-0.000 035 667 067 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 667 067(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
-0.000 035 667 067(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:

|-0.000 035 667 067| = 0.000 035 667 067


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: 0.000 035 667 067.

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) 0.000 035 667 067 × 2 = 0 + 0.000 071 334 134;
  • 2) 0.000 071 334 134 × 2 = 0 + 0.000 142 668 268;
  • 3) 0.000 142 668 268 × 2 = 0 + 0.000 285 336 536;
  • 4) 0.000 285 336 536 × 2 = 0 + 0.000 570 673 072;
  • 5) 0.000 570 673 072 × 2 = 0 + 0.001 141 346 144;
  • 6) 0.001 141 346 144 × 2 = 0 + 0.002 282 692 288;
  • 7) 0.002 282 692 288 × 2 = 0 + 0.004 565 384 576;
  • 8) 0.004 565 384 576 × 2 = 0 + 0.009 130 769 152;
  • 9) 0.009 130 769 152 × 2 = 0 + 0.018 261 538 304;
  • 10) 0.018 261 538 304 × 2 = 0 + 0.036 523 076 608;
  • 11) 0.036 523 076 608 × 2 = 0 + 0.073 046 153 216;
  • 12) 0.073 046 153 216 × 2 = 0 + 0.146 092 306 432;
  • 13) 0.146 092 306 432 × 2 = 0 + 0.292 184 612 864;
  • 14) 0.292 184 612 864 × 2 = 0 + 0.584 369 225 728;
  • 15) 0.584 369 225 728 × 2 = 1 + 0.168 738 451 456;
  • 16) 0.168 738 451 456 × 2 = 0 + 0.337 476 902 912;
  • 17) 0.337 476 902 912 × 2 = 0 + 0.674 953 805 824;
  • 18) 0.674 953 805 824 × 2 = 1 + 0.349 907 611 648;
  • 19) 0.349 907 611 648 × 2 = 0 + 0.699 815 223 296;
  • 20) 0.699 815 223 296 × 2 = 1 + 0.399 630 446 592;
  • 21) 0.399 630 446 592 × 2 = 0 + 0.799 260 893 184;
  • 22) 0.799 260 893 184 × 2 = 1 + 0.598 521 786 368;
  • 23) 0.598 521 786 368 × 2 = 1 + 0.197 043 572 736;
  • 24) 0.197 043 572 736 × 2 = 0 + 0.394 087 145 472;
  • 25) 0.394 087 145 472 × 2 = 0 + 0.788 174 290 944;
  • 26) 0.788 174 290 944 × 2 = 1 + 0.576 348 581 888;
  • 27) 0.576 348 581 888 × 2 = 1 + 0.152 697 163 776;
  • 28) 0.152 697 163 776 × 2 = 0 + 0.305 394 327 552;
  • 29) 0.305 394 327 552 × 2 = 0 + 0.610 788 655 104;
  • 30) 0.610 788 655 104 × 2 = 1 + 0.221 577 310 208;
  • 31) 0.221 577 310 208 × 2 = 0 + 0.443 154 620 416;
  • 32) 0.443 154 620 416 × 2 = 0 + 0.886 309 240 832;
  • 33) 0.886 309 240 832 × 2 = 1 + 0.772 618 481 664;
  • 34) 0.772 618 481 664 × 2 = 1 + 0.545 236 963 328;
  • 35) 0.545 236 963 328 × 2 = 1 + 0.090 473 926 656;
  • 36) 0.090 473 926 656 × 2 = 0 + 0.180 947 853 312;
  • 37) 0.180 947 853 312 × 2 = 0 + 0.361 895 706 624;
  • 38) 0.361 895 706 624 × 2 = 0 + 0.723 791 413 248;
  • 39) 0.723 791 413 248 × 2 = 1 + 0.447 582 826 496;
  • 40) 0.447 582 826 496 × 2 = 0 + 0.895 165 652 992;
  • 41) 0.895 165 652 992 × 2 = 1 + 0.790 331 305 984;
  • 42) 0.790 331 305 984 × 2 = 1 + 0.580 662 611 968;
  • 43) 0.580 662 611 968 × 2 = 1 + 0.161 325 223 936;
  • 44) 0.161 325 223 936 × 2 = 0 + 0.322 650 447 872;
  • 45) 0.322 650 447 872 × 2 = 0 + 0.645 300 895 744;
  • 46) 0.645 300 895 744 × 2 = 1 + 0.290 601 791 488;
  • 47) 0.290 601 791 488 × 2 = 0 + 0.581 203 582 976;
  • 48) 0.581 203 582 976 × 2 = 1 + 0.162 407 165 952;
  • 49) 0.162 407 165 952 × 2 = 0 + 0.324 814 331 904;
  • 50) 0.324 814 331 904 × 2 = 0 + 0.649 628 663 808;
  • 51) 0.649 628 663 808 × 2 = 1 + 0.299 257 327 616;
  • 52) 0.299 257 327 616 × 2 = 0 + 0.598 514 655 232;
  • 53) 0.598 514 655 232 × 2 = 1 + 0.197 029 310 464;
  • 54) 0.197 029 310 464 × 2 = 0 + 0.394 058 620 928;
  • 55) 0.394 058 620 928 × 2 = 0 + 0.788 117 241 856;
  • 56) 0.788 117 241 856 × 2 = 1 + 0.576 234 483 712;
  • 57) 0.576 234 483 712 × 2 = 1 + 0.152 468 967 424;
  • 58) 0.152 468 967 424 × 2 = 0 + 0.304 937 934 848;
  • 59) 0.304 937 934 848 × 2 = 0 + 0.609 875 869 696;
  • 60) 0.609 875 869 696 × 2 = 1 + 0.219 751 739 392;
  • 61) 0.219 751 739 392 × 2 = 0 + 0.439 503 478 784;
  • 62) 0.439 503 478 784 × 2 = 0 + 0.879 006 957 568;
  • 63) 0.879 006 957 568 × 2 = 1 + 0.758 013 915 136;
  • 64) 0.758 013 915 136 × 2 = 1 + 0.516 027 830 272;
  • 65) 0.516 027 830 272 × 2 = 1 + 0.032 055 660 544;
  • 66) 0.032 055 660 544 × 2 = 0 + 0.064 111 321 088;
  • 67) 0.064 111 321 088 × 2 = 0 + 0.128 222 642 176;

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:


0.000 035 667 067(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 1110 0010 1110 0101 0010 1001 1001 0011 100(2)

6. Positive number before normalization:

0.000 035 667 067(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 1110 0010 1110 0101 0010 1001 1001 0011 100(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:


0.000 035 667 067(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 1110 0010 1110 0101 0010 1001 1001 0011 100(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 1110 0010 1110 0101 0010 1001 1001 0011 100(2) × 20 =

1.0010 1011 0011 0010 0111 0001 0111 0010 1001 0100 1100 1001 1100(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 0111 0001 0111 0010 1001 0100 1100 1001 1100


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 0111 0001 0111 0010 1001 0100 1100 1001 1100 =

0010 1011 0011 0010 0111 0001 0111 0010 1001 0100 1100 1001 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 0000

Mantissa (52 bits) =
0010 1011 0011 0010 0111 0001 0111 0010 1001 0100 1100 1001 1100


Decimal number -0.000 035 667 067 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0111 0001 0111 0010 1001 0100 1100 1001 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