0.000 085 532 115 331 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 085 532 115 331 2(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
0.000 085 532 115 331 2(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: 0.000 085 532 115 331 2.

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) 0.000 085 532 115 331 2 × 2 = 0 + 0.000 171 064 230 662 4;
  • 2) 0.000 171 064 230 662 4 × 2 = 0 + 0.000 342 128 461 324 8;
  • 3) 0.000 342 128 461 324 8 × 2 = 0 + 0.000 684 256 922 649 6;
  • 4) 0.000 684 256 922 649 6 × 2 = 0 + 0.001 368 513 845 299 2;
  • 5) 0.001 368 513 845 299 2 × 2 = 0 + 0.002 737 027 690 598 4;
  • 6) 0.002 737 027 690 598 4 × 2 = 0 + 0.005 474 055 381 196 8;
  • 7) 0.005 474 055 381 196 8 × 2 = 0 + 0.010 948 110 762 393 6;
  • 8) 0.010 948 110 762 393 6 × 2 = 0 + 0.021 896 221 524 787 2;
  • 9) 0.021 896 221 524 787 2 × 2 = 0 + 0.043 792 443 049 574 4;
  • 10) 0.043 792 443 049 574 4 × 2 = 0 + 0.087 584 886 099 148 8;
  • 11) 0.087 584 886 099 148 8 × 2 = 0 + 0.175 169 772 198 297 6;
  • 12) 0.175 169 772 198 297 6 × 2 = 0 + 0.350 339 544 396 595 2;
  • 13) 0.350 339 544 396 595 2 × 2 = 0 + 0.700 679 088 793 190 4;
  • 14) 0.700 679 088 793 190 4 × 2 = 1 + 0.401 358 177 586 380 8;
  • 15) 0.401 358 177 586 380 8 × 2 = 0 + 0.802 716 355 172 761 6;
  • 16) 0.802 716 355 172 761 6 × 2 = 1 + 0.605 432 710 345 523 2;
  • 17) 0.605 432 710 345 523 2 × 2 = 1 + 0.210 865 420 691 046 4;
  • 18) 0.210 865 420 691 046 4 × 2 = 0 + 0.421 730 841 382 092 8;
  • 19) 0.421 730 841 382 092 8 × 2 = 0 + 0.843 461 682 764 185 6;
  • 20) 0.843 461 682 764 185 6 × 2 = 1 + 0.686 923 365 528 371 2;
  • 21) 0.686 923 365 528 371 2 × 2 = 1 + 0.373 846 731 056 742 4;
  • 22) 0.373 846 731 056 742 4 × 2 = 0 + 0.747 693 462 113 484 8;
  • 23) 0.747 693 462 113 484 8 × 2 = 1 + 0.495 386 924 226 969 6;
  • 24) 0.495 386 924 226 969 6 × 2 = 0 + 0.990 773 848 453 939 2;
  • 25) 0.990 773 848 453 939 2 × 2 = 1 + 0.981 547 696 907 878 4;
  • 26) 0.981 547 696 907 878 4 × 2 = 1 + 0.963 095 393 815 756 8;
  • 27) 0.963 095 393 815 756 8 × 2 = 1 + 0.926 190 787 631 513 6;
  • 28) 0.926 190 787 631 513 6 × 2 = 1 + 0.852 381 575 263 027 2;
  • 29) 0.852 381 575 263 027 2 × 2 = 1 + 0.704 763 150 526 054 4;
  • 30) 0.704 763 150 526 054 4 × 2 = 1 + 0.409 526 301 052 108 8;
  • 31) 0.409 526 301 052 108 8 × 2 = 0 + 0.819 052 602 104 217 6;
  • 32) 0.819 052 602 104 217 6 × 2 = 1 + 0.638 105 204 208 435 2;
  • 33) 0.638 105 204 208 435 2 × 2 = 1 + 0.276 210 408 416 870 4;
  • 34) 0.276 210 408 416 870 4 × 2 = 0 + 0.552 420 816 833 740 8;
  • 35) 0.552 420 816 833 740 8 × 2 = 1 + 0.104 841 633 667 481 6;
  • 36) 0.104 841 633 667 481 6 × 2 = 0 + 0.209 683 267 334 963 2;
  • 37) 0.209 683 267 334 963 2 × 2 = 0 + 0.419 366 534 669 926 4;
  • 38) 0.419 366 534 669 926 4 × 2 = 0 + 0.838 733 069 339 852 8;
  • 39) 0.838 733 069 339 852 8 × 2 = 1 + 0.677 466 138 679 705 6;
  • 40) 0.677 466 138 679 705 6 × 2 = 1 + 0.354 932 277 359 411 2;
  • 41) 0.354 932 277 359 411 2 × 2 = 0 + 0.709 864 554 718 822 4;
  • 42) 0.709 864 554 718 822 4 × 2 = 1 + 0.419 729 109 437 644 8;
  • 43) 0.419 729 109 437 644 8 × 2 = 0 + 0.839 458 218 875 289 6;
  • 44) 0.839 458 218 875 289 6 × 2 = 1 + 0.678 916 437 750 579 2;
  • 45) 0.678 916 437 750 579 2 × 2 = 1 + 0.357 832 875 501 158 4;
  • 46) 0.357 832 875 501 158 4 × 2 = 0 + 0.715 665 751 002 316 8;
  • 47) 0.715 665 751 002 316 8 × 2 = 1 + 0.431 331 502 004 633 6;
  • 48) 0.431 331 502 004 633 6 × 2 = 0 + 0.862 663 004 009 267 2;
  • 49) 0.862 663 004 009 267 2 × 2 = 1 + 0.725 326 008 018 534 4;
  • 50) 0.725 326 008 018 534 4 × 2 = 1 + 0.450 652 016 037 068 8;
  • 51) 0.450 652 016 037 068 8 × 2 = 0 + 0.901 304 032 074 137 6;
  • 52) 0.901 304 032 074 137 6 × 2 = 1 + 0.802 608 064 148 275 2;
  • 53) 0.802 608 064 148 275 2 × 2 = 1 + 0.605 216 128 296 550 4;
  • 54) 0.605 216 128 296 550 4 × 2 = 1 + 0.210 432 256 593 100 8;
  • 55) 0.210 432 256 593 100 8 × 2 = 0 + 0.420 864 513 186 201 6;
  • 56) 0.420 864 513 186 201 6 × 2 = 0 + 0.841 729 026 372 403 2;
  • 57) 0.841 729 026 372 403 2 × 2 = 1 + 0.683 458 052 744 806 4;
  • 58) 0.683 458 052 744 806 4 × 2 = 1 + 0.366 916 105 489 612 8;
  • 59) 0.366 916 105 489 612 8 × 2 = 0 + 0.733 832 210 979 225 6;
  • 60) 0.733 832 210 979 225 6 × 2 = 1 + 0.467 664 421 958 451 2;
  • 61) 0.467 664 421 958 451 2 × 2 = 0 + 0.935 328 843 916 902 4;
  • 62) 0.935 328 843 916 902 4 × 2 = 1 + 0.870 657 687 833 804 8;
  • 63) 0.870 657 687 833 804 8 × 2 = 1 + 0.741 315 375 667 609 6;
  • 64) 0.741 315 375 667 609 6 × 2 = 1 + 0.482 630 751 335 219 2;
  • 65) 0.482 630 751 335 219 2 × 2 = 0 + 0.965 261 502 670 438 4;
  • 66) 0.965 261 502 670 438 4 × 2 = 1 + 0.930 523 005 340 876 8;

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:


0.000 085 532 115 331 2(10) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1010 0011 0101 1010 1101 1100 1101 0111 01(2)

5. Positive number before normalization:

0.000 085 532 115 331 2(10) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1010 0011 0101 1010 1101 1100 1101 0111 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:


0.000 085 532 115 331 2(10) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1010 0011 0101 1010 1101 1100 1101 0111 01(2) =

0.0000 0000 0000 0101 1001 1010 1111 1101 1010 0011 0101 1010 1101 1100 1101 0111 01(2) × 20 =

1.0110 0110 1011 1111 0110 1000 1101 0110 1011 0111 0011 0101 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 0110 1011 0111 0011 0101 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 0110 1011 0111 0011 0101 1101 =

0110 0110 1011 1111 0110 1000 1101 0110 1011 0111 0011 0101 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 0110 1011 0111 0011 0101 1101


Decimal number 0.000 085 532 115 331 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0001 - 0110 0110 1011 1111 0110 1000 1101 0110 1011 0111 0011 0101 1101

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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