-0.992 685 710 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.992 685 710 3(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
-0.992 685 710 3(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:

|-0.992 685 710 3| = 0.992 685 710 3


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: 0.992 685 710 3.

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) 0.992 685 710 3 × 2 = 1 + 0.985 371 420 6;
  • 2) 0.985 371 420 6 × 2 = 1 + 0.970 742 841 2;
  • 3) 0.970 742 841 2 × 2 = 1 + 0.941 485 682 4;
  • 4) 0.941 485 682 4 × 2 = 1 + 0.882 971 364 8;
  • 5) 0.882 971 364 8 × 2 = 1 + 0.765 942 729 6;
  • 6) 0.765 942 729 6 × 2 = 1 + 0.531 885 459 2;
  • 7) 0.531 885 459 2 × 2 = 1 + 0.063 770 918 4;
  • 8) 0.063 770 918 4 × 2 = 0 + 0.127 541 836 8;
  • 9) 0.127 541 836 8 × 2 = 0 + 0.255 083 673 6;
  • 10) 0.255 083 673 6 × 2 = 0 + 0.510 167 347 2;
  • 11) 0.510 167 347 2 × 2 = 1 + 0.020 334 694 4;
  • 12) 0.020 334 694 4 × 2 = 0 + 0.040 669 388 8;
  • 13) 0.040 669 388 8 × 2 = 0 + 0.081 338 777 6;
  • 14) 0.081 338 777 6 × 2 = 0 + 0.162 677 555 2;
  • 15) 0.162 677 555 2 × 2 = 0 + 0.325 355 110 4;
  • 16) 0.325 355 110 4 × 2 = 0 + 0.650 710 220 8;
  • 17) 0.650 710 220 8 × 2 = 1 + 0.301 420 441 6;
  • 18) 0.301 420 441 6 × 2 = 0 + 0.602 840 883 2;
  • 19) 0.602 840 883 2 × 2 = 1 + 0.205 681 766 4;
  • 20) 0.205 681 766 4 × 2 = 0 + 0.411 363 532 8;
  • 21) 0.411 363 532 8 × 2 = 0 + 0.822 727 065 6;
  • 22) 0.822 727 065 6 × 2 = 1 + 0.645 454 131 2;
  • 23) 0.645 454 131 2 × 2 = 1 + 0.290 908 262 4;
  • 24) 0.290 908 262 4 × 2 = 0 + 0.581 816 524 8;
  • 25) 0.581 816 524 8 × 2 = 1 + 0.163 633 049 6;
  • 26) 0.163 633 049 6 × 2 = 0 + 0.327 266 099 2;
  • 27) 0.327 266 099 2 × 2 = 0 + 0.654 532 198 4;
  • 28) 0.654 532 198 4 × 2 = 1 + 0.309 064 396 8;
  • 29) 0.309 064 396 8 × 2 = 0 + 0.618 128 793 6;
  • 30) 0.618 128 793 6 × 2 = 1 + 0.236 257 587 2;
  • 31) 0.236 257 587 2 × 2 = 0 + 0.472 515 174 4;
  • 32) 0.472 515 174 4 × 2 = 0 + 0.945 030 348 8;
  • 33) 0.945 030 348 8 × 2 = 1 + 0.890 060 697 6;
  • 34) 0.890 060 697 6 × 2 = 1 + 0.780 121 395 2;
  • 35) 0.780 121 395 2 × 2 = 1 + 0.560 242 790 4;
  • 36) 0.560 242 790 4 × 2 = 1 + 0.120 485 580 8;
  • 37) 0.120 485 580 8 × 2 = 0 + 0.240 971 161 6;
  • 38) 0.240 971 161 6 × 2 = 0 + 0.481 942 323 2;
  • 39) 0.481 942 323 2 × 2 = 0 + 0.963 884 646 4;
  • 40) 0.963 884 646 4 × 2 = 1 + 0.927 769 292 8;
  • 41) 0.927 769 292 8 × 2 = 1 + 0.855 538 585 6;
  • 42) 0.855 538 585 6 × 2 = 1 + 0.711 077 171 2;
  • 43) 0.711 077 171 2 × 2 = 1 + 0.422 154 342 4;
  • 44) 0.422 154 342 4 × 2 = 0 + 0.844 308 684 8;
  • 45) 0.844 308 684 8 × 2 = 1 + 0.688 617 369 6;
  • 46) 0.688 617 369 6 × 2 = 1 + 0.377 234 739 2;
  • 47) 0.377 234 739 2 × 2 = 0 + 0.754 469 478 4;
  • 48) 0.754 469 478 4 × 2 = 1 + 0.508 938 956 8;
  • 49) 0.508 938 956 8 × 2 = 1 + 0.017 877 913 6;
  • 50) 0.017 877 913 6 × 2 = 0 + 0.035 755 827 2;
  • 51) 0.035 755 827 2 × 2 = 0 + 0.071 511 654 4;
  • 52) 0.071 511 654 4 × 2 = 0 + 0.143 023 308 8;
  • 53) 0.143 023 308 8 × 2 = 0 + 0.286 046 617 6;

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:


0.992 685 710 3(10) =

0.1111 1110 0010 0000 1010 0110 1001 0100 1111 0001 1110 1101 1000 0(2)

6. Positive number before normalization:

0.992 685 710 3(10) =

0.1111 1110 0010 0000 1010 0110 1001 0100 1111 0001 1110 1101 1000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.992 685 710 3(10) =

0.1111 1110 0010 0000 1010 0110 1001 0100 1111 0001 1110 1101 1000 0(2) =

0.1111 1110 0010 0000 1010 0110 1001 0100 1111 0001 1110 1101 1000 0(2) × 20 =

1.1111 1100 0100 0001 0100 1101 0010 1001 1110 0011 1101 1011 0000(2) × 2-1


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): -1

Mantissa (not normalized):
1.1111 1100 0100 0001 0100 1101 0010 1001 1110 0011 1101 1011 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-1 + 2(11-1) - 1 =

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1022(10) =

011 1111 1110(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. 1111 1100 0100 0001 0100 1101 0010 1001 1110 0011 1101 1011 0000 =

1111 1100 0100 0001 0100 1101 0010 1001 1110 0011 1101 1011 0000


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 1110

Mantissa (52 bits) =
1111 1100 0100 0001 0100 1101 0010 1001 1110 0011 1101 1011 0000


Decimal number -0.992 685 710 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1110 - 1111 1100 0100 0001 0100 1101 0010 1001 1110 0011 1101 1011 0000

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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