-0.620 527 518 458 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.620 527 518 458 5(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
-0.620 527 518 458 5(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:

|-0.620 527 518 458 5| = 0.620 527 518 458 5


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: 0.620 527 518 458 5.

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) 0.620 527 518 458 5 × 2 = 1 + 0.241 055 036 917;
  • 2) 0.241 055 036 917 × 2 = 0 + 0.482 110 073 834;
  • 3) 0.482 110 073 834 × 2 = 0 + 0.964 220 147 668;
  • 4) 0.964 220 147 668 × 2 = 1 + 0.928 440 295 336;
  • 5) 0.928 440 295 336 × 2 = 1 + 0.856 880 590 672;
  • 6) 0.856 880 590 672 × 2 = 1 + 0.713 761 181 344;
  • 7) 0.713 761 181 344 × 2 = 1 + 0.427 522 362 688;
  • 8) 0.427 522 362 688 × 2 = 0 + 0.855 044 725 376;
  • 9) 0.855 044 725 376 × 2 = 1 + 0.710 089 450 752;
  • 10) 0.710 089 450 752 × 2 = 1 + 0.420 178 901 504;
  • 11) 0.420 178 901 504 × 2 = 0 + 0.840 357 803 008;
  • 12) 0.840 357 803 008 × 2 = 1 + 0.680 715 606 016;
  • 13) 0.680 715 606 016 × 2 = 1 + 0.361 431 212 032;
  • 14) 0.361 431 212 032 × 2 = 0 + 0.722 862 424 064;
  • 15) 0.722 862 424 064 × 2 = 1 + 0.445 724 848 128;
  • 16) 0.445 724 848 128 × 2 = 0 + 0.891 449 696 256;
  • 17) 0.891 449 696 256 × 2 = 1 + 0.782 899 392 512;
  • 18) 0.782 899 392 512 × 2 = 1 + 0.565 798 785 024;
  • 19) 0.565 798 785 024 × 2 = 1 + 0.131 597 570 048;
  • 20) 0.131 597 570 048 × 2 = 0 + 0.263 195 140 096;
  • 21) 0.263 195 140 096 × 2 = 0 + 0.526 390 280 192;
  • 22) 0.526 390 280 192 × 2 = 1 + 0.052 780 560 384;
  • 23) 0.052 780 560 384 × 2 = 0 + 0.105 561 120 768;
  • 24) 0.105 561 120 768 × 2 = 0 + 0.211 122 241 536;
  • 25) 0.211 122 241 536 × 2 = 0 + 0.422 244 483 072;
  • 26) 0.422 244 483 072 × 2 = 0 + 0.844 488 966 144;
  • 27) 0.844 488 966 144 × 2 = 1 + 0.688 977 932 288;
  • 28) 0.688 977 932 288 × 2 = 1 + 0.377 955 864 576;
  • 29) 0.377 955 864 576 × 2 = 0 + 0.755 911 729 152;
  • 30) 0.755 911 729 152 × 2 = 1 + 0.511 823 458 304;
  • 31) 0.511 823 458 304 × 2 = 1 + 0.023 646 916 608;
  • 32) 0.023 646 916 608 × 2 = 0 + 0.047 293 833 216;
  • 33) 0.047 293 833 216 × 2 = 0 + 0.094 587 666 432;
  • 34) 0.094 587 666 432 × 2 = 0 + 0.189 175 332 864;
  • 35) 0.189 175 332 864 × 2 = 0 + 0.378 350 665 728;
  • 36) 0.378 350 665 728 × 2 = 0 + 0.756 701 331 456;
  • 37) 0.756 701 331 456 × 2 = 1 + 0.513 402 662 912;
  • 38) 0.513 402 662 912 × 2 = 1 + 0.026 805 325 824;
  • 39) 0.026 805 325 824 × 2 = 0 + 0.053 610 651 648;
  • 40) 0.053 610 651 648 × 2 = 0 + 0.107 221 303 296;
  • 41) 0.107 221 303 296 × 2 = 0 + 0.214 442 606 592;
  • 42) 0.214 442 606 592 × 2 = 0 + 0.428 885 213 184;
  • 43) 0.428 885 213 184 × 2 = 0 + 0.857 770 426 368;
  • 44) 0.857 770 426 368 × 2 = 1 + 0.715 540 852 736;
  • 45) 0.715 540 852 736 × 2 = 1 + 0.431 081 705 472;
  • 46) 0.431 081 705 472 × 2 = 0 + 0.862 163 410 944;
  • 47) 0.862 163 410 944 × 2 = 1 + 0.724 326 821 888;
  • 48) 0.724 326 821 888 × 2 = 1 + 0.448 653 643 776;
  • 49) 0.448 653 643 776 × 2 = 0 + 0.897 307 287 552;
  • 50) 0.897 307 287 552 × 2 = 1 + 0.794 614 575 104;
  • 51) 0.794 614 575 104 × 2 = 1 + 0.589 229 150 208;
  • 52) 0.589 229 150 208 × 2 = 1 + 0.178 458 300 416;
  • 53) 0.178 458 300 416 × 2 = 0 + 0.356 916 600 832;

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:


0.620 527 518 458 5(10) =

0.1001 1110 1101 1010 1110 0100 0011 0110 0000 1100 0001 1011 0111 0(2)

6. Positive number before normalization:

0.620 527 518 458 5(10) =

0.1001 1110 1101 1010 1110 0100 0011 0110 0000 1100 0001 1011 0111 0(2)

7. Normalize the binary representation of the number.

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


0.620 527 518 458 5(10) =

0.1001 1110 1101 1010 1110 0100 0011 0110 0000 1100 0001 1011 0111 0(2) =

0.1001 1110 1101 1010 1110 0100 0011 0110 0000 1100 0001 1011 0111 0(2) × 20 =

1.0011 1101 1011 0101 1100 1000 0110 1100 0001 1000 0011 0110 1110(2) × 2-1


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):
1.0011 1101 1011 0101 1100 1000 0110 1100 0001 1000 0011 0110 1110


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) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1022(10) =

011 1111 1110(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. 0011 1101 1011 0101 1100 1000 0110 1100 0001 1000 0011 0110 1110 =

0011 1101 1011 0101 1100 1000 0110 1100 0001 1000 0011 0110 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 1110

Mantissa (52 bits) =
0011 1101 1011 0101 1100 1000 0110 1100 0001 1000 0011 0110 1110


Decimal number -0.620 527 518 458 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1110 - 0011 1101 1011 0101 1100 1000 0110 1100 0001 1000 0011 0110 1110

» Convert decimal -0.620 527 518 452 4 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