-0.000 036 240 042 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 036 240 042(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
-0.000 036 240 042(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:

|-0.000 036 240 042| = 0.000 036 240 042


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: 0.000 036 240 042.

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) 0.000 036 240 042 × 2 = 0 + 0.000 072 480 084;
  • 2) 0.000 072 480 084 × 2 = 0 + 0.000 144 960 168;
  • 3) 0.000 144 960 168 × 2 = 0 + 0.000 289 920 336;
  • 4) 0.000 289 920 336 × 2 = 0 + 0.000 579 840 672;
  • 5) 0.000 579 840 672 × 2 = 0 + 0.001 159 681 344;
  • 6) 0.001 159 681 344 × 2 = 0 + 0.002 319 362 688;
  • 7) 0.002 319 362 688 × 2 = 0 + 0.004 638 725 376;
  • 8) 0.004 638 725 376 × 2 = 0 + 0.009 277 450 752;
  • 9) 0.009 277 450 752 × 2 = 0 + 0.018 554 901 504;
  • 10) 0.018 554 901 504 × 2 = 0 + 0.037 109 803 008;
  • 11) 0.037 109 803 008 × 2 = 0 + 0.074 219 606 016;
  • 12) 0.074 219 606 016 × 2 = 0 + 0.148 439 212 032;
  • 13) 0.148 439 212 032 × 2 = 0 + 0.296 878 424 064;
  • 14) 0.296 878 424 064 × 2 = 0 + 0.593 756 848 128;
  • 15) 0.593 756 848 128 × 2 = 1 + 0.187 513 696 256;
  • 16) 0.187 513 696 256 × 2 = 0 + 0.375 027 392 512;
  • 17) 0.375 027 392 512 × 2 = 0 + 0.750 054 785 024;
  • 18) 0.750 054 785 024 × 2 = 1 + 0.500 109 570 048;
  • 19) 0.500 109 570 048 × 2 = 1 + 0.000 219 140 096;
  • 20) 0.000 219 140 096 × 2 = 0 + 0.000 438 280 192;
  • 21) 0.000 438 280 192 × 2 = 0 + 0.000 876 560 384;
  • 22) 0.000 876 560 384 × 2 = 0 + 0.001 753 120 768;
  • 23) 0.001 753 120 768 × 2 = 0 + 0.003 506 241 536;
  • 24) 0.003 506 241 536 × 2 = 0 + 0.007 012 483 072;
  • 25) 0.007 012 483 072 × 2 = 0 + 0.014 024 966 144;
  • 26) 0.014 024 966 144 × 2 = 0 + 0.028 049 932 288;
  • 27) 0.028 049 932 288 × 2 = 0 + 0.056 099 864 576;
  • 28) 0.056 099 864 576 × 2 = 0 + 0.112 199 729 152;
  • 29) 0.112 199 729 152 × 2 = 0 + 0.224 399 458 304;
  • 30) 0.224 399 458 304 × 2 = 0 + 0.448 798 916 608;
  • 31) 0.448 798 916 608 × 2 = 0 + 0.897 597 833 216;
  • 32) 0.897 597 833 216 × 2 = 1 + 0.795 195 666 432;
  • 33) 0.795 195 666 432 × 2 = 1 + 0.590 391 332 864;
  • 34) 0.590 391 332 864 × 2 = 1 + 0.180 782 665 728;
  • 35) 0.180 782 665 728 × 2 = 0 + 0.361 565 331 456;
  • 36) 0.361 565 331 456 × 2 = 0 + 0.723 130 662 912;
  • 37) 0.723 130 662 912 × 2 = 1 + 0.446 261 325 824;
  • 38) 0.446 261 325 824 × 2 = 0 + 0.892 522 651 648;
  • 39) 0.892 522 651 648 × 2 = 1 + 0.785 045 303 296;
  • 40) 0.785 045 303 296 × 2 = 1 + 0.570 090 606 592;
  • 41) 0.570 090 606 592 × 2 = 1 + 0.140 181 213 184;
  • 42) 0.140 181 213 184 × 2 = 0 + 0.280 362 426 368;
  • 43) 0.280 362 426 368 × 2 = 0 + 0.560 724 852 736;
  • 44) 0.560 724 852 736 × 2 = 1 + 0.121 449 705 472;
  • 45) 0.121 449 705 472 × 2 = 0 + 0.242 899 410 944;
  • 46) 0.242 899 410 944 × 2 = 0 + 0.485 798 821 888;
  • 47) 0.485 798 821 888 × 2 = 0 + 0.971 597 643 776;
  • 48) 0.971 597 643 776 × 2 = 1 + 0.943 195 287 552;
  • 49) 0.943 195 287 552 × 2 = 1 + 0.886 390 575 104;
  • 50) 0.886 390 575 104 × 2 = 1 + 0.772 781 150 208;
  • 51) 0.772 781 150 208 × 2 = 1 + 0.545 562 300 416;
  • 52) 0.545 562 300 416 × 2 = 1 + 0.091 124 600 832;
  • 53) 0.091 124 600 832 × 2 = 0 + 0.182 249 201 664;
  • 54) 0.182 249 201 664 × 2 = 0 + 0.364 498 403 328;
  • 55) 0.364 498 403 328 × 2 = 0 + 0.728 996 806 656;
  • 56) 0.728 996 806 656 × 2 = 1 + 0.457 993 613 312;
  • 57) 0.457 993 613 312 × 2 = 0 + 0.915 987 226 624;
  • 58) 0.915 987 226 624 × 2 = 1 + 0.831 974 453 248;
  • 59) 0.831 974 453 248 × 2 = 1 + 0.663 948 906 496;
  • 60) 0.663 948 906 496 × 2 = 1 + 0.327 897 812 992;
  • 61) 0.327 897 812 992 × 2 = 0 + 0.655 795 625 984;
  • 62) 0.655 795 625 984 × 2 = 1 + 0.311 591 251 968;
  • 63) 0.311 591 251 968 × 2 = 0 + 0.623 182 503 936;
  • 64) 0.623 182 503 936 × 2 = 1 + 0.246 365 007 872;
  • 65) 0.246 365 007 872 × 2 = 0 + 0.492 730 015 744;
  • 66) 0.492 730 015 744 × 2 = 0 + 0.985 460 031 488;
  • 67) 0.985 460 031 488 × 2 = 1 + 0.970 920 062 976;

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:


0.000 036 240 042(10) =

0.0000 0000 0000 0010 0110 0000 0000 0001 1100 1011 1001 0001 1111 0001 0111 0101 001(2)

6. Positive number before normalization:

0.000 036 240 042(10) =

0.0000 0000 0000 0010 0110 0000 0000 0001 1100 1011 1001 0001 1111 0001 0111 0101 001(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:


0.000 036 240 042(10) =

0.0000 0000 0000 0010 0110 0000 0000 0001 1100 1011 1001 0001 1111 0001 0111 0101 001(2) =

0.0000 0000 0000 0010 0110 0000 0000 0001 1100 1011 1001 0001 1111 0001 0111 0101 001(2) × 20 =

1.0011 0000 0000 0000 1110 0101 1100 1000 1111 1000 1011 1010 1001(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):
1.0011 0000 0000 0000 1110 0101 1100 1000 1111 1000 1011 1010 1001


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. 0011 0000 0000 0000 1110 0101 1100 1000 1111 1000 1011 1010 1001 =

0011 0000 0000 0000 1110 0101 1100 1000 1111 1000 1011 1010 1001


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) =
0011 0000 0000 0000 1110 0101 1100 1000 1111 1000 1011 1010 1001


Decimal number -0.000 036 240 042 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0011 0000 0000 0000 1110 0101 1100 1000 1111 1000 1011 1010 1001

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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