-0.016 738 891 601 562 481 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 481 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
-0.016 738 891 601 562 481 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:

|-0.016 738 891 601 562 481 7| = 0.016 738 891 601 562 481 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: 0.016 738 891 601 562 481 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) 0.016 738 891 601 562 481 7 × 2 = 0 + 0.033 477 783 203 124 963 4;
  • 2) 0.033 477 783 203 124 963 4 × 2 = 0 + 0.066 955 566 406 249 926 8;
  • 3) 0.066 955 566 406 249 926 8 × 2 = 0 + 0.133 911 132 812 499 853 6;
  • 4) 0.133 911 132 812 499 853 6 × 2 = 0 + 0.267 822 265 624 999 707 2;
  • 5) 0.267 822 265 624 999 707 2 × 2 = 0 + 0.535 644 531 249 999 414 4;
  • 6) 0.535 644 531 249 999 414 4 × 2 = 1 + 0.071 289 062 499 998 828 8;
  • 7) 0.071 289 062 499 998 828 8 × 2 = 0 + 0.142 578 124 999 997 657 6;
  • 8) 0.142 578 124 999 997 657 6 × 2 = 0 + 0.285 156 249 999 995 315 2;
  • 9) 0.285 156 249 999 995 315 2 × 2 = 0 + 0.570 312 499 999 990 630 4;
  • 10) 0.570 312 499 999 990 630 4 × 2 = 1 + 0.140 624 999 999 981 260 8;
  • 11) 0.140 624 999 999 981 260 8 × 2 = 0 + 0.281 249 999 999 962 521 6;
  • 12) 0.281 249 999 999 962 521 6 × 2 = 0 + 0.562 499 999 999 925 043 2;
  • 13) 0.562 499 999 999 925 043 2 × 2 = 1 + 0.124 999 999 999 850 086 4;
  • 14) 0.124 999 999 999 850 086 4 × 2 = 0 + 0.249 999 999 999 700 172 8;
  • 15) 0.249 999 999 999 700 172 8 × 2 = 0 + 0.499 999 999 999 400 345 6;
  • 16) 0.499 999 999 999 400 345 6 × 2 = 0 + 0.999 999 999 998 800 691 2;
  • 17) 0.999 999 999 998 800 691 2 × 2 = 1 + 0.999 999 999 997 601 382 4;
  • 18) 0.999 999 999 997 601 382 4 × 2 = 1 + 0.999 999 999 995 202 764 8;
  • 19) 0.999 999 999 995 202 764 8 × 2 = 1 + 0.999 999 999 990 405 529 6;
  • 20) 0.999 999 999 990 405 529 6 × 2 = 1 + 0.999 999 999 980 811 059 2;
  • 21) 0.999 999 999 980 811 059 2 × 2 = 1 + 0.999 999 999 961 622 118 4;
  • 22) 0.999 999 999 961 622 118 4 × 2 = 1 + 0.999 999 999 923 244 236 8;
  • 23) 0.999 999 999 923 244 236 8 × 2 = 1 + 0.999 999 999 846 488 473 6;
  • 24) 0.999 999 999 846 488 473 6 × 2 = 1 + 0.999 999 999 692 976 947 2;
  • 25) 0.999 999 999 692 976 947 2 × 2 = 1 + 0.999 999 999 385 953 894 4;
  • 26) 0.999 999 999 385 953 894 4 × 2 = 1 + 0.999 999 998 771 907 788 8;
  • 27) 0.999 999 998 771 907 788 8 × 2 = 1 + 0.999 999 997 543 815 577 6;
  • 28) 0.999 999 997 543 815 577 6 × 2 = 1 + 0.999 999 995 087 631 155 2;
  • 29) 0.999 999 995 087 631 155 2 × 2 = 1 + 0.999 999 990 175 262 310 4;
  • 30) 0.999 999 990 175 262 310 4 × 2 = 1 + 0.999 999 980 350 524 620 8;
  • 31) 0.999 999 980 350 524 620 8 × 2 = 1 + 0.999 999 960 701 049 241 6;
  • 32) 0.999 999 960 701 049 241 6 × 2 = 1 + 0.999 999 921 402 098 483 2;
  • 33) 0.999 999 921 402 098 483 2 × 2 = 1 + 0.999 999 842 804 196 966 4;
  • 34) 0.999 999 842 804 196 966 4 × 2 = 1 + 0.999 999 685 608 393 932 8;
  • 35) 0.999 999 685 608 393 932 8 × 2 = 1 + 0.999 999 371 216 787 865 6;
  • 36) 0.999 999 371 216 787 865 6 × 2 = 1 + 0.999 998 742 433 575 731 2;
  • 37) 0.999 998 742 433 575 731 2 × 2 = 1 + 0.999 997 484 867 151 462 4;
  • 38) 0.999 997 484 867 151 462 4 × 2 = 1 + 0.999 994 969 734 302 924 8;
  • 39) 0.999 994 969 734 302 924 8 × 2 = 1 + 0.999 989 939 468 605 849 6;
  • 40) 0.999 989 939 468 605 849 6 × 2 = 1 + 0.999 979 878 937 211 699 2;
  • 41) 0.999 979 878 937 211 699 2 × 2 = 1 + 0.999 959 757 874 423 398 4;
  • 42) 0.999 959 757 874 423 398 4 × 2 = 1 + 0.999 919 515 748 846 796 8;
  • 43) 0.999 919 515 748 846 796 8 × 2 = 1 + 0.999 839 031 497 693 593 6;
  • 44) 0.999 839 031 497 693 593 6 × 2 = 1 + 0.999 678 062 995 387 187 2;
  • 45) 0.999 678 062 995 387 187 2 × 2 = 1 + 0.999 356 125 990 774 374 4;
  • 46) 0.999 356 125 990 774 374 4 × 2 = 1 + 0.998 712 251 981 548 748 8;
  • 47) 0.998 712 251 981 548 748 8 × 2 = 1 + 0.997 424 503 963 097 497 6;
  • 48) 0.997 424 503 963 097 497 6 × 2 = 1 + 0.994 849 007 926 194 995 2;
  • 49) 0.994 849 007 926 194 995 2 × 2 = 1 + 0.989 698 015 852 389 990 4;
  • 50) 0.989 698 015 852 389 990 4 × 2 = 1 + 0.979 396 031 704 779 980 8;
  • 51) 0.979 396 031 704 779 980 8 × 2 = 1 + 0.958 792 063 409 559 961 6;
  • 52) 0.958 792 063 409 559 961 6 × 2 = 1 + 0.917 584 126 819 119 923 2;
  • 53) 0.917 584 126 819 119 923 2 × 2 = 1 + 0.835 168 253 638 239 846 4;
  • 54) 0.835 168 253 638 239 846 4 × 2 = 1 + 0.670 336 507 276 479 692 8;
  • 55) 0.670 336 507 276 479 692 8 × 2 = 1 + 0.340 673 014 552 959 385 6;
  • 56) 0.340 673 014 552 959 385 6 × 2 = 0 + 0.681 346 029 105 918 771 2;
  • 57) 0.681 346 029 105 918 771 2 × 2 = 1 + 0.362 692 058 211 837 542 4;
  • 58) 0.362 692 058 211 837 542 4 × 2 = 0 + 0.725 384 116 423 675 084 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:


0.016 738 891 601 562 481 7(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 481 7(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 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:


0.016 738 891 601 562 481 7(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010(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 1010


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 1010 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010


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 1010


Decimal number -0.016 738 891 601 562 481 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 1010

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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