-0.016 738 891 601 562 486 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 486(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
-0.016 738 891 601 562 486(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:

|-0.016 738 891 601 562 486| = 0.016 738 891 601 562 486


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: 0.016 738 891 601 562 486.

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) 0.016 738 891 601 562 486 × 2 = 0 + 0.033 477 783 203 124 972;
  • 2) 0.033 477 783 203 124 972 × 2 = 0 + 0.066 955 566 406 249 944;
  • 3) 0.066 955 566 406 249 944 × 2 = 0 + 0.133 911 132 812 499 888;
  • 4) 0.133 911 132 812 499 888 × 2 = 0 + 0.267 822 265 624 999 776;
  • 5) 0.267 822 265 624 999 776 × 2 = 0 + 0.535 644 531 249 999 552;
  • 6) 0.535 644 531 249 999 552 × 2 = 1 + 0.071 289 062 499 999 104;
  • 7) 0.071 289 062 499 999 104 × 2 = 0 + 0.142 578 124 999 998 208;
  • 8) 0.142 578 124 999 998 208 × 2 = 0 + 0.285 156 249 999 996 416;
  • 9) 0.285 156 249 999 996 416 × 2 = 0 + 0.570 312 499 999 992 832;
  • 10) 0.570 312 499 999 992 832 × 2 = 1 + 0.140 624 999 999 985 664;
  • 11) 0.140 624 999 999 985 664 × 2 = 0 + 0.281 249 999 999 971 328;
  • 12) 0.281 249 999 999 971 328 × 2 = 0 + 0.562 499 999 999 942 656;
  • 13) 0.562 499 999 999 942 656 × 2 = 1 + 0.124 999 999 999 885 312;
  • 14) 0.124 999 999 999 885 312 × 2 = 0 + 0.249 999 999 999 770 624;
  • 15) 0.249 999 999 999 770 624 × 2 = 0 + 0.499 999 999 999 541 248;
  • 16) 0.499 999 999 999 541 248 × 2 = 0 + 0.999 999 999 999 082 496;
  • 17) 0.999 999 999 999 082 496 × 2 = 1 + 0.999 999 999 998 164 992;
  • 18) 0.999 999 999 998 164 992 × 2 = 1 + 0.999 999 999 996 329 984;
  • 19) 0.999 999 999 996 329 984 × 2 = 1 + 0.999 999 999 992 659 968;
  • 20) 0.999 999 999 992 659 968 × 2 = 1 + 0.999 999 999 985 319 936;
  • 21) 0.999 999 999 985 319 936 × 2 = 1 + 0.999 999 999 970 639 872;
  • 22) 0.999 999 999 970 639 872 × 2 = 1 + 0.999 999 999 941 279 744;
  • 23) 0.999 999 999 941 279 744 × 2 = 1 + 0.999 999 999 882 559 488;
  • 24) 0.999 999 999 882 559 488 × 2 = 1 + 0.999 999 999 765 118 976;
  • 25) 0.999 999 999 765 118 976 × 2 = 1 + 0.999 999 999 530 237 952;
  • 26) 0.999 999 999 530 237 952 × 2 = 1 + 0.999 999 999 060 475 904;
  • 27) 0.999 999 999 060 475 904 × 2 = 1 + 0.999 999 998 120 951 808;
  • 28) 0.999 999 998 120 951 808 × 2 = 1 + 0.999 999 996 241 903 616;
  • 29) 0.999 999 996 241 903 616 × 2 = 1 + 0.999 999 992 483 807 232;
  • 30) 0.999 999 992 483 807 232 × 2 = 1 + 0.999 999 984 967 614 464;
  • 31) 0.999 999 984 967 614 464 × 2 = 1 + 0.999 999 969 935 228 928;
  • 32) 0.999 999 969 935 228 928 × 2 = 1 + 0.999 999 939 870 457 856;
  • 33) 0.999 999 939 870 457 856 × 2 = 1 + 0.999 999 879 740 915 712;
  • 34) 0.999 999 879 740 915 712 × 2 = 1 + 0.999 999 759 481 831 424;
  • 35) 0.999 999 759 481 831 424 × 2 = 1 + 0.999 999 518 963 662 848;
  • 36) 0.999 999 518 963 662 848 × 2 = 1 + 0.999 999 037 927 325 696;
  • 37) 0.999 999 037 927 325 696 × 2 = 1 + 0.999 998 075 854 651 392;
  • 38) 0.999 998 075 854 651 392 × 2 = 1 + 0.999 996 151 709 302 784;
  • 39) 0.999 996 151 709 302 784 × 2 = 1 + 0.999 992 303 418 605 568;
  • 40) 0.999 992 303 418 605 568 × 2 = 1 + 0.999 984 606 837 211 136;
  • 41) 0.999 984 606 837 211 136 × 2 = 1 + 0.999 969 213 674 422 272;
  • 42) 0.999 969 213 674 422 272 × 2 = 1 + 0.999 938 427 348 844 544;
  • 43) 0.999 938 427 348 844 544 × 2 = 1 + 0.999 876 854 697 689 088;
  • 44) 0.999 876 854 697 689 088 × 2 = 1 + 0.999 753 709 395 378 176;
  • 45) 0.999 753 709 395 378 176 × 2 = 1 + 0.999 507 418 790 756 352;
  • 46) 0.999 507 418 790 756 352 × 2 = 1 + 0.999 014 837 581 512 704;
  • 47) 0.999 014 837 581 512 704 × 2 = 1 + 0.998 029 675 163 025 408;
  • 48) 0.998 029 675 163 025 408 × 2 = 1 + 0.996 059 350 326 050 816;
  • 49) 0.996 059 350 326 050 816 × 2 = 1 + 0.992 118 700 652 101 632;
  • 50) 0.992 118 700 652 101 632 × 2 = 1 + 0.984 237 401 304 203 264;
  • 51) 0.984 237 401 304 203 264 × 2 = 1 + 0.968 474 802 608 406 528;
  • 52) 0.968 474 802 608 406 528 × 2 = 1 + 0.936 949 605 216 813 056;
  • 53) 0.936 949 605 216 813 056 × 2 = 1 + 0.873 899 210 433 626 112;
  • 54) 0.873 899 210 433 626 112 × 2 = 1 + 0.747 798 420 867 252 224;
  • 55) 0.747 798 420 867 252 224 × 2 = 1 + 0.495 596 841 734 504 448;
  • 56) 0.495 596 841 734 504 448 × 2 = 0 + 0.991 193 683 469 008 896;
  • 57) 0.991 193 683 469 008 896 × 2 = 1 + 0.982 387 366 938 017 792;
  • 58) 0.982 387 366 938 017 792 × 2 = 1 + 0.964 774 733 876 035 584;

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:


0.016 738 891 601 562 486(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 486(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 11(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:


0.016 738 891 601 562 486(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1011(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 1011


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 1011 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1011


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 1011


Decimal number -0.016 738 891 601 562 486 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 1011

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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