-3.477 962 220 539 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -3.477 962 220 539(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
-3.477 962 220 539(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:

|-3.477 962 220 539| = 3.477 962 220 539


2. First, convert to binary (in base 2) the integer part: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

3(10) =

11(2)


4. Convert to binary (base 2) the fractional part: 0.477 962 220 539.

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) 0.477 962 220 539 × 2 = 0 + 0.955 924 441 078;
  • 2) 0.955 924 441 078 × 2 = 1 + 0.911 848 882 156;
  • 3) 0.911 848 882 156 × 2 = 1 + 0.823 697 764 312;
  • 4) 0.823 697 764 312 × 2 = 1 + 0.647 395 528 624;
  • 5) 0.647 395 528 624 × 2 = 1 + 0.294 791 057 248;
  • 6) 0.294 791 057 248 × 2 = 0 + 0.589 582 114 496;
  • 7) 0.589 582 114 496 × 2 = 1 + 0.179 164 228 992;
  • 8) 0.179 164 228 992 × 2 = 0 + 0.358 328 457 984;
  • 9) 0.358 328 457 984 × 2 = 0 + 0.716 656 915 968;
  • 10) 0.716 656 915 968 × 2 = 1 + 0.433 313 831 936;
  • 11) 0.433 313 831 936 × 2 = 0 + 0.866 627 663 872;
  • 12) 0.866 627 663 872 × 2 = 1 + 0.733 255 327 744;
  • 13) 0.733 255 327 744 × 2 = 1 + 0.466 510 655 488;
  • 14) 0.466 510 655 488 × 2 = 0 + 0.933 021 310 976;
  • 15) 0.933 021 310 976 × 2 = 1 + 0.866 042 621 952;
  • 16) 0.866 042 621 952 × 2 = 1 + 0.732 085 243 904;
  • 17) 0.732 085 243 904 × 2 = 1 + 0.464 170 487 808;
  • 18) 0.464 170 487 808 × 2 = 0 + 0.928 340 975 616;
  • 19) 0.928 340 975 616 × 2 = 1 + 0.856 681 951 232;
  • 20) 0.856 681 951 232 × 2 = 1 + 0.713 363 902 464;
  • 21) 0.713 363 902 464 × 2 = 1 + 0.426 727 804 928;
  • 22) 0.426 727 804 928 × 2 = 0 + 0.853 455 609 856;
  • 23) 0.853 455 609 856 × 2 = 1 + 0.706 911 219 712;
  • 24) 0.706 911 219 712 × 2 = 1 + 0.413 822 439 424;
  • 25) 0.413 822 439 424 × 2 = 0 + 0.827 644 878 848;
  • 26) 0.827 644 878 848 × 2 = 1 + 0.655 289 757 696;
  • 27) 0.655 289 757 696 × 2 = 1 + 0.310 579 515 392;
  • 28) 0.310 579 515 392 × 2 = 0 + 0.621 159 030 784;
  • 29) 0.621 159 030 784 × 2 = 1 + 0.242 318 061 568;
  • 30) 0.242 318 061 568 × 2 = 0 + 0.484 636 123 136;
  • 31) 0.484 636 123 136 × 2 = 0 + 0.969 272 246 272;
  • 32) 0.969 272 246 272 × 2 = 1 + 0.938 544 492 544;
  • 33) 0.938 544 492 544 × 2 = 1 + 0.877 088 985 088;
  • 34) 0.877 088 985 088 × 2 = 1 + 0.754 177 970 176;
  • 35) 0.754 177 970 176 × 2 = 1 + 0.508 355 940 352;
  • 36) 0.508 355 940 352 × 2 = 1 + 0.016 711 880 704;
  • 37) 0.016 711 880 704 × 2 = 0 + 0.033 423 761 408;
  • 38) 0.033 423 761 408 × 2 = 0 + 0.066 847 522 816;
  • 39) 0.066 847 522 816 × 2 = 0 + 0.133 695 045 632;
  • 40) 0.133 695 045 632 × 2 = 0 + 0.267 390 091 264;
  • 41) 0.267 390 091 264 × 2 = 0 + 0.534 780 182 528;
  • 42) 0.534 780 182 528 × 2 = 1 + 0.069 560 365 056;
  • 43) 0.069 560 365 056 × 2 = 0 + 0.139 120 730 112;
  • 44) 0.139 120 730 112 × 2 = 0 + 0.278 241 460 224;
  • 45) 0.278 241 460 224 × 2 = 0 + 0.556 482 920 448;
  • 46) 0.556 482 920 448 × 2 = 1 + 0.112 965 840 896;
  • 47) 0.112 965 840 896 × 2 = 0 + 0.225 931 681 792;
  • 48) 0.225 931 681 792 × 2 = 0 + 0.451 863 363 584;
  • 49) 0.451 863 363 584 × 2 = 0 + 0.903 726 727 168;
  • 50) 0.903 726 727 168 × 2 = 1 + 0.807 453 454 336;
  • 51) 0.807 453 454 336 × 2 = 1 + 0.614 906 908 672;
  • 52) 0.614 906 908 672 × 2 = 1 + 0.229 813 817 344;
  • 53) 0.229 813 817 344 × 2 = 0 + 0.459 627 634 688;

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:


0.477 962 220 539(10) =

0.0111 1010 0101 1011 1011 1011 0110 1001 1111 0000 0100 0100 0111 0(2)

6. Positive number before normalization:

3.477 962 220 539(10) =

11.0111 1010 0101 1011 1011 1011 0110 1001 1111 0000 0100 0100 0111 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.477 962 220 539(10) =

11.0111 1010 0101 1011 1011 1011 0110 1001 1111 0000 0100 0100 0111 0(2) =

11.0111 1010 0101 1011 1011 1011 0110 1001 1111 0000 0100 0100 0111 0(2) × 20 =

1.1011 1101 0010 1101 1101 1101 1011 0100 1111 1000 0010 0010 0011 10(2) × 21


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): 1

Mantissa (not normalized):
1.1011 1101 0010 1101 1101 1101 1011 0100 1111 1000 0010 0010 0011 10


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1011 1101 0010 1101 1101 1101 1011 0100 1111 1000 0010 0010 0011 10 =

1011 1101 0010 1101 1101 1101 1011 0100 1111 1000 0010 0010 0011


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) =
100 0000 0000

Mantissa (52 bits) =
1011 1101 0010 1101 1101 1101 1011 0100 1111 1000 0010 0010 0011


Decimal number -3.477 962 220 539 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 1011 1101 0010 1101 1101 1101 1011 0100 1111 1000 0010 0010 0011

» Convert decimal -3.477 962 220 5 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