-0.016 738 891 601 562 493 63 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 493 63(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
-0.016 738 891 601 562 493 63(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:

|-0.016 738 891 601 562 493 63| = 0.016 738 891 601 562 493 63


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: 0.016 738 891 601 562 493 63.

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) 0.016 738 891 601 562 493 63 × 2 = 0 + 0.033 477 783 203 124 987 26;
  • 2) 0.033 477 783 203 124 987 26 × 2 = 0 + 0.066 955 566 406 249 974 52;
  • 3) 0.066 955 566 406 249 974 52 × 2 = 0 + 0.133 911 132 812 499 949 04;
  • 4) 0.133 911 132 812 499 949 04 × 2 = 0 + 0.267 822 265 624 999 898 08;
  • 5) 0.267 822 265 624 999 898 08 × 2 = 0 + 0.535 644 531 249 999 796 16;
  • 6) 0.535 644 531 249 999 796 16 × 2 = 1 + 0.071 289 062 499 999 592 32;
  • 7) 0.071 289 062 499 999 592 32 × 2 = 0 + 0.142 578 124 999 999 184 64;
  • 8) 0.142 578 124 999 999 184 64 × 2 = 0 + 0.285 156 249 999 998 369 28;
  • 9) 0.285 156 249 999 998 369 28 × 2 = 0 + 0.570 312 499 999 996 738 56;
  • 10) 0.570 312 499 999 996 738 56 × 2 = 1 + 0.140 624 999 999 993 477 12;
  • 11) 0.140 624 999 999 993 477 12 × 2 = 0 + 0.281 249 999 999 986 954 24;
  • 12) 0.281 249 999 999 986 954 24 × 2 = 0 + 0.562 499 999 999 973 908 48;
  • 13) 0.562 499 999 999 973 908 48 × 2 = 1 + 0.124 999 999 999 947 816 96;
  • 14) 0.124 999 999 999 947 816 96 × 2 = 0 + 0.249 999 999 999 895 633 92;
  • 15) 0.249 999 999 999 895 633 92 × 2 = 0 + 0.499 999 999 999 791 267 84;
  • 16) 0.499 999 999 999 791 267 84 × 2 = 0 + 0.999 999 999 999 582 535 68;
  • 17) 0.999 999 999 999 582 535 68 × 2 = 1 + 0.999 999 999 999 165 071 36;
  • 18) 0.999 999 999 999 165 071 36 × 2 = 1 + 0.999 999 999 998 330 142 72;
  • 19) 0.999 999 999 998 330 142 72 × 2 = 1 + 0.999 999 999 996 660 285 44;
  • 20) 0.999 999 999 996 660 285 44 × 2 = 1 + 0.999 999 999 993 320 570 88;
  • 21) 0.999 999 999 993 320 570 88 × 2 = 1 + 0.999 999 999 986 641 141 76;
  • 22) 0.999 999 999 986 641 141 76 × 2 = 1 + 0.999 999 999 973 282 283 52;
  • 23) 0.999 999 999 973 282 283 52 × 2 = 1 + 0.999 999 999 946 564 567 04;
  • 24) 0.999 999 999 946 564 567 04 × 2 = 1 + 0.999 999 999 893 129 134 08;
  • 25) 0.999 999 999 893 129 134 08 × 2 = 1 + 0.999 999 999 786 258 268 16;
  • 26) 0.999 999 999 786 258 268 16 × 2 = 1 + 0.999 999 999 572 516 536 32;
  • 27) 0.999 999 999 572 516 536 32 × 2 = 1 + 0.999 999 999 145 033 072 64;
  • 28) 0.999 999 999 145 033 072 64 × 2 = 1 + 0.999 999 998 290 066 145 28;
  • 29) 0.999 999 998 290 066 145 28 × 2 = 1 + 0.999 999 996 580 132 290 56;
  • 30) 0.999 999 996 580 132 290 56 × 2 = 1 + 0.999 999 993 160 264 581 12;
  • 31) 0.999 999 993 160 264 581 12 × 2 = 1 + 0.999 999 986 320 529 162 24;
  • 32) 0.999 999 986 320 529 162 24 × 2 = 1 + 0.999 999 972 641 058 324 48;
  • 33) 0.999 999 972 641 058 324 48 × 2 = 1 + 0.999 999 945 282 116 648 96;
  • 34) 0.999 999 945 282 116 648 96 × 2 = 1 + 0.999 999 890 564 233 297 92;
  • 35) 0.999 999 890 564 233 297 92 × 2 = 1 + 0.999 999 781 128 466 595 84;
  • 36) 0.999 999 781 128 466 595 84 × 2 = 1 + 0.999 999 562 256 933 191 68;
  • 37) 0.999 999 562 256 933 191 68 × 2 = 1 + 0.999 999 124 513 866 383 36;
  • 38) 0.999 999 124 513 866 383 36 × 2 = 1 + 0.999 998 249 027 732 766 72;
  • 39) 0.999 998 249 027 732 766 72 × 2 = 1 + 0.999 996 498 055 465 533 44;
  • 40) 0.999 996 498 055 465 533 44 × 2 = 1 + 0.999 992 996 110 931 066 88;
  • 41) 0.999 992 996 110 931 066 88 × 2 = 1 + 0.999 985 992 221 862 133 76;
  • 42) 0.999 985 992 221 862 133 76 × 2 = 1 + 0.999 971 984 443 724 267 52;
  • 43) 0.999 971 984 443 724 267 52 × 2 = 1 + 0.999 943 968 887 448 535 04;
  • 44) 0.999 943 968 887 448 535 04 × 2 = 1 + 0.999 887 937 774 897 070 08;
  • 45) 0.999 887 937 774 897 070 08 × 2 = 1 + 0.999 775 875 549 794 140 16;
  • 46) 0.999 775 875 549 794 140 16 × 2 = 1 + 0.999 551 751 099 588 280 32;
  • 47) 0.999 551 751 099 588 280 32 × 2 = 1 + 0.999 103 502 199 176 560 64;
  • 48) 0.999 103 502 199 176 560 64 × 2 = 1 + 0.998 207 004 398 353 121 28;
  • 49) 0.998 207 004 398 353 121 28 × 2 = 1 + 0.996 414 008 796 706 242 56;
  • 50) 0.996 414 008 796 706 242 56 × 2 = 1 + 0.992 828 017 593 412 485 12;
  • 51) 0.992 828 017 593 412 485 12 × 2 = 1 + 0.985 656 035 186 824 970 24;
  • 52) 0.985 656 035 186 824 970 24 × 2 = 1 + 0.971 312 070 373 649 940 48;
  • 53) 0.971 312 070 373 649 940 48 × 2 = 1 + 0.942 624 140 747 299 880 96;
  • 54) 0.942 624 140 747 299 880 96 × 2 = 1 + 0.885 248 281 494 599 761 92;
  • 55) 0.885 248 281 494 599 761 92 × 2 = 1 + 0.770 496 562 989 199 523 84;
  • 56) 0.770 496 562 989 199 523 84 × 2 = 1 + 0.540 993 125 978 399 047 68;
  • 57) 0.540 993 125 978 399 047 68 × 2 = 1 + 0.081 986 251 956 798 095 36;
  • 58) 0.081 986 251 956 798 095 36 × 2 = 0 + 0.163 972 503 913 596 190 72;

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:


0.016 738 891 601 562 493 63(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 493 63(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:


0.016 738 891 601 562 493 63(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 -0.016 738 891 601 562 493 63 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