-0.000 164 778 450 863 07 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 164 778 450 863 07(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
-0.000 164 778 450 863 07(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. Start with the positive version of the number:

|-0.000 164 778 450 863 07| = 0.000 164 778 450 863 07


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 164 778 450 863 07.

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) 0.000 164 778 450 863 07 × 2 = 0 + 0.000 329 556 901 726 14;
  • 2) 0.000 329 556 901 726 14 × 2 = 0 + 0.000 659 113 803 452 28;
  • 3) 0.000 659 113 803 452 28 × 2 = 0 + 0.001 318 227 606 904 56;
  • 4) 0.001 318 227 606 904 56 × 2 = 0 + 0.002 636 455 213 809 12;
  • 5) 0.002 636 455 213 809 12 × 2 = 0 + 0.005 272 910 427 618 24;
  • 6) 0.005 272 910 427 618 24 × 2 = 0 + 0.010 545 820 855 236 48;
  • 7) 0.010 545 820 855 236 48 × 2 = 0 + 0.021 091 641 710 472 96;
  • 8) 0.021 091 641 710 472 96 × 2 = 0 + 0.042 183 283 420 945 92;
  • 9) 0.042 183 283 420 945 92 × 2 = 0 + 0.084 366 566 841 891 84;
  • 10) 0.084 366 566 841 891 84 × 2 = 0 + 0.168 733 133 683 783 68;
  • 11) 0.168 733 133 683 783 68 × 2 = 0 + 0.337 466 267 367 567 36;
  • 12) 0.337 466 267 367 567 36 × 2 = 0 + 0.674 932 534 735 134 72;
  • 13) 0.674 932 534 735 134 72 × 2 = 1 + 0.349 865 069 470 269 44;
  • 14) 0.349 865 069 470 269 44 × 2 = 0 + 0.699 730 138 940 538 88;
  • 15) 0.699 730 138 940 538 88 × 2 = 1 + 0.399 460 277 881 077 76;
  • 16) 0.399 460 277 881 077 76 × 2 = 0 + 0.798 920 555 762 155 52;
  • 17) 0.798 920 555 762 155 52 × 2 = 1 + 0.597 841 111 524 311 04;
  • 18) 0.597 841 111 524 311 04 × 2 = 1 + 0.195 682 223 048 622 08;
  • 19) 0.195 682 223 048 622 08 × 2 = 0 + 0.391 364 446 097 244 16;
  • 20) 0.391 364 446 097 244 16 × 2 = 0 + 0.782 728 892 194 488 32;
  • 21) 0.782 728 892 194 488 32 × 2 = 1 + 0.565 457 784 388 976 64;
  • 22) 0.565 457 784 388 976 64 × 2 = 1 + 0.130 915 568 777 953 28;
  • 23) 0.130 915 568 777 953 28 × 2 = 0 + 0.261 831 137 555 906 56;
  • 24) 0.261 831 137 555 906 56 × 2 = 0 + 0.523 662 275 111 813 12;
  • 25) 0.523 662 275 111 813 12 × 2 = 1 + 0.047 324 550 223 626 24;
  • 26) 0.047 324 550 223 626 24 × 2 = 0 + 0.094 649 100 447 252 48;
  • 27) 0.094 649 100 447 252 48 × 2 = 0 + 0.189 298 200 894 504 96;
  • 28) 0.189 298 200 894 504 96 × 2 = 0 + 0.378 596 401 789 009 92;
  • 29) 0.378 596 401 789 009 92 × 2 = 0 + 0.757 192 803 578 019 84;
  • 30) 0.757 192 803 578 019 84 × 2 = 1 + 0.514 385 607 156 039 68;
  • 31) 0.514 385 607 156 039 68 × 2 = 1 + 0.028 771 214 312 079 36;
  • 32) 0.028 771 214 312 079 36 × 2 = 0 + 0.057 542 428 624 158 72;
  • 33) 0.057 542 428 624 158 72 × 2 = 0 + 0.115 084 857 248 317 44;
  • 34) 0.115 084 857 248 317 44 × 2 = 0 + 0.230 169 714 496 634 88;
  • 35) 0.230 169 714 496 634 88 × 2 = 0 + 0.460 339 428 993 269 76;
  • 36) 0.460 339 428 993 269 76 × 2 = 0 + 0.920 678 857 986 539 52;
  • 37) 0.920 678 857 986 539 52 × 2 = 1 + 0.841 357 715 973 079 04;
  • 38) 0.841 357 715 973 079 04 × 2 = 1 + 0.682 715 431 946 158 08;
  • 39) 0.682 715 431 946 158 08 × 2 = 1 + 0.365 430 863 892 316 16;
  • 40) 0.365 430 863 892 316 16 × 2 = 0 + 0.730 861 727 784 632 32;
  • 41) 0.730 861 727 784 632 32 × 2 = 1 + 0.461 723 455 569 264 64;
  • 42) 0.461 723 455 569 264 64 × 2 = 0 + 0.923 446 911 138 529 28;
  • 43) 0.923 446 911 138 529 28 × 2 = 1 + 0.846 893 822 277 058 56;
  • 44) 0.846 893 822 277 058 56 × 2 = 1 + 0.693 787 644 554 117 12;
  • 45) 0.693 787 644 554 117 12 × 2 = 1 + 0.387 575 289 108 234 24;
  • 46) 0.387 575 289 108 234 24 × 2 = 0 + 0.775 150 578 216 468 48;
  • 47) 0.775 150 578 216 468 48 × 2 = 1 + 0.550 301 156 432 936 96;
  • 48) 0.550 301 156 432 936 96 × 2 = 1 + 0.100 602 312 865 873 92;
  • 49) 0.100 602 312 865 873 92 × 2 = 0 + 0.201 204 625 731 747 84;
  • 50) 0.201 204 625 731 747 84 × 2 = 0 + 0.402 409 251 463 495 68;
  • 51) 0.402 409 251 463 495 68 × 2 = 0 + 0.804 818 502 926 991 36;
  • 52) 0.804 818 502 926 991 36 × 2 = 1 + 0.609 637 005 853 982 72;
  • 53) 0.609 637 005 853 982 72 × 2 = 1 + 0.219 274 011 707 965 44;
  • 54) 0.219 274 011 707 965 44 × 2 = 0 + 0.438 548 023 415 930 88;
  • 55) 0.438 548 023 415 930 88 × 2 = 0 + 0.877 096 046 831 861 76;
  • 56) 0.877 096 046 831 861 76 × 2 = 1 + 0.754 192 093 663 723 52;
  • 57) 0.754 192 093 663 723 52 × 2 = 1 + 0.508 384 187 327 447 04;
  • 58) 0.508 384 187 327 447 04 × 2 = 1 + 0.016 768 374 654 894 08;
  • 59) 0.016 768 374 654 894 08 × 2 = 0 + 0.033 536 749 309 788 16;
  • 60) 0.033 536 749 309 788 16 × 2 = 0 + 0.067 073 498 619 576 32;
  • 61) 0.067 073 498 619 576 32 × 2 = 0 + 0.134 146 997 239 152 64;
  • 62) 0.134 146 997 239 152 64 × 2 = 0 + 0.268 293 994 478 305 28;
  • 63) 0.268 293 994 478 305 28 × 2 = 0 + 0.536 587 988 956 610 56;
  • 64) 0.536 587 988 956 610 56 × 2 = 1 + 0.073 175 977 913 221 12;
  • 65) 0.073 175 977 913 221 12 × 2 = 0 + 0.146 351 955 826 442 24;

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).


5. 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:


0.000 164 778 450 863 07(10) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0001 1001 1100 0001 0(2)

6. Positive number before normalization:

0.000 164 778 450 863 07(10) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0001 1001 1100 0001 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the right, so that only one non zero digit remains to the left of it:


0.000 164 778 450 863 07(10) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0001 1001 1100 0001 0(2) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0001 1001 1100 0001 0(2) × 20 =

1.0101 1001 1001 0000 1100 0001 1101 0111 0110 0011 0011 1000 0010(2) × 2-13


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): -13

Mantissa (not normalized):
1.0101 1001 1001 0000 1100 0001 1101 0111 0110 0011 0011 1000 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-13 + 2(11-1) - 1 =

(-13 + 1 023)(10) =

1 010(10)


10. 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;
  • 1 010 ÷ 2 = 505 + 0;
  • 505 ÷ 2 = 252 + 1;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1010(10) =

011 1111 0010(2)


12. 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. 0101 1001 1001 0000 1100 0001 1101 0111 0110 0011 0011 1000 0010 =

0101 1001 1001 0000 1100 0001 1101 0111 0110 0011 0011 1000 0010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0010

Mantissa (52 bits) =
0101 1001 1001 0000 1100 0001 1101 0111 0110 0011 0011 1000 0010


Decimal number -0.000 164 778 450 863 07 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0010 - 0101 1001 1001 0000 1100 0001 1101 0111 0110 0011 0011 1000 0010

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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