-0.052 242 339 538 06 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.052 242 339 538 06(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
-0.052 242 339 538 06(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:

|-0.052 242 339 538 06| = 0.052 242 339 538 06


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: 0.052 242 339 538 06.

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) 0.052 242 339 538 06 × 2 = 0 + 0.104 484 679 076 12;
  • 2) 0.104 484 679 076 12 × 2 = 0 + 0.208 969 358 152 24;
  • 3) 0.208 969 358 152 24 × 2 = 0 + 0.417 938 716 304 48;
  • 4) 0.417 938 716 304 48 × 2 = 0 + 0.835 877 432 608 96;
  • 5) 0.835 877 432 608 96 × 2 = 1 + 0.671 754 865 217 92;
  • 6) 0.671 754 865 217 92 × 2 = 1 + 0.343 509 730 435 84;
  • 7) 0.343 509 730 435 84 × 2 = 0 + 0.687 019 460 871 68;
  • 8) 0.687 019 460 871 68 × 2 = 1 + 0.374 038 921 743 36;
  • 9) 0.374 038 921 743 36 × 2 = 0 + 0.748 077 843 486 72;
  • 10) 0.748 077 843 486 72 × 2 = 1 + 0.496 155 686 973 44;
  • 11) 0.496 155 686 973 44 × 2 = 0 + 0.992 311 373 946 88;
  • 12) 0.992 311 373 946 88 × 2 = 1 + 0.984 622 747 893 76;
  • 13) 0.984 622 747 893 76 × 2 = 1 + 0.969 245 495 787 52;
  • 14) 0.969 245 495 787 52 × 2 = 1 + 0.938 490 991 575 04;
  • 15) 0.938 490 991 575 04 × 2 = 1 + 0.876 981 983 150 08;
  • 16) 0.876 981 983 150 08 × 2 = 1 + 0.753 963 966 300 16;
  • 17) 0.753 963 966 300 16 × 2 = 1 + 0.507 927 932 600 32;
  • 18) 0.507 927 932 600 32 × 2 = 1 + 0.015 855 865 200 64;
  • 19) 0.015 855 865 200 64 × 2 = 0 + 0.031 711 730 401 28;
  • 20) 0.031 711 730 401 28 × 2 = 0 + 0.063 423 460 802 56;
  • 21) 0.063 423 460 802 56 × 2 = 0 + 0.126 846 921 605 12;
  • 22) 0.126 846 921 605 12 × 2 = 0 + 0.253 693 843 210 24;
  • 23) 0.253 693 843 210 24 × 2 = 0 + 0.507 387 686 420 48;
  • 24) 0.507 387 686 420 48 × 2 = 1 + 0.014 775 372 840 96;
  • 25) 0.014 775 372 840 96 × 2 = 0 + 0.029 550 745 681 92;
  • 26) 0.029 550 745 681 92 × 2 = 0 + 0.059 101 491 363 84;
  • 27) 0.059 101 491 363 84 × 2 = 0 + 0.118 202 982 727 68;
  • 28) 0.118 202 982 727 68 × 2 = 0 + 0.236 405 965 455 36;
  • 29) 0.236 405 965 455 36 × 2 = 0 + 0.472 811 930 910 72;
  • 30) 0.472 811 930 910 72 × 2 = 0 + 0.945 623 861 821 44;
  • 31) 0.945 623 861 821 44 × 2 = 1 + 0.891 247 723 642 88;
  • 32) 0.891 247 723 642 88 × 2 = 1 + 0.782 495 447 285 76;
  • 33) 0.782 495 447 285 76 × 2 = 1 + 0.564 990 894 571 52;
  • 34) 0.564 990 894 571 52 × 2 = 1 + 0.129 981 789 143 04;
  • 35) 0.129 981 789 143 04 × 2 = 0 + 0.259 963 578 286 08;
  • 36) 0.259 963 578 286 08 × 2 = 0 + 0.519 927 156 572 16;
  • 37) 0.519 927 156 572 16 × 2 = 1 + 0.039 854 313 144 32;
  • 38) 0.039 854 313 144 32 × 2 = 0 + 0.079 708 626 288 64;
  • 39) 0.079 708 626 288 64 × 2 = 0 + 0.159 417 252 577 28;
  • 40) 0.159 417 252 577 28 × 2 = 0 + 0.318 834 505 154 56;
  • 41) 0.318 834 505 154 56 × 2 = 0 + 0.637 669 010 309 12;
  • 42) 0.637 669 010 309 12 × 2 = 1 + 0.275 338 020 618 24;
  • 43) 0.275 338 020 618 24 × 2 = 0 + 0.550 676 041 236 48;
  • 44) 0.550 676 041 236 48 × 2 = 1 + 0.101 352 082 472 96;
  • 45) 0.101 352 082 472 96 × 2 = 0 + 0.202 704 164 945 92;
  • 46) 0.202 704 164 945 92 × 2 = 0 + 0.405 408 329 891 84;
  • 47) 0.405 408 329 891 84 × 2 = 0 + 0.810 816 659 783 68;
  • 48) 0.810 816 659 783 68 × 2 = 1 + 0.621 633 319 567 36;
  • 49) 0.621 633 319 567 36 × 2 = 1 + 0.243 266 639 134 72;
  • 50) 0.243 266 639 134 72 × 2 = 0 + 0.486 533 278 269 44;
  • 51) 0.486 533 278 269 44 × 2 = 0 + 0.973 066 556 538 88;
  • 52) 0.973 066 556 538 88 × 2 = 1 + 0.946 133 113 077 76;
  • 53) 0.946 133 113 077 76 × 2 = 1 + 0.892 266 226 155 52;
  • 54) 0.892 266 226 155 52 × 2 = 1 + 0.784 532 452 311 04;
  • 55) 0.784 532 452 311 04 × 2 = 1 + 0.569 064 904 622 08;
  • 56) 0.569 064 904 622 08 × 2 = 1 + 0.138 129 809 244 16;
  • 57) 0.138 129 809 244 16 × 2 = 0 + 0.276 259 618 488 32;

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:


0.052 242 339 538 06(10) =

0.0000 1101 0101 1111 1100 0001 0000 0011 1100 1000 0101 0001 1001 1111 0(2)

6. Positive number before normalization:

0.052 242 339 538 06(10) =

0.0000 1101 0101 1111 1100 0001 0000 0011 1100 1000 0101 0001 1001 1111 0(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:


0.052 242 339 538 06(10) =

0.0000 1101 0101 1111 1100 0001 0000 0011 1100 1000 0101 0001 1001 1111 0(2) =

0.0000 1101 0101 1111 1100 0001 0000 0011 1100 1000 0101 0001 1001 1111 0(2) × 20 =

1.1010 1011 1111 1000 0010 0000 0111 1001 0000 1010 0011 0011 1110(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):
1.1010 1011 1111 1000 0010 0000 0111 1001 0000 1010 0011 0011 1110


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. 1010 1011 1111 1000 0010 0000 0111 1001 0000 1010 0011 0011 1110 =

1010 1011 1111 1000 0010 0000 0111 1001 0000 1010 0011 0011 1110


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) =
1010 1011 1111 1000 0010 0000 0111 1001 0000 1010 0011 0011 1110


Decimal number -0.052 242 339 538 06 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 1010 1011 1111 1000 0010 0000 0111 1001 0000 1010 0011 0011 1110

» Convert decimal -0.052 242 339 537 21 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