-0.016 738 891 601 562 965 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 965(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
-0.016 738 891 601 562 965(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:

|-0.016 738 891 601 562 965| = 0.016 738 891 601 562 965


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: 0.016 738 891 601 562 965.

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) 0.016 738 891 601 562 965 × 2 = 0 + 0.033 477 783 203 125 93;
  • 2) 0.033 477 783 203 125 93 × 2 = 0 + 0.066 955 566 406 251 86;
  • 3) 0.066 955 566 406 251 86 × 2 = 0 + 0.133 911 132 812 503 72;
  • 4) 0.133 911 132 812 503 72 × 2 = 0 + 0.267 822 265 625 007 44;
  • 5) 0.267 822 265 625 007 44 × 2 = 0 + 0.535 644 531 250 014 88;
  • 6) 0.535 644 531 250 014 88 × 2 = 1 + 0.071 289 062 500 029 76;
  • 7) 0.071 289 062 500 029 76 × 2 = 0 + 0.142 578 125 000 059 52;
  • 8) 0.142 578 125 000 059 52 × 2 = 0 + 0.285 156 250 000 119 04;
  • 9) 0.285 156 250 000 119 04 × 2 = 0 + 0.570 312 500 000 238 08;
  • 10) 0.570 312 500 000 238 08 × 2 = 1 + 0.140 625 000 000 476 16;
  • 11) 0.140 625 000 000 476 16 × 2 = 0 + 0.281 250 000 000 952 32;
  • 12) 0.281 250 000 000 952 32 × 2 = 0 + 0.562 500 000 001 904 64;
  • 13) 0.562 500 000 001 904 64 × 2 = 1 + 0.125 000 000 003 809 28;
  • 14) 0.125 000 000 003 809 28 × 2 = 0 + 0.250 000 000 007 618 56;
  • 15) 0.250 000 000 007 618 56 × 2 = 0 + 0.500 000 000 015 237 12;
  • 16) 0.500 000 000 015 237 12 × 2 = 1 + 0.000 000 000 030 474 24;
  • 17) 0.000 000 000 030 474 24 × 2 = 0 + 0.000 000 000 060 948 48;
  • 18) 0.000 000 000 060 948 48 × 2 = 0 + 0.000 000 000 121 896 96;
  • 19) 0.000 000 000 121 896 96 × 2 = 0 + 0.000 000 000 243 793 92;
  • 20) 0.000 000 000 243 793 92 × 2 = 0 + 0.000 000 000 487 587 84;
  • 21) 0.000 000 000 487 587 84 × 2 = 0 + 0.000 000 000 975 175 68;
  • 22) 0.000 000 000 975 175 68 × 2 = 0 + 0.000 000 001 950 351 36;
  • 23) 0.000 000 001 950 351 36 × 2 = 0 + 0.000 000 003 900 702 72;
  • 24) 0.000 000 003 900 702 72 × 2 = 0 + 0.000 000 007 801 405 44;
  • 25) 0.000 000 007 801 405 44 × 2 = 0 + 0.000 000 015 602 810 88;
  • 26) 0.000 000 015 602 810 88 × 2 = 0 + 0.000 000 031 205 621 76;
  • 27) 0.000 000 031 205 621 76 × 2 = 0 + 0.000 000 062 411 243 52;
  • 28) 0.000 000 062 411 243 52 × 2 = 0 + 0.000 000 124 822 487 04;
  • 29) 0.000 000 124 822 487 04 × 2 = 0 + 0.000 000 249 644 974 08;
  • 30) 0.000 000 249 644 974 08 × 2 = 0 + 0.000 000 499 289 948 16;
  • 31) 0.000 000 499 289 948 16 × 2 = 0 + 0.000 000 998 579 896 32;
  • 32) 0.000 000 998 579 896 32 × 2 = 0 + 0.000 001 997 159 792 64;
  • 33) 0.000 001 997 159 792 64 × 2 = 0 + 0.000 003 994 319 585 28;
  • 34) 0.000 003 994 319 585 28 × 2 = 0 + 0.000 007 988 639 170 56;
  • 35) 0.000 007 988 639 170 56 × 2 = 0 + 0.000 015 977 278 341 12;
  • 36) 0.000 015 977 278 341 12 × 2 = 0 + 0.000 031 954 556 682 24;
  • 37) 0.000 031 954 556 682 24 × 2 = 0 + 0.000 063 909 113 364 48;
  • 38) 0.000 063 909 113 364 48 × 2 = 0 + 0.000 127 818 226 728 96;
  • 39) 0.000 127 818 226 728 96 × 2 = 0 + 0.000 255 636 453 457 92;
  • 40) 0.000 255 636 453 457 92 × 2 = 0 + 0.000 511 272 906 915 84;
  • 41) 0.000 511 272 906 915 84 × 2 = 0 + 0.001 022 545 813 831 68;
  • 42) 0.001 022 545 813 831 68 × 2 = 0 + 0.002 045 091 627 663 36;
  • 43) 0.002 045 091 627 663 36 × 2 = 0 + 0.004 090 183 255 326 72;
  • 44) 0.004 090 183 255 326 72 × 2 = 0 + 0.008 180 366 510 653 44;
  • 45) 0.008 180 366 510 653 44 × 2 = 0 + 0.016 360 733 021 306 88;
  • 46) 0.016 360 733 021 306 88 × 2 = 0 + 0.032 721 466 042 613 76;
  • 47) 0.032 721 466 042 613 76 × 2 = 0 + 0.065 442 932 085 227 52;
  • 48) 0.065 442 932 085 227 52 × 2 = 0 + 0.130 885 864 170 455 04;
  • 49) 0.130 885 864 170 455 04 × 2 = 0 + 0.261 771 728 340 910 08;
  • 50) 0.261 771 728 340 910 08 × 2 = 0 + 0.523 543 456 681 820 16;
  • 51) 0.523 543 456 681 820 16 × 2 = 1 + 0.047 086 913 363 640 32;
  • 52) 0.047 086 913 363 640 32 × 2 = 0 + 0.094 173 826 727 280 64;
  • 53) 0.094 173 826 727 280 64 × 2 = 0 + 0.188 347 653 454 561 28;
  • 54) 0.188 347 653 454 561 28 × 2 = 0 + 0.376 695 306 909 122 56;
  • 55) 0.376 695 306 909 122 56 × 2 = 0 + 0.753 390 613 818 245 12;
  • 56) 0.753 390 613 818 245 12 × 2 = 1 + 0.506 781 227 636 490 24;
  • 57) 0.506 781 227 636 490 24 × 2 = 1 + 0.013 562 455 272 980 48;
  • 58) 0.013 562 455 272 980 48 × 2 = 0 + 0.027 124 910 545 960 96;

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:


0.016 738 891 601 562 965(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 965(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 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:


0.016 738 891 601 562 965(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 10(2) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0010 0001 10(2) × 20 =

1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 1000 0110(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 0100 0000 0000 0000 0000 0000 0000 0000 0000 1000 0110


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 0100 0000 0000 0000 0000 0000 0000 0000 0000 1000 0110 =

0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 1000 0110


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 0100 0000 0000 0000 0000 0000 0000 0000 0000 1000 0110


Decimal number -0.016 738 891 601 562 965 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 1000 0110

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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