-0.016 738 891 601 562 494 35 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 494 35(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
-0.016 738 891 601 562 494 35(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:

|-0.016 738 891 601 562 494 35| = 0.016 738 891 601 562 494 35


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: 0.016 738 891 601 562 494 35.

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) 0.016 738 891 601 562 494 35 × 2 = 0 + 0.033 477 783 203 124 988 7;
  • 2) 0.033 477 783 203 124 988 7 × 2 = 0 + 0.066 955 566 406 249 977 4;
  • 3) 0.066 955 566 406 249 977 4 × 2 = 0 + 0.133 911 132 812 499 954 8;
  • 4) 0.133 911 132 812 499 954 8 × 2 = 0 + 0.267 822 265 624 999 909 6;
  • 5) 0.267 822 265 624 999 909 6 × 2 = 0 + 0.535 644 531 249 999 819 2;
  • 6) 0.535 644 531 249 999 819 2 × 2 = 1 + 0.071 289 062 499 999 638 4;
  • 7) 0.071 289 062 499 999 638 4 × 2 = 0 + 0.142 578 124 999 999 276 8;
  • 8) 0.142 578 124 999 999 276 8 × 2 = 0 + 0.285 156 249 999 998 553 6;
  • 9) 0.285 156 249 999 998 553 6 × 2 = 0 + 0.570 312 499 999 997 107 2;
  • 10) 0.570 312 499 999 997 107 2 × 2 = 1 + 0.140 624 999 999 994 214 4;
  • 11) 0.140 624 999 999 994 214 4 × 2 = 0 + 0.281 249 999 999 988 428 8;
  • 12) 0.281 249 999 999 988 428 8 × 2 = 0 + 0.562 499 999 999 976 857 6;
  • 13) 0.562 499 999 999 976 857 6 × 2 = 1 + 0.124 999 999 999 953 715 2;
  • 14) 0.124 999 999 999 953 715 2 × 2 = 0 + 0.249 999 999 999 907 430 4;
  • 15) 0.249 999 999 999 907 430 4 × 2 = 0 + 0.499 999 999 999 814 860 8;
  • 16) 0.499 999 999 999 814 860 8 × 2 = 0 + 0.999 999 999 999 629 721 6;
  • 17) 0.999 999 999 999 629 721 6 × 2 = 1 + 0.999 999 999 999 259 443 2;
  • 18) 0.999 999 999 999 259 443 2 × 2 = 1 + 0.999 999 999 998 518 886 4;
  • 19) 0.999 999 999 998 518 886 4 × 2 = 1 + 0.999 999 999 997 037 772 8;
  • 20) 0.999 999 999 997 037 772 8 × 2 = 1 + 0.999 999 999 994 075 545 6;
  • 21) 0.999 999 999 994 075 545 6 × 2 = 1 + 0.999 999 999 988 151 091 2;
  • 22) 0.999 999 999 988 151 091 2 × 2 = 1 + 0.999 999 999 976 302 182 4;
  • 23) 0.999 999 999 976 302 182 4 × 2 = 1 + 0.999 999 999 952 604 364 8;
  • 24) 0.999 999 999 952 604 364 8 × 2 = 1 + 0.999 999 999 905 208 729 6;
  • 25) 0.999 999 999 905 208 729 6 × 2 = 1 + 0.999 999 999 810 417 459 2;
  • 26) 0.999 999 999 810 417 459 2 × 2 = 1 + 0.999 999 999 620 834 918 4;
  • 27) 0.999 999 999 620 834 918 4 × 2 = 1 + 0.999 999 999 241 669 836 8;
  • 28) 0.999 999 999 241 669 836 8 × 2 = 1 + 0.999 999 998 483 339 673 6;
  • 29) 0.999 999 998 483 339 673 6 × 2 = 1 + 0.999 999 996 966 679 347 2;
  • 30) 0.999 999 996 966 679 347 2 × 2 = 1 + 0.999 999 993 933 358 694 4;
  • 31) 0.999 999 993 933 358 694 4 × 2 = 1 + 0.999 999 987 866 717 388 8;
  • 32) 0.999 999 987 866 717 388 8 × 2 = 1 + 0.999 999 975 733 434 777 6;
  • 33) 0.999 999 975 733 434 777 6 × 2 = 1 + 0.999 999 951 466 869 555 2;
  • 34) 0.999 999 951 466 869 555 2 × 2 = 1 + 0.999 999 902 933 739 110 4;
  • 35) 0.999 999 902 933 739 110 4 × 2 = 1 + 0.999 999 805 867 478 220 8;
  • 36) 0.999 999 805 867 478 220 8 × 2 = 1 + 0.999 999 611 734 956 441 6;
  • 37) 0.999 999 611 734 956 441 6 × 2 = 1 + 0.999 999 223 469 912 883 2;
  • 38) 0.999 999 223 469 912 883 2 × 2 = 1 + 0.999 998 446 939 825 766 4;
  • 39) 0.999 998 446 939 825 766 4 × 2 = 1 + 0.999 996 893 879 651 532 8;
  • 40) 0.999 996 893 879 651 532 8 × 2 = 1 + 0.999 993 787 759 303 065 6;
  • 41) 0.999 993 787 759 303 065 6 × 2 = 1 + 0.999 987 575 518 606 131 2;
  • 42) 0.999 987 575 518 606 131 2 × 2 = 1 + 0.999 975 151 037 212 262 4;
  • 43) 0.999 975 151 037 212 262 4 × 2 = 1 + 0.999 950 302 074 424 524 8;
  • 44) 0.999 950 302 074 424 524 8 × 2 = 1 + 0.999 900 604 148 849 049 6;
  • 45) 0.999 900 604 148 849 049 6 × 2 = 1 + 0.999 801 208 297 698 099 2;
  • 46) 0.999 801 208 297 698 099 2 × 2 = 1 + 0.999 602 416 595 396 198 4;
  • 47) 0.999 602 416 595 396 198 4 × 2 = 1 + 0.999 204 833 190 792 396 8;
  • 48) 0.999 204 833 190 792 396 8 × 2 = 1 + 0.998 409 666 381 584 793 6;
  • 49) 0.998 409 666 381 584 793 6 × 2 = 1 + 0.996 819 332 763 169 587 2;
  • 50) 0.996 819 332 763 169 587 2 × 2 = 1 + 0.993 638 665 526 339 174 4;
  • 51) 0.993 638 665 526 339 174 4 × 2 = 1 + 0.987 277 331 052 678 348 8;
  • 52) 0.987 277 331 052 678 348 8 × 2 = 1 + 0.974 554 662 105 356 697 6;
  • 53) 0.974 554 662 105 356 697 6 × 2 = 1 + 0.949 109 324 210 713 395 2;
  • 54) 0.949 109 324 210 713 395 2 × 2 = 1 + 0.898 218 648 421 426 790 4;
  • 55) 0.898 218 648 421 426 790 4 × 2 = 1 + 0.796 437 296 842 853 580 8;
  • 56) 0.796 437 296 842 853 580 8 × 2 = 1 + 0.592 874 593 685 707 161 6;
  • 57) 0.592 874 593 685 707 161 6 × 2 = 1 + 0.185 749 187 371 414 323 2;
  • 58) 0.185 749 187 371 414 323 2 × 2 = 0 + 0.371 498 374 742 828 646 4;

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:


0.016 738 891 601 562 494 35(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 494 35(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 494 35(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 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) =

1017(10) =

011 1111 1001(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. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 494 35 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110

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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