-0.016 738 891 601 562 499 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 499 5(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
-0.016 738 891 601 562 499 5(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:

|-0.016 738 891 601 562 499 5| = 0.016 738 891 601 562 499 5


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: 0.016 738 891 601 562 499 5.

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) 0.016 738 891 601 562 499 5 × 2 = 0 + 0.033 477 783 203 124 999;
  • 2) 0.033 477 783 203 124 999 × 2 = 0 + 0.066 955 566 406 249 998;
  • 3) 0.066 955 566 406 249 998 × 2 = 0 + 0.133 911 132 812 499 996;
  • 4) 0.133 911 132 812 499 996 × 2 = 0 + 0.267 822 265 624 999 992;
  • 5) 0.267 822 265 624 999 992 × 2 = 0 + 0.535 644 531 249 999 984;
  • 6) 0.535 644 531 249 999 984 × 2 = 1 + 0.071 289 062 499 999 968;
  • 7) 0.071 289 062 499 999 968 × 2 = 0 + 0.142 578 124 999 999 936;
  • 8) 0.142 578 124 999 999 936 × 2 = 0 + 0.285 156 249 999 999 872;
  • 9) 0.285 156 249 999 999 872 × 2 = 0 + 0.570 312 499 999 999 744;
  • 10) 0.570 312 499 999 999 744 × 2 = 1 + 0.140 624 999 999 999 488;
  • 11) 0.140 624 999 999 999 488 × 2 = 0 + 0.281 249 999 999 998 976;
  • 12) 0.281 249 999 999 998 976 × 2 = 0 + 0.562 499 999 999 997 952;
  • 13) 0.562 499 999 999 997 952 × 2 = 1 + 0.124 999 999 999 995 904;
  • 14) 0.124 999 999 999 995 904 × 2 = 0 + 0.249 999 999 999 991 808;
  • 15) 0.249 999 999 999 991 808 × 2 = 0 + 0.499 999 999 999 983 616;
  • 16) 0.499 999 999 999 983 616 × 2 = 0 + 0.999 999 999 999 967 232;
  • 17) 0.999 999 999 999 967 232 × 2 = 1 + 0.999 999 999 999 934 464;
  • 18) 0.999 999 999 999 934 464 × 2 = 1 + 0.999 999 999 999 868 928;
  • 19) 0.999 999 999 999 868 928 × 2 = 1 + 0.999 999 999 999 737 856;
  • 20) 0.999 999 999 999 737 856 × 2 = 1 + 0.999 999 999 999 475 712;
  • 21) 0.999 999 999 999 475 712 × 2 = 1 + 0.999 999 999 998 951 424;
  • 22) 0.999 999 999 998 951 424 × 2 = 1 + 0.999 999 999 997 902 848;
  • 23) 0.999 999 999 997 902 848 × 2 = 1 + 0.999 999 999 995 805 696;
  • 24) 0.999 999 999 995 805 696 × 2 = 1 + 0.999 999 999 991 611 392;
  • 25) 0.999 999 999 991 611 392 × 2 = 1 + 0.999 999 999 983 222 784;
  • 26) 0.999 999 999 983 222 784 × 2 = 1 + 0.999 999 999 966 445 568;
  • 27) 0.999 999 999 966 445 568 × 2 = 1 + 0.999 999 999 932 891 136;
  • 28) 0.999 999 999 932 891 136 × 2 = 1 + 0.999 999 999 865 782 272;
  • 29) 0.999 999 999 865 782 272 × 2 = 1 + 0.999 999 999 731 564 544;
  • 30) 0.999 999 999 731 564 544 × 2 = 1 + 0.999 999 999 463 129 088;
  • 31) 0.999 999 999 463 129 088 × 2 = 1 + 0.999 999 998 926 258 176;
  • 32) 0.999 999 998 926 258 176 × 2 = 1 + 0.999 999 997 852 516 352;
  • 33) 0.999 999 997 852 516 352 × 2 = 1 + 0.999 999 995 705 032 704;
  • 34) 0.999 999 995 705 032 704 × 2 = 1 + 0.999 999 991 410 065 408;
  • 35) 0.999 999 991 410 065 408 × 2 = 1 + 0.999 999 982 820 130 816;
  • 36) 0.999 999 982 820 130 816 × 2 = 1 + 0.999 999 965 640 261 632;
  • 37) 0.999 999 965 640 261 632 × 2 = 1 + 0.999 999 931 280 523 264;
  • 38) 0.999 999 931 280 523 264 × 2 = 1 + 0.999 999 862 561 046 528;
  • 39) 0.999 999 862 561 046 528 × 2 = 1 + 0.999 999 725 122 093 056;
  • 40) 0.999 999 725 122 093 056 × 2 = 1 + 0.999 999 450 244 186 112;
  • 41) 0.999 999 450 244 186 112 × 2 = 1 + 0.999 998 900 488 372 224;
  • 42) 0.999 998 900 488 372 224 × 2 = 1 + 0.999 997 800 976 744 448;
  • 43) 0.999 997 800 976 744 448 × 2 = 1 + 0.999 995 601 953 488 896;
  • 44) 0.999 995 601 953 488 896 × 2 = 1 + 0.999 991 203 906 977 792;
  • 45) 0.999 991 203 906 977 792 × 2 = 1 + 0.999 982 407 813 955 584;
  • 46) 0.999 982 407 813 955 584 × 2 = 1 + 0.999 964 815 627 911 168;
  • 47) 0.999 964 815 627 911 168 × 2 = 1 + 0.999 929 631 255 822 336;
  • 48) 0.999 929 631 255 822 336 × 2 = 1 + 0.999 859 262 511 644 672;
  • 49) 0.999 859 262 511 644 672 × 2 = 1 + 0.999 718 525 023 289 344;
  • 50) 0.999 718 525 023 289 344 × 2 = 1 + 0.999 437 050 046 578 688;
  • 51) 0.999 437 050 046 578 688 × 2 = 1 + 0.998 874 100 093 157 376;
  • 52) 0.998 874 100 093 157 376 × 2 = 1 + 0.997 748 200 186 314 752;
  • 53) 0.997 748 200 186 314 752 × 2 = 1 + 0.995 496 400 372 629 504;
  • 54) 0.995 496 400 372 629 504 × 2 = 1 + 0.990 992 800 745 259 008;
  • 55) 0.990 992 800 745 259 008 × 2 = 1 + 0.981 985 601 490 518 016;
  • 56) 0.981 985 601 490 518 016 × 2 = 1 + 0.963 971 202 981 036 032;
  • 57) 0.963 971 202 981 036 032 × 2 = 1 + 0.927 942 405 962 072 064;
  • 58) 0.927 942 405 962 072 064 × 2 = 1 + 0.855 884 811 924 144 128;

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:


0.016 738 891 601 562 499 5(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 499 5(10) =

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


0.016 738 891 601 562 499 5(10) =

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

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

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


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 1111 =

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


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 1111


Decimal number -0.016 738 891 601 562 499 5 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 1111

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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