-0.044 122 556 551 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.044 122 556 551(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
-0.044 122 556 551(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:

|-0.044 122 556 551| = 0.044 122 556 551


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: 0.044 122 556 551.

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) 0.044 122 556 551 × 2 = 0 + 0.088 245 113 102;
  • 2) 0.088 245 113 102 × 2 = 0 + 0.176 490 226 204;
  • 3) 0.176 490 226 204 × 2 = 0 + 0.352 980 452 408;
  • 4) 0.352 980 452 408 × 2 = 0 + 0.705 960 904 816;
  • 5) 0.705 960 904 816 × 2 = 1 + 0.411 921 809 632;
  • 6) 0.411 921 809 632 × 2 = 0 + 0.823 843 619 264;
  • 7) 0.823 843 619 264 × 2 = 1 + 0.647 687 238 528;
  • 8) 0.647 687 238 528 × 2 = 1 + 0.295 374 477 056;
  • 9) 0.295 374 477 056 × 2 = 0 + 0.590 748 954 112;
  • 10) 0.590 748 954 112 × 2 = 1 + 0.181 497 908 224;
  • 11) 0.181 497 908 224 × 2 = 0 + 0.362 995 816 448;
  • 12) 0.362 995 816 448 × 2 = 0 + 0.725 991 632 896;
  • 13) 0.725 991 632 896 × 2 = 1 + 0.451 983 265 792;
  • 14) 0.451 983 265 792 × 2 = 0 + 0.903 966 531 584;
  • 15) 0.903 966 531 584 × 2 = 1 + 0.807 933 063 168;
  • 16) 0.807 933 063 168 × 2 = 1 + 0.615 866 126 336;
  • 17) 0.615 866 126 336 × 2 = 1 + 0.231 732 252 672;
  • 18) 0.231 732 252 672 × 2 = 0 + 0.463 464 505 344;
  • 19) 0.463 464 505 344 × 2 = 0 + 0.926 929 010 688;
  • 20) 0.926 929 010 688 × 2 = 1 + 0.853 858 021 376;
  • 21) 0.853 858 021 376 × 2 = 1 + 0.707 716 042 752;
  • 22) 0.707 716 042 752 × 2 = 1 + 0.415 432 085 504;
  • 23) 0.415 432 085 504 × 2 = 0 + 0.830 864 171 008;
  • 24) 0.830 864 171 008 × 2 = 1 + 0.661 728 342 016;
  • 25) 0.661 728 342 016 × 2 = 1 + 0.323 456 684 032;
  • 26) 0.323 456 684 032 × 2 = 0 + 0.646 913 368 064;
  • 27) 0.646 913 368 064 × 2 = 1 + 0.293 826 736 128;
  • 28) 0.293 826 736 128 × 2 = 0 + 0.587 653 472 256;
  • 29) 0.587 653 472 256 × 2 = 1 + 0.175 306 944 512;
  • 30) 0.175 306 944 512 × 2 = 0 + 0.350 613 889 024;
  • 31) 0.350 613 889 024 × 2 = 0 + 0.701 227 778 048;
  • 32) 0.701 227 778 048 × 2 = 1 + 0.402 455 556 096;
  • 33) 0.402 455 556 096 × 2 = 0 + 0.804 911 112 192;
  • 34) 0.804 911 112 192 × 2 = 1 + 0.609 822 224 384;
  • 35) 0.609 822 224 384 × 2 = 1 + 0.219 644 448 768;
  • 36) 0.219 644 448 768 × 2 = 0 + 0.439 288 897 536;
  • 37) 0.439 288 897 536 × 2 = 0 + 0.878 577 795 072;
  • 38) 0.878 577 795 072 × 2 = 1 + 0.757 155 590 144;
  • 39) 0.757 155 590 144 × 2 = 1 + 0.514 311 180 288;
  • 40) 0.514 311 180 288 × 2 = 1 + 0.028 622 360 576;
  • 41) 0.028 622 360 576 × 2 = 0 + 0.057 244 721 152;
  • 42) 0.057 244 721 152 × 2 = 0 + 0.114 489 442 304;
  • 43) 0.114 489 442 304 × 2 = 0 + 0.228 978 884 608;
  • 44) 0.228 978 884 608 × 2 = 0 + 0.457 957 769 216;
  • 45) 0.457 957 769 216 × 2 = 0 + 0.915 915 538 432;
  • 46) 0.915 915 538 432 × 2 = 1 + 0.831 831 076 864;
  • 47) 0.831 831 076 864 × 2 = 1 + 0.663 662 153 728;
  • 48) 0.663 662 153 728 × 2 = 1 + 0.327 324 307 456;
  • 49) 0.327 324 307 456 × 2 = 0 + 0.654 648 614 912;
  • 50) 0.654 648 614 912 × 2 = 1 + 0.309 297 229 824;
  • 51) 0.309 297 229 824 × 2 = 0 + 0.618 594 459 648;
  • 52) 0.618 594 459 648 × 2 = 1 + 0.237 188 919 296;
  • 53) 0.237 188 919 296 × 2 = 0 + 0.474 377 838 592;
  • 54) 0.474 377 838 592 × 2 = 0 + 0.948 755 677 184;
  • 55) 0.948 755 677 184 × 2 = 1 + 0.897 511 354 368;
  • 56) 0.897 511 354 368 × 2 = 1 + 0.795 022 708 736;
  • 57) 0.795 022 708 736 × 2 = 1 + 0.590 045 417 472;

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:


0.044 122 556 551(10) =

0.0000 1011 0100 1011 1001 1101 1010 1001 0110 0111 0000 0111 0101 0011 1(2)

6. Positive number before normalization:

0.044 122 556 551(10) =

0.0000 1011 0100 1011 1001 1101 1010 1001 0110 0111 0000 0111 0101 0011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.044 122 556 551(10) =

0.0000 1011 0100 1011 1001 1101 1010 1001 0110 0111 0000 0111 0101 0011 1(2) =

0.0000 1011 0100 1011 1001 1101 1010 1001 0110 0111 0000 0111 0101 0011 1(2) × 20 =

1.0110 1001 0111 0011 1011 0101 0010 1100 1110 0000 1110 1010 0111(2) × 2-5


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): -5

Mantissa (not normalized):
1.0110 1001 0111 0011 1011 0101 0010 1100 1110 0000 1110 1010 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1018(10) =

011 1111 1010(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. 0110 1001 0111 0011 1011 0101 0010 1100 1110 0000 1110 1010 0111 =

0110 1001 0111 0011 1011 0101 0010 1100 1110 0000 1110 1010 0111


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 1010

Mantissa (52 bits) =
0110 1001 0111 0011 1011 0101 0010 1100 1110 0000 1110 1010 0111


Decimal number -0.044 122 556 551 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0110 1001 0111 0011 1011 0101 0010 1100 1110 0000 1110 1010 0111

» Convert decimal -0.044 122 556 521 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