-0.016 738 891 601 561 43 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 561 43(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
-0.016 738 891 601 561 43(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:

|-0.016 738 891 601 561 43| = 0.016 738 891 601 561 43


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: 0.016 738 891 601 561 43.

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) 0.016 738 891 601 561 43 × 2 = 0 + 0.033 477 783 203 122 86;
  • 2) 0.033 477 783 203 122 86 × 2 = 0 + 0.066 955 566 406 245 72;
  • 3) 0.066 955 566 406 245 72 × 2 = 0 + 0.133 911 132 812 491 44;
  • 4) 0.133 911 132 812 491 44 × 2 = 0 + 0.267 822 265 624 982 88;
  • 5) 0.267 822 265 624 982 88 × 2 = 0 + 0.535 644 531 249 965 76;
  • 6) 0.535 644 531 249 965 76 × 2 = 1 + 0.071 289 062 499 931 52;
  • 7) 0.071 289 062 499 931 52 × 2 = 0 + 0.142 578 124 999 863 04;
  • 8) 0.142 578 124 999 863 04 × 2 = 0 + 0.285 156 249 999 726 08;
  • 9) 0.285 156 249 999 726 08 × 2 = 0 + 0.570 312 499 999 452 16;
  • 10) 0.570 312 499 999 452 16 × 2 = 1 + 0.140 624 999 998 904 32;
  • 11) 0.140 624 999 998 904 32 × 2 = 0 + 0.281 249 999 997 808 64;
  • 12) 0.281 249 999 997 808 64 × 2 = 0 + 0.562 499 999 995 617 28;
  • 13) 0.562 499 999 995 617 28 × 2 = 1 + 0.124 999 999 991 234 56;
  • 14) 0.124 999 999 991 234 56 × 2 = 0 + 0.249 999 999 982 469 12;
  • 15) 0.249 999 999 982 469 12 × 2 = 0 + 0.499 999 999 964 938 24;
  • 16) 0.499 999 999 964 938 24 × 2 = 0 + 0.999 999 999 929 876 48;
  • 17) 0.999 999 999 929 876 48 × 2 = 1 + 0.999 999 999 859 752 96;
  • 18) 0.999 999 999 859 752 96 × 2 = 1 + 0.999 999 999 719 505 92;
  • 19) 0.999 999 999 719 505 92 × 2 = 1 + 0.999 999 999 439 011 84;
  • 20) 0.999 999 999 439 011 84 × 2 = 1 + 0.999 999 998 878 023 68;
  • 21) 0.999 999 998 878 023 68 × 2 = 1 + 0.999 999 997 756 047 36;
  • 22) 0.999 999 997 756 047 36 × 2 = 1 + 0.999 999 995 512 094 72;
  • 23) 0.999 999 995 512 094 72 × 2 = 1 + 0.999 999 991 024 189 44;
  • 24) 0.999 999 991 024 189 44 × 2 = 1 + 0.999 999 982 048 378 88;
  • 25) 0.999 999 982 048 378 88 × 2 = 1 + 0.999 999 964 096 757 76;
  • 26) 0.999 999 964 096 757 76 × 2 = 1 + 0.999 999 928 193 515 52;
  • 27) 0.999 999 928 193 515 52 × 2 = 1 + 0.999 999 856 387 031 04;
  • 28) 0.999 999 856 387 031 04 × 2 = 1 + 0.999 999 712 774 062 08;
  • 29) 0.999 999 712 774 062 08 × 2 = 1 + 0.999 999 425 548 124 16;
  • 30) 0.999 999 425 548 124 16 × 2 = 1 + 0.999 998 851 096 248 32;
  • 31) 0.999 998 851 096 248 32 × 2 = 1 + 0.999 997 702 192 496 64;
  • 32) 0.999 997 702 192 496 64 × 2 = 1 + 0.999 995 404 384 993 28;
  • 33) 0.999 995 404 384 993 28 × 2 = 1 + 0.999 990 808 769 986 56;
  • 34) 0.999 990 808 769 986 56 × 2 = 1 + 0.999 981 617 539 973 12;
  • 35) 0.999 981 617 539 973 12 × 2 = 1 + 0.999 963 235 079 946 24;
  • 36) 0.999 963 235 079 946 24 × 2 = 1 + 0.999 926 470 159 892 48;
  • 37) 0.999 926 470 159 892 48 × 2 = 1 + 0.999 852 940 319 784 96;
  • 38) 0.999 852 940 319 784 96 × 2 = 1 + 0.999 705 880 639 569 92;
  • 39) 0.999 705 880 639 569 92 × 2 = 1 + 0.999 411 761 279 139 84;
  • 40) 0.999 411 761 279 139 84 × 2 = 1 + 0.998 823 522 558 279 68;
  • 41) 0.998 823 522 558 279 68 × 2 = 1 + 0.997 647 045 116 559 36;
  • 42) 0.997 647 045 116 559 36 × 2 = 1 + 0.995 294 090 233 118 72;
  • 43) 0.995 294 090 233 118 72 × 2 = 1 + 0.990 588 180 466 237 44;
  • 44) 0.990 588 180 466 237 44 × 2 = 1 + 0.981 176 360 932 474 88;
  • 45) 0.981 176 360 932 474 88 × 2 = 1 + 0.962 352 721 864 949 76;
  • 46) 0.962 352 721 864 949 76 × 2 = 1 + 0.924 705 443 729 899 52;
  • 47) 0.924 705 443 729 899 52 × 2 = 1 + 0.849 410 887 459 799 04;
  • 48) 0.849 410 887 459 799 04 × 2 = 1 + 0.698 821 774 919 598 08;
  • 49) 0.698 821 774 919 598 08 × 2 = 1 + 0.397 643 549 839 196 16;
  • 50) 0.397 643 549 839 196 16 × 2 = 0 + 0.795 287 099 678 392 32;
  • 51) 0.795 287 099 678 392 32 × 2 = 1 + 0.590 574 199 356 784 64;
  • 52) 0.590 574 199 356 784 64 × 2 = 1 + 0.181 148 398 713 569 28;
  • 53) 0.181 148 398 713 569 28 × 2 = 0 + 0.362 296 797 427 138 56;
  • 54) 0.362 296 797 427 138 56 × 2 = 0 + 0.724 593 594 854 277 12;
  • 55) 0.724 593 594 854 277 12 × 2 = 1 + 0.449 187 189 708 554 24;
  • 56) 0.449 187 189 708 554 24 × 2 = 0 + 0.898 374 379 417 108 48;
  • 57) 0.898 374 379 417 108 48 × 2 = 1 + 0.796 748 758 834 216 96;
  • 58) 0.796 748 758 834 216 96 × 2 = 1 + 0.593 497 517 668 433 92;

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:


0.016 738 891 601 561 43(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 0010 11(2)

6. Positive number before normalization:

0.016 738 891 601 561 43(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 0010 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:


0.016 738 891 601 561 43(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 0010 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1011 0010 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1110 1100 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 1110 1100 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 1110 1100 1011 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1110 1100 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 1110 1100 1011


Decimal number -0.016 738 891 601 561 43 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 1110 1100 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