-0.016 738 891 601 562 496 525 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 525(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
-0.016 738 891 601 562 496 525(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:

|-0.016 738 891 601 562 496 525| = 0.016 738 891 601 562 496 525


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: 0.016 738 891 601 562 496 525.

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) 0.016 738 891 601 562 496 525 × 2 = 0 + 0.033 477 783 203 124 993 05;
  • 2) 0.033 477 783 203 124 993 05 × 2 = 0 + 0.066 955 566 406 249 986 1;
  • 3) 0.066 955 566 406 249 986 1 × 2 = 0 + 0.133 911 132 812 499 972 2;
  • 4) 0.133 911 132 812 499 972 2 × 2 = 0 + 0.267 822 265 624 999 944 4;
  • 5) 0.267 822 265 624 999 944 4 × 2 = 0 + 0.535 644 531 249 999 888 8;
  • 6) 0.535 644 531 249 999 888 8 × 2 = 1 + 0.071 289 062 499 999 777 6;
  • 7) 0.071 289 062 499 999 777 6 × 2 = 0 + 0.142 578 124 999 999 555 2;
  • 8) 0.142 578 124 999 999 555 2 × 2 = 0 + 0.285 156 249 999 999 110 4;
  • 9) 0.285 156 249 999 999 110 4 × 2 = 0 + 0.570 312 499 999 998 220 8;
  • 10) 0.570 312 499 999 998 220 8 × 2 = 1 + 0.140 624 999 999 996 441 6;
  • 11) 0.140 624 999 999 996 441 6 × 2 = 0 + 0.281 249 999 999 992 883 2;
  • 12) 0.281 249 999 999 992 883 2 × 2 = 0 + 0.562 499 999 999 985 766 4;
  • 13) 0.562 499 999 999 985 766 4 × 2 = 1 + 0.124 999 999 999 971 532 8;
  • 14) 0.124 999 999 999 971 532 8 × 2 = 0 + 0.249 999 999 999 943 065 6;
  • 15) 0.249 999 999 999 943 065 6 × 2 = 0 + 0.499 999 999 999 886 131 2;
  • 16) 0.499 999 999 999 886 131 2 × 2 = 0 + 0.999 999 999 999 772 262 4;
  • 17) 0.999 999 999 999 772 262 4 × 2 = 1 + 0.999 999 999 999 544 524 8;
  • 18) 0.999 999 999 999 544 524 8 × 2 = 1 + 0.999 999 999 999 089 049 6;
  • 19) 0.999 999 999 999 089 049 6 × 2 = 1 + 0.999 999 999 998 178 099 2;
  • 20) 0.999 999 999 998 178 099 2 × 2 = 1 + 0.999 999 999 996 356 198 4;
  • 21) 0.999 999 999 996 356 198 4 × 2 = 1 + 0.999 999 999 992 712 396 8;
  • 22) 0.999 999 999 992 712 396 8 × 2 = 1 + 0.999 999 999 985 424 793 6;
  • 23) 0.999 999 999 985 424 793 6 × 2 = 1 + 0.999 999 999 970 849 587 2;
  • 24) 0.999 999 999 970 849 587 2 × 2 = 1 + 0.999 999 999 941 699 174 4;
  • 25) 0.999 999 999 941 699 174 4 × 2 = 1 + 0.999 999 999 883 398 348 8;
  • 26) 0.999 999 999 883 398 348 8 × 2 = 1 + 0.999 999 999 766 796 697 6;
  • 27) 0.999 999 999 766 796 697 6 × 2 = 1 + 0.999 999 999 533 593 395 2;
  • 28) 0.999 999 999 533 593 395 2 × 2 = 1 + 0.999 999 999 067 186 790 4;
  • 29) 0.999 999 999 067 186 790 4 × 2 = 1 + 0.999 999 998 134 373 580 8;
  • 30) 0.999 999 998 134 373 580 8 × 2 = 1 + 0.999 999 996 268 747 161 6;
  • 31) 0.999 999 996 268 747 161 6 × 2 = 1 + 0.999 999 992 537 494 323 2;
  • 32) 0.999 999 992 537 494 323 2 × 2 = 1 + 0.999 999 985 074 988 646 4;
  • 33) 0.999 999 985 074 988 646 4 × 2 = 1 + 0.999 999 970 149 977 292 8;
  • 34) 0.999 999 970 149 977 292 8 × 2 = 1 + 0.999 999 940 299 954 585 6;
  • 35) 0.999 999 940 299 954 585 6 × 2 = 1 + 0.999 999 880 599 909 171 2;
  • 36) 0.999 999 880 599 909 171 2 × 2 = 1 + 0.999 999 761 199 818 342 4;
  • 37) 0.999 999 761 199 818 342 4 × 2 = 1 + 0.999 999 522 399 636 684 8;
  • 38) 0.999 999 522 399 636 684 8 × 2 = 1 + 0.999 999 044 799 273 369 6;
  • 39) 0.999 999 044 799 273 369 6 × 2 = 1 + 0.999 998 089 598 546 739 2;
  • 40) 0.999 998 089 598 546 739 2 × 2 = 1 + 0.999 996 179 197 093 478 4;
  • 41) 0.999 996 179 197 093 478 4 × 2 = 1 + 0.999 992 358 394 186 956 8;
  • 42) 0.999 992 358 394 186 956 8 × 2 = 1 + 0.999 984 716 788 373 913 6;
  • 43) 0.999 984 716 788 373 913 6 × 2 = 1 + 0.999 969 433 576 747 827 2;
  • 44) 0.999 969 433 576 747 827 2 × 2 = 1 + 0.999 938 867 153 495 654 4;
  • 45) 0.999 938 867 153 495 654 4 × 2 = 1 + 0.999 877 734 306 991 308 8;
  • 46) 0.999 877 734 306 991 308 8 × 2 = 1 + 0.999 755 468 613 982 617 6;
  • 47) 0.999 755 468 613 982 617 6 × 2 = 1 + 0.999 510 937 227 965 235 2;
  • 48) 0.999 510 937 227 965 235 2 × 2 = 1 + 0.999 021 874 455 930 470 4;
  • 49) 0.999 021 874 455 930 470 4 × 2 = 1 + 0.998 043 748 911 860 940 8;
  • 50) 0.998 043 748 911 860 940 8 × 2 = 1 + 0.996 087 497 823 721 881 6;
  • 51) 0.996 087 497 823 721 881 6 × 2 = 1 + 0.992 174 995 647 443 763 2;
  • 52) 0.992 174 995 647 443 763 2 × 2 = 1 + 0.984 349 991 294 887 526 4;
  • 53) 0.984 349 991 294 887 526 4 × 2 = 1 + 0.968 699 982 589 775 052 8;
  • 54) 0.968 699 982 589 775 052 8 × 2 = 1 + 0.937 399 965 179 550 105 6;
  • 55) 0.937 399 965 179 550 105 6 × 2 = 1 + 0.874 799 930 359 100 211 2;
  • 56) 0.874 799 930 359 100 211 2 × 2 = 1 + 0.749 599 860 718 200 422 4;
  • 57) 0.749 599 860 718 200 422 4 × 2 = 1 + 0.499 199 721 436 400 844 8;
  • 58) 0.499 199 721 436 400 844 8 × 2 = 0 + 0.998 399 442 872 801 689 6;

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:


0.016 738 891 601 562 496 525(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 525(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:


0.016 738 891 601 562 496 525(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 -0.016 738 891 601 562 496 525 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