-0.016 738 891 601 562 496 528 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 528 8(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
-0.016 738 891 601 562 496 528 8(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:

|-0.016 738 891 601 562 496 528 8| = 0.016 738 891 601 562 496 528 8


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: 0.016 738 891 601 562 496 528 8.

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) 0.016 738 891 601 562 496 528 8 × 2 = 0 + 0.033 477 783 203 124 993 057 6;
  • 2) 0.033 477 783 203 124 993 057 6 × 2 = 0 + 0.066 955 566 406 249 986 115 2;
  • 3) 0.066 955 566 406 249 986 115 2 × 2 = 0 + 0.133 911 132 812 499 972 230 4;
  • 4) 0.133 911 132 812 499 972 230 4 × 2 = 0 + 0.267 822 265 624 999 944 460 8;
  • 5) 0.267 822 265 624 999 944 460 8 × 2 = 0 + 0.535 644 531 249 999 888 921 6;
  • 6) 0.535 644 531 249 999 888 921 6 × 2 = 1 + 0.071 289 062 499 999 777 843 2;
  • 7) 0.071 289 062 499 999 777 843 2 × 2 = 0 + 0.142 578 124 999 999 555 686 4;
  • 8) 0.142 578 124 999 999 555 686 4 × 2 = 0 + 0.285 156 249 999 999 111 372 8;
  • 9) 0.285 156 249 999 999 111 372 8 × 2 = 0 + 0.570 312 499 999 998 222 745 6;
  • 10) 0.570 312 499 999 998 222 745 6 × 2 = 1 + 0.140 624 999 999 996 445 491 2;
  • 11) 0.140 624 999 999 996 445 491 2 × 2 = 0 + 0.281 249 999 999 992 890 982 4;
  • 12) 0.281 249 999 999 992 890 982 4 × 2 = 0 + 0.562 499 999 999 985 781 964 8;
  • 13) 0.562 499 999 999 985 781 964 8 × 2 = 1 + 0.124 999 999 999 971 563 929 6;
  • 14) 0.124 999 999 999 971 563 929 6 × 2 = 0 + 0.249 999 999 999 943 127 859 2;
  • 15) 0.249 999 999 999 943 127 859 2 × 2 = 0 + 0.499 999 999 999 886 255 718 4;
  • 16) 0.499 999 999 999 886 255 718 4 × 2 = 0 + 0.999 999 999 999 772 511 436 8;
  • 17) 0.999 999 999 999 772 511 436 8 × 2 = 1 + 0.999 999 999 999 545 022 873 6;
  • 18) 0.999 999 999 999 545 022 873 6 × 2 = 1 + 0.999 999 999 999 090 045 747 2;
  • 19) 0.999 999 999 999 090 045 747 2 × 2 = 1 + 0.999 999 999 998 180 091 494 4;
  • 20) 0.999 999 999 998 180 091 494 4 × 2 = 1 + 0.999 999 999 996 360 182 988 8;
  • 21) 0.999 999 999 996 360 182 988 8 × 2 = 1 + 0.999 999 999 992 720 365 977 6;
  • 22) 0.999 999 999 992 720 365 977 6 × 2 = 1 + 0.999 999 999 985 440 731 955 2;
  • 23) 0.999 999 999 985 440 731 955 2 × 2 = 1 + 0.999 999 999 970 881 463 910 4;
  • 24) 0.999 999 999 970 881 463 910 4 × 2 = 1 + 0.999 999 999 941 762 927 820 8;
  • 25) 0.999 999 999 941 762 927 820 8 × 2 = 1 + 0.999 999 999 883 525 855 641 6;
  • 26) 0.999 999 999 883 525 855 641 6 × 2 = 1 + 0.999 999 999 767 051 711 283 2;
  • 27) 0.999 999 999 767 051 711 283 2 × 2 = 1 + 0.999 999 999 534 103 422 566 4;
  • 28) 0.999 999 999 534 103 422 566 4 × 2 = 1 + 0.999 999 999 068 206 845 132 8;
  • 29) 0.999 999 999 068 206 845 132 8 × 2 = 1 + 0.999 999 998 136 413 690 265 6;
  • 30) 0.999 999 998 136 413 690 265 6 × 2 = 1 + 0.999 999 996 272 827 380 531 2;
  • 31) 0.999 999 996 272 827 380 531 2 × 2 = 1 + 0.999 999 992 545 654 761 062 4;
  • 32) 0.999 999 992 545 654 761 062 4 × 2 = 1 + 0.999 999 985 091 309 522 124 8;
  • 33) 0.999 999 985 091 309 522 124 8 × 2 = 1 + 0.999 999 970 182 619 044 249 6;
  • 34) 0.999 999 970 182 619 044 249 6 × 2 = 1 + 0.999 999 940 365 238 088 499 2;
  • 35) 0.999 999 940 365 238 088 499 2 × 2 = 1 + 0.999 999 880 730 476 176 998 4;
  • 36) 0.999 999 880 730 476 176 998 4 × 2 = 1 + 0.999 999 761 460 952 353 996 8;
  • 37) 0.999 999 761 460 952 353 996 8 × 2 = 1 + 0.999 999 522 921 904 707 993 6;
  • 38) 0.999 999 522 921 904 707 993 6 × 2 = 1 + 0.999 999 045 843 809 415 987 2;
  • 39) 0.999 999 045 843 809 415 987 2 × 2 = 1 + 0.999 998 091 687 618 831 974 4;
  • 40) 0.999 998 091 687 618 831 974 4 × 2 = 1 + 0.999 996 183 375 237 663 948 8;
  • 41) 0.999 996 183 375 237 663 948 8 × 2 = 1 + 0.999 992 366 750 475 327 897 6;
  • 42) 0.999 992 366 750 475 327 897 6 × 2 = 1 + 0.999 984 733 500 950 655 795 2;
  • 43) 0.999 984 733 500 950 655 795 2 × 2 = 1 + 0.999 969 467 001 901 311 590 4;
  • 44) 0.999 969 467 001 901 311 590 4 × 2 = 1 + 0.999 938 934 003 802 623 180 8;
  • 45) 0.999 938 934 003 802 623 180 8 × 2 = 1 + 0.999 877 868 007 605 246 361 6;
  • 46) 0.999 877 868 007 605 246 361 6 × 2 = 1 + 0.999 755 736 015 210 492 723 2;
  • 47) 0.999 755 736 015 210 492 723 2 × 2 = 1 + 0.999 511 472 030 420 985 446 4;
  • 48) 0.999 511 472 030 420 985 446 4 × 2 = 1 + 0.999 022 944 060 841 970 892 8;
  • 49) 0.999 022 944 060 841 970 892 8 × 2 = 1 + 0.998 045 888 121 683 941 785 6;
  • 50) 0.998 045 888 121 683 941 785 6 × 2 = 1 + 0.996 091 776 243 367 883 571 2;
  • 51) 0.996 091 776 243 367 883 571 2 × 2 = 1 + 0.992 183 552 486 735 767 142 4;
  • 52) 0.992 183 552 486 735 767 142 4 × 2 = 1 + 0.984 367 104 973 471 534 284 8;
  • 53) 0.984 367 104 973 471 534 284 8 × 2 = 1 + 0.968 734 209 946 943 068 569 6;
  • 54) 0.968 734 209 946 943 068 569 6 × 2 = 1 + 0.937 468 419 893 886 137 139 2;
  • 55) 0.937 468 419 893 886 137 139 2 × 2 = 1 + 0.874 936 839 787 772 274 278 4;
  • 56) 0.874 936 839 787 772 274 278 4 × 2 = 1 + 0.749 873 679 575 544 548 556 8;
  • 57) 0.749 873 679 575 544 548 556 8 × 2 = 1 + 0.499 747 359 151 089 097 113 6;
  • 58) 0.499 747 359 151 089 097 113 6 × 2 = 0 + 0.999 494 718 302 178 194 227 2;

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:


0.016 738 891 601 562 496 528 8(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 528 8(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:


0.016 738 891 601 562 496 528 8(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 -0.016 738 891 601 562 496 528 8 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