-0.016 738 891 601 562 496 93 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 93(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
-0.016 738 891 601 562 496 93(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:

|-0.016 738 891 601 562 496 93| = 0.016 738 891 601 562 496 93


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: 0.016 738 891 601 562 496 93.

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) 0.016 738 891 601 562 496 93 × 2 = 0 + 0.033 477 783 203 124 993 86;
  • 2) 0.033 477 783 203 124 993 86 × 2 = 0 + 0.066 955 566 406 249 987 72;
  • 3) 0.066 955 566 406 249 987 72 × 2 = 0 + 0.133 911 132 812 499 975 44;
  • 4) 0.133 911 132 812 499 975 44 × 2 = 0 + 0.267 822 265 624 999 950 88;
  • 5) 0.267 822 265 624 999 950 88 × 2 = 0 + 0.535 644 531 249 999 901 76;
  • 6) 0.535 644 531 249 999 901 76 × 2 = 1 + 0.071 289 062 499 999 803 52;
  • 7) 0.071 289 062 499 999 803 52 × 2 = 0 + 0.142 578 124 999 999 607 04;
  • 8) 0.142 578 124 999 999 607 04 × 2 = 0 + 0.285 156 249 999 999 214 08;
  • 9) 0.285 156 249 999 999 214 08 × 2 = 0 + 0.570 312 499 999 998 428 16;
  • 10) 0.570 312 499 999 998 428 16 × 2 = 1 + 0.140 624 999 999 996 856 32;
  • 11) 0.140 624 999 999 996 856 32 × 2 = 0 + 0.281 249 999 999 993 712 64;
  • 12) 0.281 249 999 999 993 712 64 × 2 = 0 + 0.562 499 999 999 987 425 28;
  • 13) 0.562 499 999 999 987 425 28 × 2 = 1 + 0.124 999 999 999 974 850 56;
  • 14) 0.124 999 999 999 974 850 56 × 2 = 0 + 0.249 999 999 999 949 701 12;
  • 15) 0.249 999 999 999 949 701 12 × 2 = 0 + 0.499 999 999 999 899 402 24;
  • 16) 0.499 999 999 999 899 402 24 × 2 = 0 + 0.999 999 999 999 798 804 48;
  • 17) 0.999 999 999 999 798 804 48 × 2 = 1 + 0.999 999 999 999 597 608 96;
  • 18) 0.999 999 999 999 597 608 96 × 2 = 1 + 0.999 999 999 999 195 217 92;
  • 19) 0.999 999 999 999 195 217 92 × 2 = 1 + 0.999 999 999 998 390 435 84;
  • 20) 0.999 999 999 998 390 435 84 × 2 = 1 + 0.999 999 999 996 780 871 68;
  • 21) 0.999 999 999 996 780 871 68 × 2 = 1 + 0.999 999 999 993 561 743 36;
  • 22) 0.999 999 999 993 561 743 36 × 2 = 1 + 0.999 999 999 987 123 486 72;
  • 23) 0.999 999 999 987 123 486 72 × 2 = 1 + 0.999 999 999 974 246 973 44;
  • 24) 0.999 999 999 974 246 973 44 × 2 = 1 + 0.999 999 999 948 493 946 88;
  • 25) 0.999 999 999 948 493 946 88 × 2 = 1 + 0.999 999 999 896 987 893 76;
  • 26) 0.999 999 999 896 987 893 76 × 2 = 1 + 0.999 999 999 793 975 787 52;
  • 27) 0.999 999 999 793 975 787 52 × 2 = 1 + 0.999 999 999 587 951 575 04;
  • 28) 0.999 999 999 587 951 575 04 × 2 = 1 + 0.999 999 999 175 903 150 08;
  • 29) 0.999 999 999 175 903 150 08 × 2 = 1 + 0.999 999 998 351 806 300 16;
  • 30) 0.999 999 998 351 806 300 16 × 2 = 1 + 0.999 999 996 703 612 600 32;
  • 31) 0.999 999 996 703 612 600 32 × 2 = 1 + 0.999 999 993 407 225 200 64;
  • 32) 0.999 999 993 407 225 200 64 × 2 = 1 + 0.999 999 986 814 450 401 28;
  • 33) 0.999 999 986 814 450 401 28 × 2 = 1 + 0.999 999 973 628 900 802 56;
  • 34) 0.999 999 973 628 900 802 56 × 2 = 1 + 0.999 999 947 257 801 605 12;
  • 35) 0.999 999 947 257 801 605 12 × 2 = 1 + 0.999 999 894 515 603 210 24;
  • 36) 0.999 999 894 515 603 210 24 × 2 = 1 + 0.999 999 789 031 206 420 48;
  • 37) 0.999 999 789 031 206 420 48 × 2 = 1 + 0.999 999 578 062 412 840 96;
  • 38) 0.999 999 578 062 412 840 96 × 2 = 1 + 0.999 999 156 124 825 681 92;
  • 39) 0.999 999 156 124 825 681 92 × 2 = 1 + 0.999 998 312 249 651 363 84;
  • 40) 0.999 998 312 249 651 363 84 × 2 = 1 + 0.999 996 624 499 302 727 68;
  • 41) 0.999 996 624 499 302 727 68 × 2 = 1 + 0.999 993 248 998 605 455 36;
  • 42) 0.999 993 248 998 605 455 36 × 2 = 1 + 0.999 986 497 997 210 910 72;
  • 43) 0.999 986 497 997 210 910 72 × 2 = 1 + 0.999 972 995 994 421 821 44;
  • 44) 0.999 972 995 994 421 821 44 × 2 = 1 + 0.999 945 991 988 843 642 88;
  • 45) 0.999 945 991 988 843 642 88 × 2 = 1 + 0.999 891 983 977 687 285 76;
  • 46) 0.999 891 983 977 687 285 76 × 2 = 1 + 0.999 783 967 955 374 571 52;
  • 47) 0.999 783 967 955 374 571 52 × 2 = 1 + 0.999 567 935 910 749 143 04;
  • 48) 0.999 567 935 910 749 143 04 × 2 = 1 + 0.999 135 871 821 498 286 08;
  • 49) 0.999 135 871 821 498 286 08 × 2 = 1 + 0.998 271 743 642 996 572 16;
  • 50) 0.998 271 743 642 996 572 16 × 2 = 1 + 0.996 543 487 285 993 144 32;
  • 51) 0.996 543 487 285 993 144 32 × 2 = 1 + 0.993 086 974 571 986 288 64;
  • 52) 0.993 086 974 571 986 288 64 × 2 = 1 + 0.986 173 949 143 972 577 28;
  • 53) 0.986 173 949 143 972 577 28 × 2 = 1 + 0.972 347 898 287 945 154 56;
  • 54) 0.972 347 898 287 945 154 56 × 2 = 1 + 0.944 695 796 575 890 309 12;
  • 55) 0.944 695 796 575 890 309 12 × 2 = 1 + 0.889 391 593 151 780 618 24;
  • 56) 0.889 391 593 151 780 618 24 × 2 = 1 + 0.778 783 186 303 561 236 48;
  • 57) 0.778 783 186 303 561 236 48 × 2 = 1 + 0.557 566 372 607 122 472 96;
  • 58) 0.557 566 372 607 122 472 96 × 2 = 1 + 0.115 132 745 214 244 945 92;

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:


0.016 738 891 601 562 496 93(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 93(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:


0.016 738 891 601 562 496 93(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 -0.016 738 891 601 562 496 93 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