64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -0.062 209 657 741 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -0.062 209 657 741(10) converted and written in 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:

|-0.062 209 657 741| = 0.062 209 657 741

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: 0.062 209 657 741.

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) 0.062 209 657 741 × 2 = 0 + 0.124 419 315 482;
  • 2) 0.124 419 315 482 × 2 = 0 + 0.248 838 630 964;
  • 3) 0.248 838 630 964 × 2 = 0 + 0.497 677 261 928;
  • 4) 0.497 677 261 928 × 2 = 0 + 0.995 354 523 856;
  • 5) 0.995 354 523 856 × 2 = 1 + 0.990 709 047 712;
  • 6) 0.990 709 047 712 × 2 = 1 + 0.981 418 095 424;
  • 7) 0.981 418 095 424 × 2 = 1 + 0.962 836 190 848;
  • 8) 0.962 836 190 848 × 2 = 1 + 0.925 672 381 696;
  • 9) 0.925 672 381 696 × 2 = 1 + 0.851 344 763 392;
  • 10) 0.851 344 763 392 × 2 = 1 + 0.702 689 526 784;
  • 11) 0.702 689 526 784 × 2 = 1 + 0.405 379 053 568;
  • 12) 0.405 379 053 568 × 2 = 0 + 0.810 758 107 136;
  • 13) 0.810 758 107 136 × 2 = 1 + 0.621 516 214 272;
  • 14) 0.621 516 214 272 × 2 = 1 + 0.243 032 428 544;
  • 15) 0.243 032 428 544 × 2 = 0 + 0.486 064 857 088;
  • 16) 0.486 064 857 088 × 2 = 0 + 0.972 129 714 176;
  • 17) 0.972 129 714 176 × 2 = 1 + 0.944 259 428 352;
  • 18) 0.944 259 428 352 × 2 = 1 + 0.888 518 856 704;
  • 19) 0.888 518 856 704 × 2 = 1 + 0.777 037 713 408;
  • 20) 0.777 037 713 408 × 2 = 1 + 0.554 075 426 816;
  • 21) 0.554 075 426 816 × 2 = 1 + 0.108 150 853 632;
  • 22) 0.108 150 853 632 × 2 = 0 + 0.216 301 707 264;
  • 23) 0.216 301 707 264 × 2 = 0 + 0.432 603 414 528;
  • 24) 0.432 603 414 528 × 2 = 0 + 0.865 206 829 056;
  • 25) 0.865 206 829 056 × 2 = 1 + 0.730 413 658 112;
  • 26) 0.730 413 658 112 × 2 = 1 + 0.460 827 316 224;
  • 27) 0.460 827 316 224 × 2 = 0 + 0.921 654 632 448;
  • 28) 0.921 654 632 448 × 2 = 1 + 0.843 309 264 896;
  • 29) 0.843 309 264 896 × 2 = 1 + 0.686 618 529 792;
  • 30) 0.686 618 529 792 × 2 = 1 + 0.373 237 059 584;
  • 31) 0.373 237 059 584 × 2 = 0 + 0.746 474 119 168;
  • 32) 0.746 474 119 168 × 2 = 1 + 0.492 948 238 336;
  • 33) 0.492 948 238 336 × 2 = 0 + 0.985 896 476 672;
  • 34) 0.985 896 476 672 × 2 = 1 + 0.971 792 953 344;
  • 35) 0.971 792 953 344 × 2 = 1 + 0.943 585 906 688;
  • 36) 0.943 585 906 688 × 2 = 1 + 0.887 171 813 376;
  • 37) 0.887 171 813 376 × 2 = 1 + 0.774 343 626 752;
  • 38) 0.774 343 626 752 × 2 = 1 + 0.548 687 253 504;
  • 39) 0.548 687 253 504 × 2 = 1 + 0.097 374 507 008;
  • 40) 0.097 374 507 008 × 2 = 0 + 0.194 749 014 016;
  • 41) 0.194 749 014 016 × 2 = 0 + 0.389 498 028 032;
  • 42) 0.389 498 028 032 × 2 = 0 + 0.778 996 056 064;
  • 43) 0.778 996 056 064 × 2 = 1 + 0.557 992 112 128;
  • 44) 0.557 992 112 128 × 2 = 1 + 0.115 984 224 256;
  • 45) 0.115 984 224 256 × 2 = 0 + 0.231 968 448 512;
  • 46) 0.231 968 448 512 × 2 = 0 + 0.463 936 897 024;
  • 47) 0.463 936 897 024 × 2 = 0 + 0.927 873 794 048;
  • 48) 0.927 873 794 048 × 2 = 1 + 0.855 747 588 096;
  • 49) 0.855 747 588 096 × 2 = 1 + 0.711 495 176 192;
  • 50) 0.711 495 176 192 × 2 = 1 + 0.422 990 352 384;
  • 51) 0.422 990 352 384 × 2 = 0 + 0.845 980 704 768;
  • 52) 0.845 980 704 768 × 2 = 1 + 0.691 961 409 536;
  • 53) 0.691 961 409 536 × 2 = 1 + 0.383 922 819 072;
  • 54) 0.383 922 819 072 × 2 = 0 + 0.767 845 638 144;
  • 55) 0.767 845 638 144 × 2 = 1 + 0.535 691 276 288;
  • 56) 0.535 691 276 288 × 2 = 1 + 0.071 382 552 576;
  • 57) 0.071 382 552 576 × 2 = 0 + 0.142 765 105 152;

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...)


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:


0.062 209 657 741(10) =

0.0000 1111 1110 1100 1111 1000 1101 1101 0111 1110 0011 0001 1101 1011 0(2)


6. Positive number before normalization:

0.062 209 657 741(10) =

0.0000 1111 1110 1100 1111 1000 1101 1101 0111 1110 0011 0001 1101 1011 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.062 209 657 741(10) =

0.0000 1111 1110 1100 1111 1000 1101 1101 0111 1110 0011 0001 1101 1011 0(2) =

0.0000 1111 1110 1100 1111 1000 1101 1101 0111 1110 0011 0001 1101 1011 0(2) × 20 =

1.1111 1101 1001 1111 0001 1011 1010 1111 1100 0110 0011 1011 0110(2) × 2-5


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): -5

Mantissa (not normalized):
1.1111 1101 1001 1111 0001 1011 1010 1111 1100 0110 0011 1011 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1018(10) =

011 1111 1010(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. 1111 1101 1001 1111 0001 1011 1010 1111 1100 0110 0011 1011 0110 =

1111 1101 1001 1111 0001 1011 1010 1111 1100 0110 0011 1011 0110


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 1010

Mantissa (52 bits) =
1111 1101 1001 1111 0001 1011 1010 1111 1100 0110 0011 1011 0110


The base ten decimal number -0.062 209 657 741 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 011 1111 1010 - 1111 1101 1001 1111 0001 1011 1010 1111 1100 0110 0011 1011 0110

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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