-1.833 880 943 28 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -1.833 880 943 28(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
-1.833 880 943 28(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:

|-1.833 880 943 28| = 1.833 880 943 28


2. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

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.

1(10) =

1(2)


4. Convert to binary (base 2) the fractional part: 0.833 880 943 28.

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) 0.833 880 943 28 × 2 = 1 + 0.667 761 886 56;
  • 2) 0.667 761 886 56 × 2 = 1 + 0.335 523 773 12;
  • 3) 0.335 523 773 12 × 2 = 0 + 0.671 047 546 24;
  • 4) 0.671 047 546 24 × 2 = 1 + 0.342 095 092 48;
  • 5) 0.342 095 092 48 × 2 = 0 + 0.684 190 184 96;
  • 6) 0.684 190 184 96 × 2 = 1 + 0.368 380 369 92;
  • 7) 0.368 380 369 92 × 2 = 0 + 0.736 760 739 84;
  • 8) 0.736 760 739 84 × 2 = 1 + 0.473 521 479 68;
  • 9) 0.473 521 479 68 × 2 = 0 + 0.947 042 959 36;
  • 10) 0.947 042 959 36 × 2 = 1 + 0.894 085 918 72;
  • 11) 0.894 085 918 72 × 2 = 1 + 0.788 171 837 44;
  • 12) 0.788 171 837 44 × 2 = 1 + 0.576 343 674 88;
  • 13) 0.576 343 674 88 × 2 = 1 + 0.152 687 349 76;
  • 14) 0.152 687 349 76 × 2 = 0 + 0.305 374 699 52;
  • 15) 0.305 374 699 52 × 2 = 0 + 0.610 749 399 04;
  • 16) 0.610 749 399 04 × 2 = 1 + 0.221 498 798 08;
  • 17) 0.221 498 798 08 × 2 = 0 + 0.442 997 596 16;
  • 18) 0.442 997 596 16 × 2 = 0 + 0.885 995 192 32;
  • 19) 0.885 995 192 32 × 2 = 1 + 0.771 990 384 64;
  • 20) 0.771 990 384 64 × 2 = 1 + 0.543 980 769 28;
  • 21) 0.543 980 769 28 × 2 = 1 + 0.087 961 538 56;
  • 22) 0.087 961 538 56 × 2 = 0 + 0.175 923 077 12;
  • 23) 0.175 923 077 12 × 2 = 0 + 0.351 846 154 24;
  • 24) 0.351 846 154 24 × 2 = 0 + 0.703 692 308 48;
  • 25) 0.703 692 308 48 × 2 = 1 + 0.407 384 616 96;
  • 26) 0.407 384 616 96 × 2 = 0 + 0.814 769 233 92;
  • 27) 0.814 769 233 92 × 2 = 1 + 0.629 538 467 84;
  • 28) 0.629 538 467 84 × 2 = 1 + 0.259 076 935 68;
  • 29) 0.259 076 935 68 × 2 = 0 + 0.518 153 871 36;
  • 30) 0.518 153 871 36 × 2 = 1 + 0.036 307 742 72;
  • 31) 0.036 307 742 72 × 2 = 0 + 0.072 615 485 44;
  • 32) 0.072 615 485 44 × 2 = 0 + 0.145 230 970 88;
  • 33) 0.145 230 970 88 × 2 = 0 + 0.290 461 941 76;
  • 34) 0.290 461 941 76 × 2 = 0 + 0.580 923 883 52;
  • 35) 0.580 923 883 52 × 2 = 1 + 0.161 847 767 04;
  • 36) 0.161 847 767 04 × 2 = 0 + 0.323 695 534 08;
  • 37) 0.323 695 534 08 × 2 = 0 + 0.647 391 068 16;
  • 38) 0.647 391 068 16 × 2 = 1 + 0.294 782 136 32;
  • 39) 0.294 782 136 32 × 2 = 0 + 0.589 564 272 64;
  • 40) 0.589 564 272 64 × 2 = 1 + 0.179 128 545 28;
  • 41) 0.179 128 545 28 × 2 = 0 + 0.358 257 090 56;
  • 42) 0.358 257 090 56 × 2 = 0 + 0.716 514 181 12;
  • 43) 0.716 514 181 12 × 2 = 1 + 0.433 028 362 24;
  • 44) 0.433 028 362 24 × 2 = 0 + 0.866 056 724 48;
  • 45) 0.866 056 724 48 × 2 = 1 + 0.732 113 448 96;
  • 46) 0.732 113 448 96 × 2 = 1 + 0.464 226 897 92;
  • 47) 0.464 226 897 92 × 2 = 0 + 0.928 453 795 84;
  • 48) 0.928 453 795 84 × 2 = 1 + 0.856 907 591 68;
  • 49) 0.856 907 591 68 × 2 = 1 + 0.713 815 183 36;
  • 50) 0.713 815 183 36 × 2 = 1 + 0.427 630 366 72;
  • 51) 0.427 630 366 72 × 2 = 0 + 0.855 260 733 44;
  • 52) 0.855 260 733 44 × 2 = 1 + 0.710 521 466 88;
  • 53) 0.710 521 466 88 × 2 = 1 + 0.421 042 933 76;

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:


0.833 880 943 28(10) =

0.1101 0101 0111 1001 0011 1000 1011 0100 0010 0101 0010 1101 1101 1(2)

6. Positive number before normalization:

1.833 880 943 28(10) =

1.1101 0101 0111 1001 0011 1000 1011 0100 0010 0101 0010 1101 1101 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.833 880 943 28(10) =

1.1101 0101 0111 1001 0011 1000 1011 0100 0010 0101 0010 1101 1101 1(2) =

1.1101 0101 0111 1001 0011 1000 1011 0100 0010 0101 0010 1101 1101 1(2) × 20


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): 0

Mantissa (not normalized):
1.1101 0101 0111 1001 0011 1000 1011 0100 0010 0101 0010 1101 1101 1


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 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) =

1023(10) =

011 1111 1111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1101 0101 0111 1001 0011 1000 1011 0100 0010 0101 0010 1101 1101 1 =

1101 0101 0111 1001 0011 1000 1011 0100 0010 0101 0010 1101 1101


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 1111

Mantissa (52 bits) =
1101 0101 0111 1001 0011 1000 1011 0100 0010 0101 0010 1101 1101


Decimal number -1.833 880 943 28 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1111 - 1101 0101 0111 1001 0011 1000 1011 0100 0010 0101 0010 1101 1101

» Convert decimal -1.833 880 942 52 to 64 bit double precision IEEE 754 binary floating point representation standard

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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