-0.016 738 891 601 562 495 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 495 3(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
-0.016 738 891 601 562 495 3(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:

|-0.016 738 891 601 562 495 3| = 0.016 738 891 601 562 495 3


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: 0.016 738 891 601 562 495 3.

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) 0.016 738 891 601 562 495 3 × 2 = 0 + 0.033 477 783 203 124 990 6;
  • 2) 0.033 477 783 203 124 990 6 × 2 = 0 + 0.066 955 566 406 249 981 2;
  • 3) 0.066 955 566 406 249 981 2 × 2 = 0 + 0.133 911 132 812 499 962 4;
  • 4) 0.133 911 132 812 499 962 4 × 2 = 0 + 0.267 822 265 624 999 924 8;
  • 5) 0.267 822 265 624 999 924 8 × 2 = 0 + 0.535 644 531 249 999 849 6;
  • 6) 0.535 644 531 249 999 849 6 × 2 = 1 + 0.071 289 062 499 999 699 2;
  • 7) 0.071 289 062 499 999 699 2 × 2 = 0 + 0.142 578 124 999 999 398 4;
  • 8) 0.142 578 124 999 999 398 4 × 2 = 0 + 0.285 156 249 999 998 796 8;
  • 9) 0.285 156 249 999 998 796 8 × 2 = 0 + 0.570 312 499 999 997 593 6;
  • 10) 0.570 312 499 999 997 593 6 × 2 = 1 + 0.140 624 999 999 995 187 2;
  • 11) 0.140 624 999 999 995 187 2 × 2 = 0 + 0.281 249 999 999 990 374 4;
  • 12) 0.281 249 999 999 990 374 4 × 2 = 0 + 0.562 499 999 999 980 748 8;
  • 13) 0.562 499 999 999 980 748 8 × 2 = 1 + 0.124 999 999 999 961 497 6;
  • 14) 0.124 999 999 999 961 497 6 × 2 = 0 + 0.249 999 999 999 922 995 2;
  • 15) 0.249 999 999 999 922 995 2 × 2 = 0 + 0.499 999 999 999 845 990 4;
  • 16) 0.499 999 999 999 845 990 4 × 2 = 0 + 0.999 999 999 999 691 980 8;
  • 17) 0.999 999 999 999 691 980 8 × 2 = 1 + 0.999 999 999 999 383 961 6;
  • 18) 0.999 999 999 999 383 961 6 × 2 = 1 + 0.999 999 999 998 767 923 2;
  • 19) 0.999 999 999 998 767 923 2 × 2 = 1 + 0.999 999 999 997 535 846 4;
  • 20) 0.999 999 999 997 535 846 4 × 2 = 1 + 0.999 999 999 995 071 692 8;
  • 21) 0.999 999 999 995 071 692 8 × 2 = 1 + 0.999 999 999 990 143 385 6;
  • 22) 0.999 999 999 990 143 385 6 × 2 = 1 + 0.999 999 999 980 286 771 2;
  • 23) 0.999 999 999 980 286 771 2 × 2 = 1 + 0.999 999 999 960 573 542 4;
  • 24) 0.999 999 999 960 573 542 4 × 2 = 1 + 0.999 999 999 921 147 084 8;
  • 25) 0.999 999 999 921 147 084 8 × 2 = 1 + 0.999 999 999 842 294 169 6;
  • 26) 0.999 999 999 842 294 169 6 × 2 = 1 + 0.999 999 999 684 588 339 2;
  • 27) 0.999 999 999 684 588 339 2 × 2 = 1 + 0.999 999 999 369 176 678 4;
  • 28) 0.999 999 999 369 176 678 4 × 2 = 1 + 0.999 999 998 738 353 356 8;
  • 29) 0.999 999 998 738 353 356 8 × 2 = 1 + 0.999 999 997 476 706 713 6;
  • 30) 0.999 999 997 476 706 713 6 × 2 = 1 + 0.999 999 994 953 413 427 2;
  • 31) 0.999 999 994 953 413 427 2 × 2 = 1 + 0.999 999 989 906 826 854 4;
  • 32) 0.999 999 989 906 826 854 4 × 2 = 1 + 0.999 999 979 813 653 708 8;
  • 33) 0.999 999 979 813 653 708 8 × 2 = 1 + 0.999 999 959 627 307 417 6;
  • 34) 0.999 999 959 627 307 417 6 × 2 = 1 + 0.999 999 919 254 614 835 2;
  • 35) 0.999 999 919 254 614 835 2 × 2 = 1 + 0.999 999 838 509 229 670 4;
  • 36) 0.999 999 838 509 229 670 4 × 2 = 1 + 0.999 999 677 018 459 340 8;
  • 37) 0.999 999 677 018 459 340 8 × 2 = 1 + 0.999 999 354 036 918 681 6;
  • 38) 0.999 999 354 036 918 681 6 × 2 = 1 + 0.999 998 708 073 837 363 2;
  • 39) 0.999 998 708 073 837 363 2 × 2 = 1 + 0.999 997 416 147 674 726 4;
  • 40) 0.999 997 416 147 674 726 4 × 2 = 1 + 0.999 994 832 295 349 452 8;
  • 41) 0.999 994 832 295 349 452 8 × 2 = 1 + 0.999 989 664 590 698 905 6;
  • 42) 0.999 989 664 590 698 905 6 × 2 = 1 + 0.999 979 329 181 397 811 2;
  • 43) 0.999 979 329 181 397 811 2 × 2 = 1 + 0.999 958 658 362 795 622 4;
  • 44) 0.999 958 658 362 795 622 4 × 2 = 1 + 0.999 917 316 725 591 244 8;
  • 45) 0.999 917 316 725 591 244 8 × 2 = 1 + 0.999 834 633 451 182 489 6;
  • 46) 0.999 834 633 451 182 489 6 × 2 = 1 + 0.999 669 266 902 364 979 2;
  • 47) 0.999 669 266 902 364 979 2 × 2 = 1 + 0.999 338 533 804 729 958 4;
  • 48) 0.999 338 533 804 729 958 4 × 2 = 1 + 0.998 677 067 609 459 916 8;
  • 49) 0.998 677 067 609 459 916 8 × 2 = 1 + 0.997 354 135 218 919 833 6;
  • 50) 0.997 354 135 218 919 833 6 × 2 = 1 + 0.994 708 270 437 839 667 2;
  • 51) 0.994 708 270 437 839 667 2 × 2 = 1 + 0.989 416 540 875 679 334 4;
  • 52) 0.989 416 540 875 679 334 4 × 2 = 1 + 0.978 833 081 751 358 668 8;
  • 53) 0.978 833 081 751 358 668 8 × 2 = 1 + 0.957 666 163 502 717 337 6;
  • 54) 0.957 666 163 502 717 337 6 × 2 = 1 + 0.915 332 327 005 434 675 2;
  • 55) 0.915 332 327 005 434 675 2 × 2 = 1 + 0.830 664 654 010 869 350 4;
  • 56) 0.830 664 654 010 869 350 4 × 2 = 1 + 0.661 329 308 021 738 700 8;
  • 57) 0.661 329 308 021 738 700 8 × 2 = 1 + 0.322 658 616 043 477 401 6;
  • 58) 0.322 658 616 043 477 401 6 × 2 = 0 + 0.645 317 232 086 954 803 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:


0.016 738 891 601 562 495 3(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 495 3(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:


0.016 738 891 601 562 495 3(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 -0.016 738 891 601 562 495 3 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