 = 1 + 0.519 088 205 824;
  • 35) 0.519 088 205 824 × 2 = 1 + 0.038 176 411 648;
  • 36) 0.038 176 411 648 × 2 = 0 + 0.076 352 823 296;
  • 37) 0.076 352 823 296 × 2 = 0 + 0.152 705 646 592;
  • 38) 0.152 705 646 592 × 2 = 0 + 0.305 411 293 184;
  • 39) 0.305 411 293 184 × 2 = 0 + 0.610 822 586 368;
  • 40) 0.610 822 586 368 × 2 = 1 + 0.221 645 172 736;
  • 41) 0.221 645 172 736 × 2 = 0 + 0.443 290 345 472;
  • 42) 0.443 290 345 472 × 2 = 0 + 0.886 580 690 944;
  • 43) 0.886 580 690 944 × 2 = 1 + 0.773 161 381 888;
  • 44) 0.773 161 381 888 × 2 = 1 + 0.546 322 763 776;
  • 45) 0.546 322 763 776 × 2 = 1 + 0.092 645 527 552;
  • 46) 0.092 645 527 552 × 2 = 0 + 0.185 291 055 104;
  • 47) 0.185 291 055 104 × 2 = 0 + 0.370 582 110 208;
  • 48) 0.370 582 110 208 × 2 = 0 + 0.741 164 220 416;
  • 49) 0.741 164 220 416 × 2 = 1 + 0.482 328 440 832;
  • 50) 0.482 328 440 832 × 2 = 0 + 0.964 656 881 664;
  • 51) 0.964 656 881 664 × 2 = 1 + 0.929 313 763 328;
  • 52) 0.929 313 763 328 × 2 = 1 + 0.858 627 526 656;
  • 53) 0.858 627 526 656 × 2 = 1 + 0.717 255 053 312;
  • 54) 0.717 255 053 312 × 2 = 1 + 0.434 510 106 624;
  • 55) 0.434 510 106 624 × 2 = 0 + 0.869 020 213 248;
  • 56) 0.869 020 213 248 × 2 = 1 + 0.738 040 426 496;
  • 57) 0.738 040 426 496 × 2 = 1 + 0.476 080 852 992;
  • 58) 0.476 080 852 992 × 2 = 0 + 0.952 161 705 984;
  • 59) 0.952 161 705 984 × 2 = 1 + 0.904 323 411 968;
  • 60) 0.904 323 411 968 × 2 = 1 + 0.808 646 823 936;
  • 61) 0.808 646 823 936 × 2 = 1 + 0.617 293 647 872;
  • 62) 0.617 293 647 872 × 2 = 1 + 0.234 587 295 744;
  • 63) 0.234 587 295 744 × 2 = 0 + 0.469 174 591 488;
  • 64) 0.469 174 591 488 × 2 = 0 + 0.938 349 182 976;

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:


0.000 282 005 961(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 0001 0011 1000 1011 1101 1011 1100(2)

6. Positive number before normalization:

0.000 282 005 961(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 0001 0011 1000 1011 1101 1011 1100(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:


0.000 282 005 961(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 0001 0011 1000 1011 1101 1011 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 0001 0011 1000 1011 1101 1011 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0110 0001 0011 1000 1011 1101 1011 1100(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 0110 0001 0011 1000 1011 1101 1011 1100


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 0110 0001 0011 1000 1011 1101 1011 1100 =

0010 0111 1011 0100 0110 0110 0001 0011 1000 1011 1101 1011 1100


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 0110 0001 0011 1000 1011 1101 1011 1100


Decimal number -0.000 282 005 961 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0110 0001 0011 1000 1011 1101 1011 1100

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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