-0.016 738 891 601 562 497 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 497 1(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
-0.016 738 891 601 562 497 1(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:

|-0.016 738 891 601 562 497 1| = 0.016 738 891 601 562 497 1


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: 0.016 738 891 601 562 497 1.

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) 0.016 738 891 601 562 497 1 × 2 = 0 + 0.033 477 783 203 124 994 2;
  • 2) 0.033 477 783 203 124 994 2 × 2 = 0 + 0.066 955 566 406 249 988 4;
  • 3) 0.066 955 566 406 249 988 4 × 2 = 0 + 0.133 911 132 812 499 976 8;
  • 4) 0.133 911 132 812 499 976 8 × 2 = 0 + 0.267 822 265 624 999 953 6;
  • 5) 0.267 822 265 624 999 953 6 × 2 = 0 + 0.535 644 531 249 999 907 2;
  • 6) 0.535 644 531 249 999 907 2 × 2 = 1 + 0.071 289 062 499 999 814 4;
  • 7) 0.071 289 062 499 999 814 4 × 2 = 0 + 0.142 578 124 999 999 628 8;
  • 8) 0.142 578 124 999 999 628 8 × 2 = 0 + 0.285 156 249 999 999 257 6;
  • 9) 0.285 156 249 999 999 257 6 × 2 = 0 + 0.570 312 499 999 998 515 2;
  • 10) 0.570 312 499 999 998 515 2 × 2 = 1 + 0.140 624 999 999 997 030 4;
  • 11) 0.140 624 999 999 997 030 4 × 2 = 0 + 0.281 249 999 999 994 060 8;
  • 12) 0.281 249 999 999 994 060 8 × 2 = 0 + 0.562 499 999 999 988 121 6;
  • 13) 0.562 499 999 999 988 121 6 × 2 = 1 + 0.124 999 999 999 976 243 2;
  • 14) 0.124 999 999 999 976 243 2 × 2 = 0 + 0.249 999 999 999 952 486 4;
  • 15) 0.249 999 999 999 952 486 4 × 2 = 0 + 0.499 999 999 999 904 972 8;
  • 16) 0.499 999 999 999 904 972 8 × 2 = 0 + 0.999 999 999 999 809 945 6;
  • 17) 0.999 999 999 999 809 945 6 × 2 = 1 + 0.999 999 999 999 619 891 2;
  • 18) 0.999 999 999 999 619 891 2 × 2 = 1 + 0.999 999 999 999 239 782 4;
  • 19) 0.999 999 999 999 239 782 4 × 2 = 1 + 0.999 999 999 998 479 564 8;
  • 20) 0.999 999 999 998 479 564 8 × 2 = 1 + 0.999 999 999 996 959 129 6;
  • 21) 0.999 999 999 996 959 129 6 × 2 = 1 + 0.999 999 999 993 918 259 2;
  • 22) 0.999 999 999 993 918 259 2 × 2 = 1 + 0.999 999 999 987 836 518 4;
  • 23) 0.999 999 999 987 836 518 4 × 2 = 1 + 0.999 999 999 975 673 036 8;
  • 24) 0.999 999 999 975 673 036 8 × 2 = 1 + 0.999 999 999 951 346 073 6;
  • 25) 0.999 999 999 951 346 073 6 × 2 = 1 + 0.999 999 999 902 692 147 2;
  • 26) 0.999 999 999 902 692 147 2 × 2 = 1 + 0.999 999 999 805 384 294 4;
  • 27) 0.999 999 999 805 384 294 4 × 2 = 1 + 0.999 999 999 610 768 588 8;
  • 28) 0.999 999 999 610 768 588 8 × 2 = 1 + 0.999 999 999 221 537 177 6;
  • 29) 0.999 999 999 221 537 177 6 × 2 = 1 + 0.999 999 998 443 074 355 2;
  • 30) 0.999 999 998 443 074 355 2 × 2 = 1 + 0.999 999 996 886 148 710 4;
  • 31) 0.999 999 996 886 148 710 4 × 2 = 1 + 0.999 999 993 772 297 420 8;
  • 32) 0.999 999 993 772 297 420 8 × 2 = 1 + 0.999 999 987 544 594 841 6;
  • 33) 0.999 999 987 544 594 841 6 × 2 = 1 + 0.999 999 975 089 189 683 2;
  • 34) 0.999 999 975 089 189 683 2 × 2 = 1 + 0.999 999 950 178 379 366 4;
  • 35) 0.999 999 950 178 379 366 4 × 2 = 1 + 0.999 999 900 356 758 732 8;
  • 36) 0.999 999 900 356 758 732 8 × 2 = 1 + 0.999 999 800 713 517 465 6;
  • 37) 0.999 999 800 713 517 465 6 × 2 = 1 + 0.999 999 601 427 034 931 2;
  • 38) 0.999 999 601 427 034 931 2 × 2 = 1 + 0.999 999 202 854 069 862 4;
  • 39) 0.999 999 202 854 069 862 4 × 2 = 1 + 0.999 998 405 708 139 724 8;
  • 40) 0.999 998 405 708 139 724 8 × 2 = 1 + 0.999 996 811 416 279 449 6;
  • 41) 0.999 996 811 416 279 449 6 × 2 = 1 + 0.999 993 622 832 558 899 2;
  • 42) 0.999 993 622 832 558 899 2 × 2 = 1 + 0.999 987 245 665 117 798 4;
  • 43) 0.999 987 245 665 117 798 4 × 2 = 1 + 0.999 974 491 330 235 596 8;
  • 44) 0.999 974 491 330 235 596 8 × 2 = 1 + 0.999 948 982 660 471 193 6;
  • 45) 0.999 948 982 660 471 193 6 × 2 = 1 + 0.999 897 965 320 942 387 2;
  • 46) 0.999 897 965 320 942 387 2 × 2 = 1 + 0.999 795 930 641 884 774 4;
  • 47) 0.999 795 930 641 884 774 4 × 2 = 1 + 0.999 591 861 283 769 548 8;
  • 48) 0.999 591 861 283 769 548 8 × 2 = 1 + 0.999 183 722 567 539 097 6;
  • 49) 0.999 183 722 567 539 097 6 × 2 = 1 + 0.998 367 445 135 078 195 2;
  • 50) 0.998 367 445 135 078 195 2 × 2 = 1 + 0.996 734 890 270 156 390 4;
  • 51) 0.996 734 890 270 156 390 4 × 2 = 1 + 0.993 469 780 540 312 780 8;
  • 52) 0.993 469 780 540 312 780 8 × 2 = 1 + 0.986 939 561 080 625 561 6;
  • 53) 0.986 939 561 080 625 561 6 × 2 = 1 + 0.973 879 122 161 251 123 2;
  • 54) 0.973 879 122 161 251 123 2 × 2 = 1 + 0.947 758 244 322 502 246 4;
  • 55) 0.947 758 244 322 502 246 4 × 2 = 1 + 0.895 516 488 645 004 492 8;
  • 56) 0.895 516 488 645 004 492 8 × 2 = 1 + 0.791 032 977 290 008 985 6;
  • 57) 0.791 032 977 290 008 985 6 × 2 = 1 + 0.582 065 954 580 017 971 2;
  • 58) 0.582 065 954 580 017 971 2 × 2 = 1 + 0.164 131 909 160 035 942 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:


0.016 738 891 601 562 497 1(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 497 1(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 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:


0.016 738 891 601 562 497 1(10) =

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

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

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


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 1111 =

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


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 1111


Decimal number -0.016 738 891 601 562 497 1 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 1111

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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