-0.000 035 666 971 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 971(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
-0.000 035 666 971(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:

|-0.000 035 666 971| = 0.000 035 666 971


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: 0.000 035 666 971.

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) 0.000 035 666 971 × 2 = 0 + 0.000 071 333 942;
  • 2) 0.000 071 333 942 × 2 = 0 + 0.000 142 667 884;
  • 3) 0.000 142 667 884 × 2 = 0 + 0.000 285 335 768;
  • 4) 0.000 285 335 768 × 2 = 0 + 0.000 570 671 536;
  • 5) 0.000 570 671 536 × 2 = 0 + 0.001 141 343 072;
  • 6) 0.001 141 343 072 × 2 = 0 + 0.002 282 686 144;
  • 7) 0.002 282 686 144 × 2 = 0 + 0.004 565 372 288;
  • 8) 0.004 565 372 288 × 2 = 0 + 0.009 130 744 576;
  • 9) 0.009 130 744 576 × 2 = 0 + 0.018 261 489 152;
  • 10) 0.018 261 489 152 × 2 = 0 + 0.036 522 978 304;
  • 11) 0.036 522 978 304 × 2 = 0 + 0.073 045 956 608;
  • 12) 0.073 045 956 608 × 2 = 0 + 0.146 091 913 216;
  • 13) 0.146 091 913 216 × 2 = 0 + 0.292 183 826 432;
  • 14) 0.292 183 826 432 × 2 = 0 + 0.584 367 652 864;
  • 15) 0.584 367 652 864 × 2 = 1 + 0.168 735 305 728;
  • 16) 0.168 735 305 728 × 2 = 0 + 0.337 470 611 456;
  • 17) 0.337 470 611 456 × 2 = 0 + 0.674 941 222 912;
  • 18) 0.674 941 222 912 × 2 = 1 + 0.349 882 445 824;
  • 19) 0.349 882 445 824 × 2 = 0 + 0.699 764 891 648;
  • 20) 0.699 764 891 648 × 2 = 1 + 0.399 529 783 296;
  • 21) 0.399 529 783 296 × 2 = 0 + 0.799 059 566 592;
  • 22) 0.799 059 566 592 × 2 = 1 + 0.598 119 133 184;
  • 23) 0.598 119 133 184 × 2 = 1 + 0.196 238 266 368;
  • 24) 0.196 238 266 368 × 2 = 0 + 0.392 476 532 736;
  • 25) 0.392 476 532 736 × 2 = 0 + 0.784 953 065 472;
  • 26) 0.784 953 065 472 × 2 = 1 + 0.569 906 130 944;
  • 27) 0.569 906 130 944 × 2 = 1 + 0.139 812 261 888;
  • 28) 0.139 812 261 888 × 2 = 0 + 0.279 624 523 776;
  • 29) 0.279 624 523 776 × 2 = 0 + 0.559 249 047 552;
  • 30) 0.559 249 047 552 × 2 = 1 + 0.118 498 095 104;
  • 31) 0.118 498 095 104 × 2 = 0 + 0.236 996 190 208;
  • 32) 0.236 996 190 208 × 2 = 0 + 0.473 992 380 416;
  • 33) 0.473 992 380 416 × 2 = 0 + 0.947 984 760 832;
  • 34) 0.947 984 760 832 × 2 = 1 + 0.895 969 521 664;
  • 35) 0.895 969 521 664 × 2 = 1 + 0.791 939 043 328;
  • 36) 0.791 939 043 328 × 2 = 1 + 0.583 878 086 656;
  • 37) 0.583 878 086 656 × 2 = 1 + 0.167 756 173 312;
  • 38) 0.167 756 173 312 × 2 = 0 + 0.335 512 346 624;
  • 39) 0.335 512 346 624 × 2 = 0 + 0.671 024 693 248;
  • 40) 0.671 024 693 248 × 2 = 1 + 0.342 049 386 496;
  • 41) 0.342 049 386 496 × 2 = 0 + 0.684 098 772 992;
  • 42) 0.684 098 772 992 × 2 = 1 + 0.368 197 545 984;
  • 43) 0.368 197 545 984 × 2 = 0 + 0.736 395 091 968;
  • 44) 0.736 395 091 968 × 2 = 1 + 0.472 790 183 936;
  • 45) 0.472 790 183 936 × 2 = 0 + 0.945 580 367 872;
  • 46) 0.945 580 367 872 × 2 = 1 + 0.891 160 735 744;
  • 47) 0.891 160 735 744 × 2 = 1 + 0.782 321 471 488;
  • 48) 0.782 321 471 488 × 2 = 1 + 0.564 642 942 976;
  • 49) 0.564 642 942 976 × 2 = 1 + 0.129 285 885 952;
  • 50) 0.129 285 885 952 × 2 = 0 + 0.258 571 771 904;
  • 51) 0.258 571 771 904 × 2 = 0 + 0.517 143 543 808;
  • 52) 0.517 143 543 808 × 2 = 1 + 0.034 287 087 616;
  • 53) 0.034 287 087 616 × 2 = 0 + 0.068 574 175 232;
  • 54) 0.068 574 175 232 × 2 = 0 + 0.137 148 350 464;
  • 55) 0.137 148 350 464 × 2 = 0 + 0.274 296 700 928;
  • 56) 0.274 296 700 928 × 2 = 0 + 0.548 593 401 856;
  • 57) 0.548 593 401 856 × 2 = 1 + 0.097 186 803 712;
  • 58) 0.097 186 803 712 × 2 = 0 + 0.194 373 607 424;
  • 59) 0.194 373 607 424 × 2 = 0 + 0.388 747 214 848;
  • 60) 0.388 747 214 848 × 2 = 0 + 0.777 494 429 696;
  • 61) 0.777 494 429 696 × 2 = 1 + 0.554 988 859 392;
  • 62) 0.554 988 859 392 × 2 = 1 + 0.109 977 718 784;
  • 63) 0.109 977 718 784 × 2 = 0 + 0.219 955 437 568;
  • 64) 0.219 955 437 568 × 2 = 0 + 0.439 910 875 136;
  • 65) 0.439 910 875 136 × 2 = 0 + 0.879 821 750 272;
  • 66) 0.879 821 750 272 × 2 = 1 + 0.759 643 500 544;
  • 67) 0.759 643 500 544 × 2 = 1 + 0.519 287 001 088;

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:


0.000 035 666 971(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0111 1001 0101 0111 1001 0000 1000 1100 011(2)

6. Positive number before normalization:

0.000 035 666 971(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0111 1001 0101 0111 1001 0000 1000 1100 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 035 666 971(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0111 1001 0101 0111 1001 0000 1000 1100 011(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0111 1001 0101 0111 1001 0000 1000 1100 011(2) × 20 =

1.0010 1011 0011 0010 0011 1100 1010 1011 1100 1000 0100 0110 0011(2) × 2-15


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): -15

Mantissa (not normalized):
1.0010 1011 0011 0010 0011 1100 1010 1011 1100 1000 0100 0110 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 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) =

1008(10) =

011 1111 0000(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 1011 0011 0010 0011 1100 1010 1011 1100 1000 0100 0110 0011 =

0010 1011 0011 0010 0011 1100 1010 1011 1100 1000 0100 0110 0011


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 0000

Mantissa (52 bits) =
0010 1011 0011 0010 0011 1100 1010 1011 1100 1000 0100 0110 0011


Decimal number -0.000 035 666 971 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0011 1100 1010 1011 1100 1000 0100 0110 0011

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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