0.000 085 532 115 337 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 085 532 115 337 1(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
0.000 085 532 115 337 1(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: 0.000 085 532 115 337 1.

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) 0.000 085 532 115 337 1 × 2 = 0 + 0.000 171 064 230 674 2;
  • 2) 0.000 171 064 230 674 2 × 2 = 0 + 0.000 342 128 461 348 4;
  • 3) 0.000 342 128 461 348 4 × 2 = 0 + 0.000 684 256 922 696 8;
  • 4) 0.000 684 256 922 696 8 × 2 = 0 + 0.001 368 513 845 393 6;
  • 5) 0.001 368 513 845 393 6 × 2 = 0 + 0.002 737 027 690 787 2;
  • 6) 0.002 737 027 690 787 2 × 2 = 0 + 0.005 474 055 381 574 4;
  • 7) 0.005 474 055 381 574 4 × 2 = 0 + 0.010 948 110 763 148 8;
  • 8) 0.010 948 110 763 148 8 × 2 = 0 + 0.021 896 221 526 297 6;
  • 9) 0.021 896 221 526 297 6 × 2 = 0 + 0.043 792 443 052 595 2;
  • 10) 0.043 792 443 052 595 2 × 2 = 0 + 0.087 584 886 105 190 4;
  • 11) 0.087 584 886 105 190 4 × 2 = 0 + 0.175 169 772 210 380 8;
  • 12) 0.175 169 772 210 380 8 × 2 = 0 + 0.350 339 544 420 761 6;
  • 13) 0.350 339 544 420 761 6 × 2 = 0 + 0.700 679 088 841 523 2;
  • 14) 0.700 679 088 841 523 2 × 2 = 1 + 0.401 358 177 683 046 4;
  • 15) 0.401 358 177 683 046 4 × 2 = 0 + 0.802 716 355 366 092 8;
  • 16) 0.802 716 355 366 092 8 × 2 = 1 + 0.605 432 710 732 185 6;
  • 17) 0.605 432 710 732 185 6 × 2 = 1 + 0.210 865 421 464 371 2;
  • 18) 0.210 865 421 464 371 2 × 2 = 0 + 0.421 730 842 928 742 4;
  • 19) 0.421 730 842 928 742 4 × 2 = 0 + 0.843 461 685 857 484 8;
  • 20) 0.843 461 685 857 484 8 × 2 = 1 + 0.686 923 371 714 969 6;
  • 21) 0.686 923 371 714 969 6 × 2 = 1 + 0.373 846 743 429 939 2;
  • 22) 0.373 846 743 429 939 2 × 2 = 0 + 0.747 693 486 859 878 4;
  • 23) 0.747 693 486 859 878 4 × 2 = 1 + 0.495 386 973 719 756 8;
  • 24) 0.495 386 973 719 756 8 × 2 = 0 + 0.990 773 947 439 513 6;
  • 25) 0.990 773 947 439 513 6 × 2 = 1 + 0.981 547 894 879 027 2;
  • 26) 0.981 547 894 879 027 2 × 2 = 1 + 0.963 095 789 758 054 4;
  • 27) 0.963 095 789 758 054 4 × 2 = 1 + 0.926 191 579 516 108 8;
  • 28) 0.926 191 579 516 108 8 × 2 = 1 + 0.852 383 159 032 217 6;
  • 29) 0.852 383 159 032 217 6 × 2 = 1 + 0.704 766 318 064 435 2;
  • 30) 0.704 766 318 064 435 2 × 2 = 1 + 0.409 532 636 128 870 4;
  • 31) 0.409 532 636 128 870 4 × 2 = 0 + 0.819 065 272 257 740 8;
  • 32) 0.819 065 272 257 740 8 × 2 = 1 + 0.638 130 544 515 481 6;
  • 33) 0.638 130 544 515 481 6 × 2 = 1 + 0.276 261 089 030 963 2;
  • 34) 0.276 261 089 030 963 2 × 2 = 0 + 0.552 522 178 061 926 4;
  • 35) 0.552 522 178 061 926 4 × 2 = 1 + 0.105 044 356 123 852 8;
  • 36) 0.105 044 356 123 852 8 × 2 = 0 + 0.210 088 712 247 705 6;
  • 37) 0.210 088 712 247 705 6 × 2 = 0 + 0.420 177 424 495 411 2;
  • 38) 0.420 177 424 495 411 2 × 2 = 0 + 0.840 354 848 990 822 4;
  • 39) 0.840 354 848 990 822 4 × 2 = 1 + 0.680 709 697 981 644 8;
  • 40) 0.680 709 697 981 644 8 × 2 = 1 + 0.361 419 395 963 289 6;
  • 41) 0.361 419 395 963 289 6 × 2 = 0 + 0.722 838 791 926 579 2;
  • 42) 0.722 838 791 926 579 2 × 2 = 1 + 0.445 677 583 853 158 4;
  • 43) 0.445 677 583 853 158 4 × 2 = 0 + 0.891 355 167 706 316 8;
  • 44) 0.891 355 167 706 316 8 × 2 = 1 + 0.782 710 335 412 633 6;
  • 45) 0.782 710 335 412 633 6 × 2 = 1 + 0.565 420 670 825 267 2;
  • 46) 0.565 420 670 825 267 2 × 2 = 1 + 0.130 841 341 650 534 4;
  • 47) 0.130 841 341 650 534 4 × 2 = 0 + 0.261 682 683 301 068 8;
  • 48) 0.261 682 683 301 068 8 × 2 = 0 + 0.523 365 366 602 137 6;
  • 49) 0.523 365 366 602 137 6 × 2 = 1 + 0.046 730 733 204 275 2;
  • 50) 0.046 730 733 204 275 2 × 2 = 0 + 0.093 461 466 408 550 4;
  • 51) 0.093 461 466 408 550 4 × 2 = 0 + 0.186 922 932 817 100 8;
  • 52) 0.186 922 932 817 100 8 × 2 = 0 + 0.373 845 865 634 201 6;
  • 53) 0.373 845 865 634 201 6 × 2 = 0 + 0.747 691 731 268 403 2;
  • 54) 0.747 691 731 268 403 2 × 2 = 1 + 0.495 383 462 536 806 4;
  • 55) 0.495 383 462 536 806 4 × 2 = 0 + 0.990 766 925 073 612 8;
  • 56) 0.990 766 925 073 612 8 × 2 = 1 + 0.981 533 850 147 225 6;
  • 57) 0.981 533 850 147 225 6 × 2 = 1 + 0.963 067 700 294 451 2;
  • 58) 0.963 067 700 294 451 2 × 2 = 1 + 0.926 135 400 588 902 4;
  • 59) 0.926 135 400 588 902 4 × 2 = 1 + 0.852 270 801 177 804 8;
  • 60) 0.852 270 801 177 804 8 × 2 = 1 + 0.704 541 602 355 609 6;
  • 61) 0.704 541 602 355 609 6 × 2 = 1 + 0.409 083 204 711 219 2;
  • 62) 0.409 083 204 711 219 2 × 2 = 0 + 0.818 166 409 422 438 4;
  • 63) 0.818 166 409 422 438 4 × 2 = 1 + 0.636 332 818 844 876 8;
  • 64) 0.636 332 818 844 876 8 × 2 = 1 + 0.272 665 637 689 753 6;
  • 65) 0.272 665 637 689 753 6 × 2 = 0 + 0.545 331 275 379 507 2;
  • 66) 0.545 331 275 379 507 2 × 2 = 1 + 0.090 662 550 759 014 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:


0.000 085 532 115 337 1(10) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1010 0011 0101 1100 1000 0101 1111 1011 01(2)

5. Positive number before normalization:

0.000 085 532 115 337 1(10) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1010 0011 0101 1100 1000 0101 1111 1011 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 085 532 115 337 1(10) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1010 0011 0101 1100 1000 0101 1111 1011 01(2) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1010 0011 0101 1100 1000 0101 1111 1011 01(2) × 20 =

1.0110 0110 1011 1111 0110 1000 1101 0111 0010 0001 0111 1110 1101(2) × 2-14


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): -14

Mantissa (not normalized):
1.0110 0110 1011 1111 0110 1000 1101 0111 0010 0001 0111 1110 1101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-14 + 2(11-1) - 1 =

(-14 + 1 023)(10) =

1 009(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;
  • 1 009 ÷ 2 = 504 + 1;
  • 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;

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) =

1009(10) =

011 1111 0001(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 0110 1011 1111 0110 1000 1101 0111 0010 0001 0111 1110 1101 =

0110 0110 1011 1111 0110 1000 1101 0111 0010 0001 0111 1110 1101


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 0001

Mantissa (52 bits) =
0110 0110 1011 1111 0110 1000 1101 0111 0010 0001 0111 1110 1101


Decimal number 0.000 085 532 115 337 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0001 - 0110 0110 1011 1111 0110 1000 1101 0111 0010 0001 0111 1110 1101

» Convert decimal 0.000 085 532 115 331 2 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