-0.062 209 657 594 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.062 209 657 594(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
-0.062 209 657 594(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:

|-0.062 209 657 594| = 0.062 209 657 594


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: 0.062 209 657 594.

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) 0.062 209 657 594 × 2 = 0 + 0.124 419 315 188;
  • 2) 0.124 419 315 188 × 2 = 0 + 0.248 838 630 376;
  • 3) 0.248 838 630 376 × 2 = 0 + 0.497 677 260 752;
  • 4) 0.497 677 260 752 × 2 = 0 + 0.995 354 521 504;
  • 5) 0.995 354 521 504 × 2 = 1 + 0.990 709 043 008;
  • 6) 0.990 709 043 008 × 2 = 1 + 0.981 418 086 016;
  • 7) 0.981 418 086 016 × 2 = 1 + 0.962 836 172 032;
  • 8) 0.962 836 172 032 × 2 = 1 + 0.925 672 344 064;
  • 9) 0.925 672 344 064 × 2 = 1 + 0.851 344 688 128;
  • 10) 0.851 344 688 128 × 2 = 1 + 0.702 689 376 256;
  • 11) 0.702 689 376 256 × 2 = 1 + 0.405 378 752 512;
  • 12) 0.405 378 752 512 × 2 = 0 + 0.810 757 505 024;
  • 13) 0.810 757 505 024 × 2 = 1 + 0.621 515 010 048;
  • 14) 0.621 515 010 048 × 2 = 1 + 0.243 030 020 096;
  • 15) 0.243 030 020 096 × 2 = 0 + 0.486 060 040 192;
  • 16) 0.486 060 040 192 × 2 = 0 + 0.972 120 080 384;
  • 17) 0.972 120 080 384 × 2 = 1 + 0.944 240 160 768;
  • 18) 0.944 240 160 768 × 2 = 1 + 0.888 480 321 536;
  • 19) 0.888 480 321 536 × 2 = 1 + 0.776 960 643 072;
  • 20) 0.776 960 643 072 × 2 = 1 + 0.553 921 286 144;
  • 21) 0.553 921 286 144 × 2 = 1 + 0.107 842 572 288;
  • 22) 0.107 842 572 288 × 2 = 0 + 0.215 685 144 576;
  • 23) 0.215 685 144 576 × 2 = 0 + 0.431 370 289 152;
  • 24) 0.431 370 289 152 × 2 = 0 + 0.862 740 578 304;
  • 25) 0.862 740 578 304 × 2 = 1 + 0.725 481 156 608;
  • 26) 0.725 481 156 608 × 2 = 1 + 0.450 962 313 216;
  • 27) 0.450 962 313 216 × 2 = 0 + 0.901 924 626 432;
  • 28) 0.901 924 626 432 × 2 = 1 + 0.803 849 252 864;
  • 29) 0.803 849 252 864 × 2 = 1 + 0.607 698 505 728;
  • 30) 0.607 698 505 728 × 2 = 1 + 0.215 397 011 456;
  • 31) 0.215 397 011 456 × 2 = 0 + 0.430 794 022 912;
  • 32) 0.430 794 022 912 × 2 = 0 + 0.861 588 045 824;
  • 33) 0.861 588 045 824 × 2 = 1 + 0.723 176 091 648;
  • 34) 0.723 176 091 648 × 2 = 1 + 0.446 352 183 296;
  • 35) 0.446 352 183 296 × 2 = 0 + 0.892 704 366 592;
  • 36) 0.892 704 366 592 × 2 = 1 + 0.785 408 733 184;
  • 37) 0.785 408 733 184 × 2 = 1 + 0.570 817 466 368;
  • 38) 0.570 817 466 368 × 2 = 1 + 0.141 634 932 736;
  • 39) 0.141 634 932 736 × 2 = 0 + 0.283 269 865 472;
  • 40) 0.283 269 865 472 × 2 = 0 + 0.566 539 730 944;
  • 41) 0.566 539 730 944 × 2 = 1 + 0.133 079 461 888;
  • 42) 0.133 079 461 888 × 2 = 0 + 0.266 158 923 776;
  • 43) 0.266 158 923 776 × 2 = 0 + 0.532 317 847 552;
  • 44) 0.532 317 847 552 × 2 = 1 + 0.064 635 695 104;
  • 45) 0.064 635 695 104 × 2 = 0 + 0.129 271 390 208;
  • 46) 0.129 271 390 208 × 2 = 0 + 0.258 542 780 416;
  • 47) 0.258 542 780 416 × 2 = 0 + 0.517 085 560 832;
  • 48) 0.517 085 560 832 × 2 = 1 + 0.034 171 121 664;
  • 49) 0.034 171 121 664 × 2 = 0 + 0.068 342 243 328;
  • 50) 0.068 342 243 328 × 2 = 0 + 0.136 684 486 656;
  • 51) 0.136 684 486 656 × 2 = 0 + 0.273 368 973 312;
  • 52) 0.273 368 973 312 × 2 = 0 + 0.546 737 946 624;
  • 53) 0.546 737 946 624 × 2 = 1 + 0.093 475 893 248;
  • 54) 0.093 475 893 248 × 2 = 0 + 0.186 951 786 496;
  • 55) 0.186 951 786 496 × 2 = 0 + 0.373 903 572 992;
  • 56) 0.373 903 572 992 × 2 = 0 + 0.747 807 145 984;
  • 57) 0.747 807 145 984 × 2 = 1 + 0.495 614 291 968;

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:


0.062 209 657 594(10) =

0.0000 1111 1110 1100 1111 1000 1101 1100 1101 1100 1001 0001 0000 1000 1(2)

6. Positive number before normalization:

0.062 209 657 594(10) =

0.0000 1111 1110 1100 1111 1000 1101 1100 1101 1100 1001 0001 0000 1000 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:


0.062 209 657 594(10) =

0.0000 1111 1110 1100 1111 1000 1101 1100 1101 1100 1001 0001 0000 1000 1(2) =

0.0000 1111 1110 1100 1111 1000 1101 1100 1101 1100 1001 0001 0000 1000 1(2) × 20 =

1.1111 1101 1001 1111 0001 1011 1001 1011 1001 0010 0010 0001 0001(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):
1.1111 1101 1001 1111 0001 1011 1001 1011 1001 0010 0010 0001 0001


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. 1111 1101 1001 1111 0001 1011 1001 1011 1001 0010 0010 0001 0001 =

1111 1101 1001 1111 0001 1011 1001 1011 1001 0010 0010 0001 0001


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) =
1111 1101 1001 1111 0001 1011 1001 1011 1001 0010 0010 0001 0001


Decimal number -0.062 209 657 594 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 1111 1101 1001 1111 0001 1011 1001 1011 1001 0010 0010 0001 0001

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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