0.000 684 924 130 754 05 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 684 924 130 754 05(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
0.000 684 924 130 754 05(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: 0.000 684 924 130 754 05.

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) 0.000 684 924 130 754 05 × 2 = 0 + 0.001 369 848 261 508 1;
  • 2) 0.001 369 848 261 508 1 × 2 = 0 + 0.002 739 696 523 016 2;
  • 3) 0.002 739 696 523 016 2 × 2 = 0 + 0.005 479 393 046 032 4;
  • 4) 0.005 479 393 046 032 4 × 2 = 0 + 0.010 958 786 092 064 8;
  • 5) 0.010 958 786 092 064 8 × 2 = 0 + 0.021 917 572 184 129 6;
  • 6) 0.021 917 572 184 129 6 × 2 = 0 + 0.043 835 144 368 259 2;
  • 7) 0.043 835 144 368 259 2 × 2 = 0 + 0.087 670 288 736 518 4;
  • 8) 0.087 670 288 736 518 4 × 2 = 0 + 0.175 340 577 473 036 8;
  • 9) 0.175 340 577 473 036 8 × 2 = 0 + 0.350 681 154 946 073 6;
  • 10) 0.350 681 154 946 073 6 × 2 = 0 + 0.701 362 309 892 147 2;
  • 11) 0.701 362 309 892 147 2 × 2 = 1 + 0.402 724 619 784 294 4;
  • 12) 0.402 724 619 784 294 4 × 2 = 0 + 0.805 449 239 568 588 8;
  • 13) 0.805 449 239 568 588 8 × 2 = 1 + 0.610 898 479 137 177 6;
  • 14) 0.610 898 479 137 177 6 × 2 = 1 + 0.221 796 958 274 355 2;
  • 15) 0.221 796 958 274 355 2 × 2 = 0 + 0.443 593 916 548 710 4;
  • 16) 0.443 593 916 548 710 4 × 2 = 0 + 0.887 187 833 097 420 8;
  • 17) 0.887 187 833 097 420 8 × 2 = 1 + 0.774 375 666 194 841 6;
  • 18) 0.774 375 666 194 841 6 × 2 = 1 + 0.548 751 332 389 683 2;
  • 19) 0.548 751 332 389 683 2 × 2 = 1 + 0.097 502 664 779 366 4;
  • 20) 0.097 502 664 779 366 4 × 2 = 0 + 0.195 005 329 558 732 8;
  • 21) 0.195 005 329 558 732 8 × 2 = 0 + 0.390 010 659 117 465 6;
  • 22) 0.390 010 659 117 465 6 × 2 = 0 + 0.780 021 318 234 931 2;
  • 23) 0.780 021 318 234 931 2 × 2 = 1 + 0.560 042 636 469 862 4;
  • 24) 0.560 042 636 469 862 4 × 2 = 1 + 0.120 085 272 939 724 8;
  • 25) 0.120 085 272 939 724 8 × 2 = 0 + 0.240 170 545 879 449 6;
  • 26) 0.240 170 545 879 449 6 × 2 = 0 + 0.480 341 091 758 899 2;
  • 27) 0.480 341 091 758 899 2 × 2 = 0 + 0.960 682 183 517 798 4;
  • 28) 0.960 682 183 517 798 4 × 2 = 1 + 0.921 364 367 035 596 8;
  • 29) 0.921 364 367 035 596 8 × 2 = 1 + 0.842 728 734 071 193 6;
  • 30) 0.842 728 734 071 193 6 × 2 = 1 + 0.685 457 468 142 387 2;
  • 31) 0.685 457 468 142 387 2 × 2 = 1 + 0.370 914 936 284 774 4;
  • 32) 0.370 914 936 284 774 4 × 2 = 0 + 0.741 829 872 569 548 8;
  • 33) 0.741 829 872 569 548 8 × 2 = 1 + 0.483 659 745 139 097 6;
  • 34) 0.483 659 745 139 097 6 × 2 = 0 + 0.967 319 490 278 195 2;
  • 35) 0.967 319 490 278 195 2 × 2 = 1 + 0.934 638 980 556 390 4;
  • 36) 0.934 638 980 556 390 4 × 2 = 1 + 0.869 277 961 112 780 8;
  • 37) 0.869 277 961 112 780 8 × 2 = 1 + 0.738 555 922 225 561 6;
  • 38) 0.738 555 922 225 561 6 × 2 = 1 + 0.477 111 844 451 123 2;
  • 39) 0.477 111 844 451 123 2 × 2 = 0 + 0.954 223 688 902 246 4;
  • 40) 0.954 223 688 902 246 4 × 2 = 1 + 0.908 447 377 804 492 8;
  • 41) 0.908 447 377 804 492 8 × 2 = 1 + 0.816 894 755 608 985 6;
  • 42) 0.816 894 755 608 985 6 × 2 = 1 + 0.633 789 511 217 971 2;
  • 43) 0.633 789 511 217 971 2 × 2 = 1 + 0.267 579 022 435 942 4;
  • 44) 0.267 579 022 435 942 4 × 2 = 0 + 0.535 158 044 871 884 8;
  • 45) 0.535 158 044 871 884 8 × 2 = 1 + 0.070 316 089 743 769 6;
  • 46) 0.070 316 089 743 769 6 × 2 = 0 + 0.140 632 179 487 539 2;
  • 47) 0.140 632 179 487 539 2 × 2 = 0 + 0.281 264 358 975 078 4;
  • 48) 0.281 264 358 975 078 4 × 2 = 0 + 0.562 528 717 950 156 8;
  • 49) 0.562 528 717 950 156 8 × 2 = 1 + 0.125 057 435 900 313 6;
  • 50) 0.125 057 435 900 313 6 × 2 = 0 + 0.250 114 871 800 627 2;
  • 51) 0.250 114 871 800 627 2 × 2 = 0 + 0.500 229 743 601 254 4;
  • 52) 0.500 229 743 601 254 4 × 2 = 1 + 0.000 459 487 202 508 8;
  • 53) 0.000 459 487 202 508 8 × 2 = 0 + 0.000 918 974 405 017 6;
  • 54) 0.000 918 974 405 017 6 × 2 = 0 + 0.001 837 948 810 035 2;
  • 55) 0.001 837 948 810 035 2 × 2 = 0 + 0.003 675 897 620 070 4;
  • 56) 0.003 675 897 620 070 4 × 2 = 0 + 0.007 351 795 240 140 8;
  • 57) 0.007 351 795 240 140 8 × 2 = 0 + 0.014 703 590 480 281 6;
  • 58) 0.014 703 590 480 281 6 × 2 = 0 + 0.029 407 180 960 563 2;
  • 59) 0.029 407 180 960 563 2 × 2 = 0 + 0.058 814 361 921 126 4;
  • 60) 0.058 814 361 921 126 4 × 2 = 0 + 0.117 628 723 842 252 8;
  • 61) 0.117 628 723 842 252 8 × 2 = 0 + 0.235 257 447 684 505 6;
  • 62) 0.235 257 447 684 505 6 × 2 = 0 + 0.470 514 895 369 011 2;
  • 63) 0.470 514 895 369 011 2 × 2 = 0 + 0.941 029 790 738 022 4;

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:


0.000 684 924 130 754 05(10) =

0.0000 0000 0010 1100 1110 0011 0001 1110 1011 1101 1110 1000 1001 0000 0000 000(2)

5. Positive number before normalization:

0.000 684 924 130 754 05(10) =

0.0000 0000 0010 1100 1110 0011 0001 1110 1011 1101 1110 1000 1001 0000 0000 000(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:


0.000 684 924 130 754 05(10) =

0.0000 0000 0010 1100 1110 0011 0001 1110 1011 1101 1110 1000 1001 0000 0000 000(2) =

0.0000 0000 0010 1100 1110 0011 0001 1110 1011 1101 1110 1000 1001 0000 0000 000(2) × 20 =

1.0110 0111 0001 1000 1111 0101 1110 1111 0100 0100 1000 0000 0000(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 0100 0100 1000 0000 0000


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 0100 0100 1000 0000 0000 =

0110 0111 0001 1000 1111 0101 1110 1111 0100 0100 1000 0000 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 1111 0100

Mantissa (52 bits) =
0110 0111 0001 1000 1111 0101 1110 1111 0100 0100 1000 0000 0000


Decimal number 0.000 684 924 130 754 05 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0100 - 0110 0111 0001 1000 1111 0101 1110 1111 0100 0100 1000 0000 0000

» Convert decimal 0.000 684 924 130 753 41 to 64 bit double precision IEEE 754 binary floating point representation standard

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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