0.000 001 300 900 768 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 001 300 900 768(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
0.000 001 300 900 768(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 001 300 900 768.

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) 0.000 001 300 900 768 × 2 = 0 + 0.000 002 601 801 536;
  • 2) 0.000 002 601 801 536 × 2 = 0 + 0.000 005 203 603 072;
  • 3) 0.000 005 203 603 072 × 2 = 0 + 0.000 010 407 206 144;
  • 4) 0.000 010 407 206 144 × 2 = 0 + 0.000 020 814 412 288;
  • 5) 0.000 020 814 412 288 × 2 = 0 + 0.000 041 628 824 576;
  • 6) 0.000 041 628 824 576 × 2 = 0 + 0.000 083 257 649 152;
  • 7) 0.000 083 257 649 152 × 2 = 0 + 0.000 166 515 298 304;
  • 8) 0.000 166 515 298 304 × 2 = 0 + 0.000 333 030 596 608;
  • 9) 0.000 333 030 596 608 × 2 = 0 + 0.000 666 061 193 216;
  • 10) 0.000 666 061 193 216 × 2 = 0 + 0.001 332 122 386 432;
  • 11) 0.001 332 122 386 432 × 2 = 0 + 0.002 664 244 772 864;
  • 12) 0.002 664 244 772 864 × 2 = 0 + 0.005 328 489 545 728;
  • 13) 0.005 328 489 545 728 × 2 = 0 + 0.010 656 979 091 456;
  • 14) 0.010 656 979 091 456 × 2 = 0 + 0.021 313 958 182 912;
  • 15) 0.021 313 958 182 912 × 2 = 0 + 0.042 627 916 365 824;
  • 16) 0.042 627 916 365 824 × 2 = 0 + 0.085 255 832 731 648;
  • 17) 0.085 255 832 731 648 × 2 = 0 + 0.170 511 665 463 296;
  • 18) 0.170 511 665 463 296 × 2 = 0 + 0.341 023 330 926 592;
  • 19) 0.341 023 330 926 592 × 2 = 0 + 0.682 046 661 853 184;
  • 20) 0.682 046 661 853 184 × 2 = 1 + 0.364 093 323 706 368;
  • 21) 0.364 093 323 706 368 × 2 = 0 + 0.728 186 647 412 736;
  • 22) 0.728 186 647 412 736 × 2 = 1 + 0.456 373 294 825 472;
  • 23) 0.456 373 294 825 472 × 2 = 0 + 0.912 746 589 650 944;
  • 24) 0.912 746 589 650 944 × 2 = 1 + 0.825 493 179 301 888;
  • 25) 0.825 493 179 301 888 × 2 = 1 + 0.650 986 358 603 776;
  • 26) 0.650 986 358 603 776 × 2 = 1 + 0.301 972 717 207 552;
  • 27) 0.301 972 717 207 552 × 2 = 0 + 0.603 945 434 415 104;
  • 28) 0.603 945 434 415 104 × 2 = 1 + 0.207 890 868 830 208;
  • 29) 0.207 890 868 830 208 × 2 = 0 + 0.415 781 737 660 416;
  • 30) 0.415 781 737 660 416 × 2 = 0 + 0.831 563 475 320 832;
  • 31) 0.831 563 475 320 832 × 2 = 1 + 0.663 126 950 641 664;
  • 32) 0.663 126 950 641 664 × 2 = 1 + 0.326 253 901 283 328;
  • 33) 0.326 253 901 283 328 × 2 = 0 + 0.652 507 802 566 656;
  • 34) 0.652 507 802 566 656 × 2 = 1 + 0.305 015 605 133 312;
  • 35) 0.305 015 605 133 312 × 2 = 0 + 0.610 031 210 266 624;
  • 36) 0.610 031 210 266 624 × 2 = 1 + 0.220 062 420 533 248;
  • 37) 0.220 062 420 533 248 × 2 = 0 + 0.440 124 841 066 496;
  • 38) 0.440 124 841 066 496 × 2 = 0 + 0.880 249 682 132 992;
  • 39) 0.880 249 682 132 992 × 2 = 1 + 0.760 499 364 265 984;
  • 40) 0.760 499 364 265 984 × 2 = 1 + 0.520 998 728 531 968;
  • 41) 0.520 998 728 531 968 × 2 = 1 + 0.041 997 457 063 936;
  • 42) 0.041 997 457 063 936 × 2 = 0 + 0.083 994 914 127 872;
  • 43) 0.083 994 914 127 872 × 2 = 0 + 0.167 989 828 255 744;
  • 44) 0.167 989 828 255 744 × 2 = 0 + 0.335 979 656 511 488;
  • 45) 0.335 979 656 511 488 × 2 = 0 + 0.671 959 313 022 976;
  • 46) 0.671 959 313 022 976 × 2 = 1 + 0.343 918 626 045 952;
  • 47) 0.343 918 626 045 952 × 2 = 0 + 0.687 837 252 091 904;
  • 48) 0.687 837 252 091 904 × 2 = 1 + 0.375 674 504 183 808;
  • 49) 0.375 674 504 183 808 × 2 = 0 + 0.751 349 008 367 616;
  • 50) 0.751 349 008 367 616 × 2 = 1 + 0.502 698 016 735 232;
  • 51) 0.502 698 016 735 232 × 2 = 1 + 0.005 396 033 470 464;
  • 52) 0.005 396 033 470 464 × 2 = 0 + 0.010 792 066 940 928;
  • 53) 0.010 792 066 940 928 × 2 = 0 + 0.021 584 133 881 856;
  • 54) 0.021 584 133 881 856 × 2 = 0 + 0.043 168 267 763 712;
  • 55) 0.043 168 267 763 712 × 2 = 0 + 0.086 336 535 527 424;
  • 56) 0.086 336 535 527 424 × 2 = 0 + 0.172 673 071 054 848;
  • 57) 0.172 673 071 054 848 × 2 = 0 + 0.345 346 142 109 696;
  • 58) 0.345 346 142 109 696 × 2 = 0 + 0.690 692 284 219 392;
  • 59) 0.690 692 284 219 392 × 2 = 1 + 0.381 384 568 438 784;
  • 60) 0.381 384 568 438 784 × 2 = 0 + 0.762 769 136 877 568;
  • 61) 0.762 769 136 877 568 × 2 = 1 + 0.525 538 273 755 136;
  • 62) 0.525 538 273 755 136 × 2 = 1 + 0.051 076 547 510 272;
  • 63) 0.051 076 547 510 272 × 2 = 0 + 0.102 153 095 020 544;
  • 64) 0.102 153 095 020 544 × 2 = 0 + 0.204 306 190 041 088;
  • 65) 0.204 306 190 041 088 × 2 = 0 + 0.408 612 380 082 176;
  • 66) 0.408 612 380 082 176 × 2 = 0 + 0.817 224 760 164 352;
  • 67) 0.817 224 760 164 352 × 2 = 1 + 0.634 449 520 328 704;
  • 68) 0.634 449 520 328 704 × 2 = 1 + 0.268 899 040 657 408;
  • 69) 0.268 899 040 657 408 × 2 = 0 + 0.537 798 081 314 816;
  • 70) 0.537 798 081 314 816 × 2 = 1 + 0.075 596 162 629 632;
  • 71) 0.075 596 162 629 632 × 2 = 0 + 0.151 192 325 259 264;
  • 72) 0.151 192 325 259 264 × 2 = 0 + 0.302 384 650 518 528;

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).


4. 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:


0.000 001 300 900 768(10) =

0.0000 0000 0000 0000 0001 0101 1101 0011 0101 0011 1000 0101 0110 0000 0010 1100 0011 0100(2)

5. Positive number before normalization:

0.000 001 300 900 768(10) =

0.0000 0000 0000 0000 0001 0101 1101 0011 0101 0011 1000 0101 0110 0000 0010 1100 0011 0100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 300 900 768(10) =

0.0000 0000 0000 0000 0001 0101 1101 0011 0101 0011 1000 0101 0110 0000 0010 1100 0011 0100(2) =

0.0000 0000 0000 0000 0001 0101 1101 0011 0101 0011 1000 0101 0110 0000 0010 1100 0011 0100(2) × 20 =

1.0101 1101 0011 0101 0011 1000 0101 0110 0000 0010 1100 0011 0100(2) × 2-20


7. 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 0 (a positive number)

Exponent (unadjusted): -20

Mantissa (not normalized):
1.0101 1101 0011 0101 0011 1000 0101 0110 0000 0010 1100 0011 0100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(11-1) - 1 =

(-20 + 1 023)(10) =

1 003(10)


9. 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;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1003(10) =

011 1110 1011(2)


11. 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 1101 0011 0101 0011 1000 0101 0110 0000 0010 1100 0011 0100 =

0101 1101 0011 0101 0011 1000 0101 0110 0000 0010 1100 0011 0100


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1110 1011

Mantissa (52 bits) =
0101 1101 0011 0101 0011 1000 0101 0110 0000 0010 1100 0011 0100


Decimal number 0.000 001 300 900 768 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1011 - 0101 1101 0011 0101 0011 1000 0101 0110 0000 0010 1100 0011 0100

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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