Decimal to 64 Bit IEEE 754 Binary: Convert Number -0.164 999 999 999 964 5 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.164 999 999 999 964 5(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.164 999 999 999 964 5| = 0.164 999 999 999 964 5


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: 0.164 999 999 999 964 5.

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) 0.164 999 999 999 964 5 × 2 = 0 + 0.329 999 999 999 929;
  • 2) 0.329 999 999 999 929 × 2 = 0 + 0.659 999 999 999 858;
  • 3) 0.659 999 999 999 858 × 2 = 1 + 0.319 999 999 999 716;
  • 4) 0.319 999 999 999 716 × 2 = 0 + 0.639 999 999 999 432;
  • 5) 0.639 999 999 999 432 × 2 = 1 + 0.279 999 999 998 864;
  • 6) 0.279 999 999 998 864 × 2 = 0 + 0.559 999 999 997 728;
  • 7) 0.559 999 999 997 728 × 2 = 1 + 0.119 999 999 995 456;
  • 8) 0.119 999 999 995 456 × 2 = 0 + 0.239 999 999 990 912;
  • 9) 0.239 999 999 990 912 × 2 = 0 + 0.479 999 999 981 824;
  • 10) 0.479 999 999 981 824 × 2 = 0 + 0.959 999 999 963 648;
  • 11) 0.959 999 999 963 648 × 2 = 1 + 0.919 999 999 927 296;
  • 12) 0.919 999 999 927 296 × 2 = 1 + 0.839 999 999 854 592;
  • 13) 0.839 999 999 854 592 × 2 = 1 + 0.679 999 999 709 184;
  • 14) 0.679 999 999 709 184 × 2 = 1 + 0.359 999 999 418 368;
  • 15) 0.359 999 999 418 368 × 2 = 0 + 0.719 999 998 836 736;
  • 16) 0.719 999 998 836 736 × 2 = 1 + 0.439 999 997 673 472;
  • 17) 0.439 999 997 673 472 × 2 = 0 + 0.879 999 995 346 944;
  • 18) 0.879 999 995 346 944 × 2 = 1 + 0.759 999 990 693 888;
  • 19) 0.759 999 990 693 888 × 2 = 1 + 0.519 999 981 387 776;
  • 20) 0.519 999 981 387 776 × 2 = 1 + 0.039 999 962 775 552;
  • 21) 0.039 999 962 775 552 × 2 = 0 + 0.079 999 925 551 104;
  • 22) 0.079 999 925 551 104 × 2 = 0 + 0.159 999 851 102 208;
  • 23) 0.159 999 851 102 208 × 2 = 0 + 0.319 999 702 204 416;
  • 24) 0.319 999 702 204 416 × 2 = 0 + 0.639 999 404 408 832;
  • 25) 0.639 999 404 408 832 × 2 = 1 + 0.279 998 808 817 664;
  • 26) 0.279 998 808 817 664 × 2 = 0 + 0.559 997 617 635 328;
  • 27) 0.559 997 617 635 328 × 2 = 1 + 0.119 995 235 270 656;
  • 28) 0.119 995 235 270 656 × 2 = 0 + 0.239 990 470 541 312;
  • 29) 0.239 990 470 541 312 × 2 = 0 + 0.479 980 941 082 624;
  • 30) 0.479 980 941 082 624 × 2 = 0 + 0.959 961 882 165 248;
  • 31) 0.959 961 882 165 248 × 2 = 1 + 0.919 923 764 330 496;
  • 32) 0.919 923 764 330 496 × 2 = 1 + 0.839 847 528 660 992;
  • 33) 0.839 847 528 660 992 × 2 = 1 + 0.679 695 057 321 984;
  • 34) 0.679 695 057 321 984 × 2 = 1 + 0.359 390 114 643 968;
  • 35) 0.359 390 114 643 968 × 2 = 0 + 0.718 780 229 287 936;
  • 36) 0.718 780 229 287 936 × 2 = 1 + 0.437 560 458 575 872;
  • 37) 0.437 560 458 575 872 × 2 = 0 + 0.875 120 917 151 744;
  • 38) 0.875 120 917 151 744 × 2 = 1 + 0.750 241 834 303 488;
  • 39) 0.750 241 834 303 488 × 2 = 1 + 0.500 483 668 606 976;
  • 40) 0.500 483 668 606 976 × 2 = 1 + 0.000 967 337 213 952;
  • 41) 0.000 967 337 213 952 × 2 = 0 + 0.001 934 674 427 904;
  • 42) 0.001 934 674 427 904 × 2 = 0 + 0.003 869 348 855 808;
  • 43) 0.003 869 348 855 808 × 2 = 0 + 0.007 738 697 711 616;
  • 44) 0.007 738 697 711 616 × 2 = 0 + 0.015 477 395 423 232;
  • 45) 0.015 477 395 423 232 × 2 = 0 + 0.030 954 790 846 464;
  • 46) 0.030 954 790 846 464 × 2 = 0 + 0.061 909 581 692 928;
  • 47) 0.061 909 581 692 928 × 2 = 0 + 0.123 819 163 385 856;
  • 48) 0.123 819 163 385 856 × 2 = 0 + 0.247 638 326 771 712;
  • 49) 0.247 638 326 771 712 × 2 = 0 + 0.495 276 653 543 424;
  • 50) 0.495 276 653 543 424 × 2 = 0 + 0.990 553 307 086 848;
  • 51) 0.990 553 307 086 848 × 2 = 1 + 0.981 106 614 173 696;
  • 52) 0.981 106 614 173 696 × 2 = 1 + 0.962 213 228 347 392;
  • 53) 0.962 213 228 347 392 × 2 = 1 + 0.924 426 456 694 784;
  • 54) 0.924 426 456 694 784 × 2 = 1 + 0.848 852 913 389 568;
  • 55) 0.848 852 913 389 568 × 2 = 1 + 0.697 705 826 779 136;

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:


0.164 999 999 999 964 5(10) =


0.0010 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000 0011 111(2)

6. Positive number before normalization:

0.164 999 999 999 964 5(10) =


0.0010 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000 0011 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.164 999 999 999 964 5(10) =


0.0010 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000 0011 111(2) =


0.0010 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000 0011 111(2) × 20 =


1.0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0001 1111(2) × 2-3


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): -3


Mantissa (not normalized):
1.0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0001 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


-3 + 2(11-1) - 1 =


(-3 + 1 023)(10) =


1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 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) =


1020(10) =


011 1111 1100(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. 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0001 1111 =


0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0001 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 1100


Mantissa (52 bits) =
0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0001 1111


The base ten decimal number -0.164 999 999 999 964 5 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0101 0001 1110 1011 1000 0101 0001 1110 1011 1000 0000 0001 1111

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