0.000 684 924 130 745 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 684 924 130 745 9(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
0.000 684 924 130 745 9(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: 0.000 684 924 130 745 9.

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) 0.000 684 924 130 745 9 × 2 = 0 + 0.001 369 848 261 491 8;
  • 2) 0.001 369 848 261 491 8 × 2 = 0 + 0.002 739 696 522 983 6;
  • 3) 0.002 739 696 522 983 6 × 2 = 0 + 0.005 479 393 045 967 2;
  • 4) 0.005 479 393 045 967 2 × 2 = 0 + 0.010 958 786 091 934 4;
  • 5) 0.010 958 786 091 934 4 × 2 = 0 + 0.021 917 572 183 868 8;
  • 6) 0.021 917 572 183 868 8 × 2 = 0 + 0.043 835 144 367 737 6;
  • 7) 0.043 835 144 367 737 6 × 2 = 0 + 0.087 670 288 735 475 2;
  • 8) 0.087 670 288 735 475 2 × 2 = 0 + 0.175 340 577 470 950 4;
  • 9) 0.175 340 577 470 950 4 × 2 = 0 + 0.350 681 154 941 900 8;
  • 10) 0.350 681 154 941 900 8 × 2 = 0 + 0.701 362 309 883 801 6;
  • 11) 0.701 362 309 883 801 6 × 2 = 1 + 0.402 724 619 767 603 2;
  • 12) 0.402 724 619 767 603 2 × 2 = 0 + 0.805 449 239 535 206 4;
  • 13) 0.805 449 239 535 206 4 × 2 = 1 + 0.610 898 479 070 412 8;
  • 14) 0.610 898 479 070 412 8 × 2 = 1 + 0.221 796 958 140 825 6;
  • 15) 0.221 796 958 140 825 6 × 2 = 0 + 0.443 593 916 281 651 2;
  • 16) 0.443 593 916 281 651 2 × 2 = 0 + 0.887 187 832 563 302 4;
  • 17) 0.887 187 832 563 302 4 × 2 = 1 + 0.774 375 665 126 604 8;
  • 18) 0.774 375 665 126 604 8 × 2 = 1 + 0.548 751 330 253 209 6;
  • 19) 0.548 751 330 253 209 6 × 2 = 1 + 0.097 502 660 506 419 2;
  • 20) 0.097 502 660 506 419 2 × 2 = 0 + 0.195 005 321 012 838 4;
  • 21) 0.195 005 321 012 838 4 × 2 = 0 + 0.390 010 642 025 676 8;
  • 22) 0.390 010 642 025 676 8 × 2 = 0 + 0.780 021 284 051 353 6;
  • 23) 0.780 021 284 051 353 6 × 2 = 1 + 0.560 042 568 102 707 2;
  • 24) 0.560 042 568 102 707 2 × 2 = 1 + 0.120 085 136 205 414 4;
  • 25) 0.120 085 136 205 414 4 × 2 = 0 + 0.240 170 272 410 828 8;
  • 26) 0.240 170 272 410 828 8 × 2 = 0 + 0.480 340 544 821 657 6;
  • 27) 0.480 340 544 821 657 6 × 2 = 0 + 0.960 681 089 643 315 2;
  • 28) 0.960 681 089 643 315 2 × 2 = 1 + 0.921 362 179 286 630 4;
  • 29) 0.921 362 179 286 630 4 × 2 = 1 + 0.842 724 358 573 260 8;
  • 30) 0.842 724 358 573 260 8 × 2 = 1 + 0.685 448 717 146 521 6;
  • 31) 0.685 448 717 146 521 6 × 2 = 1 + 0.370 897 434 293 043 2;
  • 32) 0.370 897 434 293 043 2 × 2 = 0 + 0.741 794 868 586 086 4;
  • 33) 0.741 794 868 586 086 4 × 2 = 1 + 0.483 589 737 172 172 8;
  • 34) 0.483 589 737 172 172 8 × 2 = 0 + 0.967 179 474 344 345 6;
  • 35) 0.967 179 474 344 345 6 × 2 = 1 + 0.934 358 948 688 691 2;
  • 36) 0.934 358 948 688 691 2 × 2 = 1 + 0.868 717 897 377 382 4;
  • 37) 0.868 717 897 377 382 4 × 2 = 1 + 0.737 435 794 754 764 8;
  • 38) 0.737 435 794 754 764 8 × 2 = 1 + 0.474 871 589 509 529 6;
  • 39) 0.474 871 589 509 529 6 × 2 = 0 + 0.949 743 179 019 059 2;
  • 40) 0.949 743 179 019 059 2 × 2 = 1 + 0.899 486 358 038 118 4;
  • 41) 0.899 486 358 038 118 4 × 2 = 1 + 0.798 972 716 076 236 8;
  • 42) 0.798 972 716 076 236 8 × 2 = 1 + 0.597 945 432 152 473 6;
  • 43) 0.597 945 432 152 473 6 × 2 = 1 + 0.195 890 864 304 947 2;
  • 44) 0.195 890 864 304 947 2 × 2 = 0 + 0.391 781 728 609 894 4;
  • 45) 0.391 781 728 609 894 4 × 2 = 0 + 0.783 563 457 219 788 8;
  • 46) 0.783 563 457 219 788 8 × 2 = 1 + 0.567 126 914 439 577 6;
  • 47) 0.567 126 914 439 577 6 × 2 = 1 + 0.134 253 828 879 155 2;
  • 48) 0.134 253 828 879 155 2 × 2 = 0 + 0.268 507 657 758 310 4;
  • 49) 0.268 507 657 758 310 4 × 2 = 0 + 0.537 015 315 516 620 8;
  • 50) 0.537 015 315 516 620 8 × 2 = 1 + 0.074 030 631 033 241 6;
  • 51) 0.074 030 631 033 241 6 × 2 = 0 + 0.148 061 262 066 483 2;
  • 52) 0.148 061 262 066 483 2 × 2 = 0 + 0.296 122 524 132 966 4;
  • 53) 0.296 122 524 132 966 4 × 2 = 0 + 0.592 245 048 265 932 8;
  • 54) 0.592 245 048 265 932 8 × 2 = 1 + 0.184 490 096 531 865 6;
  • 55) 0.184 490 096 531 865 6 × 2 = 0 + 0.368 980 193 063 731 2;
  • 56) 0.368 980 193 063 731 2 × 2 = 0 + 0.737 960 386 127 462 4;
  • 57) 0.737 960 386 127 462 4 × 2 = 1 + 0.475 920 772 254 924 8;
  • 58) 0.475 920 772 254 924 8 × 2 = 0 + 0.951 841 544 509 849 6;
  • 59) 0.951 841 544 509 849 6 × 2 = 1 + 0.903 683 089 019 699 2;
  • 60) 0.903 683 089 019 699 2 × 2 = 1 + 0.807 366 178 039 398 4;
  • 61) 0.807 366 178 039 398 4 × 2 = 1 + 0.614 732 356 078 796 8;
  • 62) 0.614 732 356 078 796 8 × 2 = 1 + 0.229 464 712 157 593 6;
  • 63) 0.229 464 712 157 593 6 × 2 = 0 + 0.458 929 424 315 187 2;

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:


0.000 684 924 130 745 9(10) =

0.0000 0000 0010 1100 1110 0011 0001 1110 1011 1101 1110 0110 0100 0100 1011 110(2)

5. Positive number before normalization:

0.000 684 924 130 745 9(10) =

0.0000 0000 0010 1100 1110 0011 0001 1110 1011 1101 1110 0110 0100 0100 1011 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 684 924 130 745 9(10) =

0.0000 0000 0010 1100 1110 0011 0001 1110 1011 1101 1110 0110 0100 0100 1011 110(2) =

0.0000 0000 0010 1100 1110 0011 0001 1110 1011 1101 1110 0110 0100 0100 1011 110(2) × 20 =

1.0110 0111 0001 1000 1111 0101 1110 1111 0011 0010 0010 0101 1110(2) × 2-11


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): -11

Mantissa (not normalized):
1.0110 0111 0001 1000 1111 0101 1110 1111 0011 0010 0010 0101 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-11 + 1 023)(10) =

1 012(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;
  • 1 012 ÷ 2 = 506 + 0;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 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;

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) =

1012(10) =

011 1111 0100(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. 0110 0111 0001 1000 1111 0101 1110 1111 0011 0010 0010 0101 1110 =

0110 0111 0001 1000 1111 0101 1110 1111 0011 0010 0010 0101 1110


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 1111 0100

Mantissa (52 bits) =
0110 0111 0001 1000 1111 0101 1110 1111 0011 0010 0010 0101 1110


Decimal number 0.000 684 924 130 745 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0100 - 0110 0111 0001 1000 1111 0101 1110 1111 0011 0010 0010 0101 1110

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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