0.000 000 347 328 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 347 328(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
0.000 000 347 328(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: 0.000 000 347 328.

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) 0.000 000 347 328 × 2 = 0 + 0.000 000 694 656;
  • 2) 0.000 000 694 656 × 2 = 0 + 0.000 001 389 312;
  • 3) 0.000 001 389 312 × 2 = 0 + 0.000 002 778 624;
  • 4) 0.000 002 778 624 × 2 = 0 + 0.000 005 557 248;
  • 5) 0.000 005 557 248 × 2 = 0 + 0.000 011 114 496;
  • 6) 0.000 011 114 496 × 2 = 0 + 0.000 022 228 992;
  • 7) 0.000 022 228 992 × 2 = 0 + 0.000 044 457 984;
  • 8) 0.000 044 457 984 × 2 = 0 + 0.000 088 915 968;
  • 9) 0.000 088 915 968 × 2 = 0 + 0.000 177 831 936;
  • 10) 0.000 177 831 936 × 2 = 0 + 0.000 355 663 872;
  • 11) 0.000 355 663 872 × 2 = 0 + 0.000 711 327 744;
  • 12) 0.000 711 327 744 × 2 = 0 + 0.001 422 655 488;
  • 13) 0.001 422 655 488 × 2 = 0 + 0.002 845 310 976;
  • 14) 0.002 845 310 976 × 2 = 0 + 0.005 690 621 952;
  • 15) 0.005 690 621 952 × 2 = 0 + 0.011 381 243 904;
  • 16) 0.011 381 243 904 × 2 = 0 + 0.022 762 487 808;
  • 17) 0.022 762 487 808 × 2 = 0 + 0.045 524 975 616;
  • 18) 0.045 524 975 616 × 2 = 0 + 0.091 049 951 232;
  • 19) 0.091 049 951 232 × 2 = 0 + 0.182 099 902 464;
  • 20) 0.182 099 902 464 × 2 = 0 + 0.364 199 804 928;
  • 21) 0.364 199 804 928 × 2 = 0 + 0.728 399 609 856;
  • 22) 0.728 399 609 856 × 2 = 1 + 0.456 799 219 712;
  • 23) 0.456 799 219 712 × 2 = 0 + 0.913 598 439 424;
  • 24) 0.913 598 439 424 × 2 = 1 + 0.827 196 878 848;
  • 25) 0.827 196 878 848 × 2 = 1 + 0.654 393 757 696;
  • 26) 0.654 393 757 696 × 2 = 1 + 0.308 787 515 392;
  • 27) 0.308 787 515 392 × 2 = 0 + 0.617 575 030 784;
  • 28) 0.617 575 030 784 × 2 = 1 + 0.235 150 061 568;
  • 29) 0.235 150 061 568 × 2 = 0 + 0.470 300 123 136;
  • 30) 0.470 300 123 136 × 2 = 0 + 0.940 600 246 272;
  • 31) 0.940 600 246 272 × 2 = 1 + 0.881 200 492 544;
  • 32) 0.881 200 492 544 × 2 = 1 + 0.762 400 985 088;
  • 33) 0.762 400 985 088 × 2 = 1 + 0.524 801 970 176;
  • 34) 0.524 801 970 176 × 2 = 1 + 0.049 603 940 352;
  • 35) 0.049 603 940 352 × 2 = 0 + 0.099 207 880 704;
  • 36) 0.099 207 880 704 × 2 = 0 + 0.198 415 761 408;
  • 37) 0.198 415 761 408 × 2 = 0 + 0.396 831 522 816;
  • 38) 0.396 831 522 816 × 2 = 0 + 0.793 663 045 632;
  • 39) 0.793 663 045 632 × 2 = 1 + 0.587 326 091 264;
  • 40) 0.587 326 091 264 × 2 = 1 + 0.174 652 182 528;
  • 41) 0.174 652 182 528 × 2 = 0 + 0.349 304 365 056;
  • 42) 0.349 304 365 056 × 2 = 0 + 0.698 608 730 112;
  • 43) 0.698 608 730 112 × 2 = 1 + 0.397 217 460 224;
  • 44) 0.397 217 460 224 × 2 = 0 + 0.794 434 920 448;
  • 45) 0.794 434 920 448 × 2 = 1 + 0.588 869 840 896;
  • 46) 0.588 869 840 896 × 2 = 1 + 0.177 739 681 792;
  • 47) 0.177 739 681 792 × 2 = 0 + 0.355 479 363 584;
  • 48) 0.355 479 363 584 × 2 = 0 + 0.710 958 727 168;
  • 49) 0.710 958 727 168 × 2 = 1 + 0.421 917 454 336;
  • 50) 0.421 917 454 336 × 2 = 0 + 0.843 834 908 672;
  • 51) 0.843 834 908 672 × 2 = 1 + 0.687 669 817 344;
  • 52) 0.687 669 817 344 × 2 = 1 + 0.375 339 634 688;
  • 53) 0.375 339 634 688 × 2 = 0 + 0.750 679 269 376;
  • 54) 0.750 679 269 376 × 2 = 1 + 0.501 358 538 752;
  • 55) 0.501 358 538 752 × 2 = 1 + 0.002 717 077 504;
  • 56) 0.002 717 077 504 × 2 = 0 + 0.005 434 155 008;
  • 57) 0.005 434 155 008 × 2 = 0 + 0.010 868 310 016;
  • 58) 0.010 868 310 016 × 2 = 0 + 0.021 736 620 032;
  • 59) 0.021 736 620 032 × 2 = 0 + 0.043 473 240 064;
  • 60) 0.043 473 240 064 × 2 = 0 + 0.086 946 480 128;
  • 61) 0.086 946 480 128 × 2 = 0 + 0.173 892 960 256;
  • 62) 0.173 892 960 256 × 2 = 0 + 0.347 785 920 512;
  • 63) 0.347 785 920 512 × 2 = 0 + 0.695 571 841 024;
  • 64) 0.695 571 841 024 × 2 = 1 + 0.391 143 682 048;
  • 65) 0.391 143 682 048 × 2 = 0 + 0.782 287 364 096;
  • 66) 0.782 287 364 096 × 2 = 1 + 0.564 574 728 192;
  • 67) 0.564 574 728 192 × 2 = 1 + 0.129 149 456 384;
  • 68) 0.129 149 456 384 × 2 = 0 + 0.258 298 912 768;
  • 69) 0.258 298 912 768 × 2 = 0 + 0.516 597 825 536;
  • 70) 0.516 597 825 536 × 2 = 1 + 0.033 195 651 072;
  • 71) 0.033 195 651 072 × 2 = 0 + 0.066 391 302 144;
  • 72) 0.066 391 302 144 × 2 = 0 + 0.132 782 604 288;
  • 73) 0.132 782 604 288 × 2 = 0 + 0.265 565 208 576;
  • 74) 0.265 565 208 576 × 2 = 0 + 0.531 130 417 152;

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:


0.000 000 347 328(10) =

0.0000 0000 0000 0000 0000 0101 1101 0011 1100 0011 0010 1100 1011 0110 0000 0001 0110 0100 00(2)

5. Positive number before normalization:

0.000 000 347 328(10) =

0.0000 0000 0000 0000 0000 0101 1101 0011 1100 0011 0010 1100 1011 0110 0000 0001 0110 0100 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 347 328(10) =

0.0000 0000 0000 0000 0000 0101 1101 0011 1100 0011 0010 1100 1011 0110 0000 0001 0110 0100 00(2) =

0.0000 0000 0000 0000 0000 0101 1101 0011 1100 0011 0010 1100 1011 0110 0000 0001 0110 0100 00(2) × 20 =

1.0111 0100 1111 0000 1100 1011 0010 1101 1000 0000 0101 1001 0000(2) × 2-22


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): -22

Mantissa (not normalized):
1.0111 0100 1111 0000 1100 1011 0010 1101 1000 0000 0101 1001 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(11-1) - 1 =

(-22 + 1 023)(10) =

1 001(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;
  • 1 001 ÷ 2 = 500 + 1;
  • 500 ÷ 2 = 250 + 0;
  • 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) =

1001(10) =

011 1110 1001(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. 0111 0100 1111 0000 1100 1011 0010 1101 1000 0000 0101 1001 0000 =

0111 0100 1111 0000 1100 1011 0010 1101 1000 0000 0101 1001 0000


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 1001

Mantissa (52 bits) =
0111 0100 1111 0000 1100 1011 0010 1101 1000 0000 0101 1001 0000


Decimal number 0.000 000 347 328 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1001 - 0111 0100 1111 0000 1100 1011 0010 1101 1000 0000 0101 1001 0000

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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