-0.016 738 891 601 562 495 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 495 7(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 7(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 7| = 0.016 738 891 601 562 495 7


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 7.

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 7 × 2 = 0 + 0.033 477 783 203 124 991 4;
  • 2) 0.033 477 783 203 124 991 4 × 2 = 0 + 0.066 955 566 406 249 982 8;
  • 3) 0.066 955 566 406 249 982 8 × 2 = 0 + 0.133 911 132 812 499 965 6;
  • 4) 0.133 911 132 812 499 965 6 × 2 = 0 + 0.267 822 265 624 999 931 2;
  • 5) 0.267 822 265 624 999 931 2 × 2 = 0 + 0.535 644 531 249 999 862 4;
  • 6) 0.535 644 531 249 999 862 4 × 2 = 1 + 0.071 289 062 499 999 724 8;
  • 7) 0.071 289 062 499 999 724 8 × 2 = 0 + 0.142 578 124 999 999 449 6;
  • 8) 0.142 578 124 999 999 449 6 × 2 = 0 + 0.285 156 249 999 998 899 2;
  • 9) 0.285 156 249 999 998 899 2 × 2 = 0 + 0.570 312 499 999 997 798 4;
  • 10) 0.570 312 499 999 997 798 4 × 2 = 1 + 0.140 624 999 999 995 596 8;
  • 11) 0.140 624 999 999 995 596 8 × 2 = 0 + 0.281 249 999 999 991 193 6;
  • 12) 0.281 249 999 999 991 193 6 × 2 = 0 + 0.562 499 999 999 982 387 2;
  • 13) 0.562 499 999 999 982 387 2 × 2 = 1 + 0.124 999 999 999 964 774 4;
  • 14) 0.124 999 999 999 964 774 4 × 2 = 0 + 0.249 999 999 999 929 548 8;
  • 15) 0.249 999 999 999 929 548 8 × 2 = 0 + 0.499 999 999 999 859 097 6;
  • 16) 0.499 999 999 999 859 097 6 × 2 = 0 + 0.999 999 999 999 718 195 2;
  • 17) 0.999 999 999 999 718 195 2 × 2 = 1 + 0.999 999 999 999 436 390 4;
  • 18) 0.999 999 999 999 436 390 4 × 2 = 1 + 0.999 999 999 998 872 780 8;
  • 19) 0.999 999 999 998 872 780 8 × 2 = 1 + 0.999 999 999 997 745 561 6;
  • 20) 0.999 999 999 997 745 561 6 × 2 = 1 + 0.999 999 999 995 491 123 2;
  • 21) 0.999 999 999 995 491 123 2 × 2 = 1 + 0.999 999 999 990 982 246 4;
  • 22) 0.999 999 999 990 982 246 4 × 2 = 1 + 0.999 999 999 981 964 492 8;
  • 23) 0.999 999 999 981 964 492 8 × 2 = 1 + 0.999 999 999 963 928 985 6;
  • 24) 0.999 999 999 963 928 985 6 × 2 = 1 + 0.999 999 999 927 857 971 2;
  • 25) 0.999 999 999 927 857 971 2 × 2 = 1 + 0.999 999 999 855 715 942 4;
  • 26) 0.999 999 999 855 715 942 4 × 2 = 1 + 0.999 999 999 711 431 884 8;
  • 27) 0.999 999 999 711 431 884 8 × 2 = 1 + 0.999 999 999 422 863 769 6;
  • 28) 0.999 999 999 422 863 769 6 × 2 = 1 + 0.999 999 998 845 727 539 2;
  • 29) 0.999 999 998 845 727 539 2 × 2 = 1 + 0.999 999 997 691 455 078 4;
  • 30) 0.999 999 997 691 455 078 4 × 2 = 1 + 0.999 999 995 382 910 156 8;
  • 31) 0.999 999 995 382 910 156 8 × 2 = 1 + 0.999 999 990 765 820 313 6;
  • 32) 0.999 999 990 765 820 313 6 × 2 = 1 + 0.999 999 981 531 640 627 2;
  • 33) 0.999 999 981 531 640 627 2 × 2 = 1 + 0.999 999 963 063 281 254 4;
  • 34) 0.999 999 963 063 281 254 4 × 2 = 1 + 0.999 999 926 126 562 508 8;
  • 35) 0.999 999 926 126 562 508 8 × 2 = 1 + 0.999 999 852 253 125 017 6;
  • 36) 0.999 999 852 253 125 017 6 × 2 = 1 + 0.999 999 704 506 250 035 2;
  • 37) 0.999 999 704 506 250 035 2 × 2 = 1 + 0.999 999 409 012 500 070 4;
  • 38) 0.999 999 409 012 500 070 4 × 2 = 1 + 0.999 998 818 025 000 140 8;
  • 39) 0.999 998 818 025 000 140 8 × 2 = 1 + 0.999 997 636 050 000 281 6;
  • 40) 0.999 997 636 050 000 281 6 × 2 = 1 + 0.999 995 272 100 000 563 2;
  • 41) 0.999 995 272 100 000 563 2 × 2 = 1 + 0.999 990 544 200 001 126 4;
  • 42) 0.999 990 544 200 001 126 4 × 2 = 1 + 0.999 981 088 400 002 252 8;
  • 43) 0.999 981 088 400 002 252 8 × 2 = 1 + 0.999 962 176 800 004 505 6;
  • 44) 0.999 962 176 800 004 505 6 × 2 = 1 + 0.999 924 353 600 009 011 2;
  • 45) 0.999 924 353 600 009 011 2 × 2 = 1 + 0.999 848 707 200 018 022 4;
  • 46) 0.999 848 707 200 018 022 4 × 2 = 1 + 0.999 697 414 400 036 044 8;
  • 47) 0.999 697 414 400 036 044 8 × 2 = 1 + 0.999 394 828 800 072 089 6;
  • 48) 0.999 394 828 800 072 089 6 × 2 = 1 + 0.998 789 657 600 144 179 2;
  • 49) 0.998 789 657 600 144 179 2 × 2 = 1 + 0.997 579 315 200 288 358 4;
  • 50) 0.997 579 315 200 288 358 4 × 2 = 1 + 0.995 158 630 400 576 716 8;
  • 51) 0.995 158 630 400 576 716 8 × 2 = 1 + 0.990 317 260 801 153 433 6;
  • 52) 0.990 317 260 801 153 433 6 × 2 = 1 + 0.980 634 521 602 306 867 2;
  • 53) 0.980 634 521 602 306 867 2 × 2 = 1 + 0.961 269 043 204 613 734 4;
  • 54) 0.961 269 043 204 613 734 4 × 2 = 1 + 0.922 538 086 409 227 468 8;
  • 55) 0.922 538 086 409 227 468 8 × 2 = 1 + 0.845 076 172 818 454 937 6;
  • 56) 0.845 076 172 818 454 937 6 × 2 = 1 + 0.690 152 345 636 909 875 2;
  • 57) 0.690 152 345 636 909 875 2 × 2 = 1 + 0.380 304 691 273 819 750 4;
  • 58) 0.380 304 691 273 819 750 4 × 2 = 0 + 0.760 609 382 547 639 500 8;

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 7(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 7(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 7(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 7 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