-1.833 880 942 96 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -1.833 880 942 96(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
-1.833 880 942 96(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:

|-1.833 880 942 96| = 1.833 880 942 96


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: 0.833 880 942 96.

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) 0.833 880 942 96 × 2 = 1 + 0.667 761 885 92;
  • 2) 0.667 761 885 92 × 2 = 1 + 0.335 523 771 84;
  • 3) 0.335 523 771 84 × 2 = 0 + 0.671 047 543 68;
  • 4) 0.671 047 543 68 × 2 = 1 + 0.342 095 087 36;
  • 5) 0.342 095 087 36 × 2 = 0 + 0.684 190 174 72;
  • 6) 0.684 190 174 72 × 2 = 1 + 0.368 380 349 44;
  • 7) 0.368 380 349 44 × 2 = 0 + 0.736 760 698 88;
  • 8) 0.736 760 698 88 × 2 = 1 + 0.473 521 397 76;
  • 9) 0.473 521 397 76 × 2 = 0 + 0.947 042 795 52;
  • 10) 0.947 042 795 52 × 2 = 1 + 0.894 085 591 04;
  • 11) 0.894 085 591 04 × 2 = 1 + 0.788 171 182 08;
  • 12) 0.788 171 182 08 × 2 = 1 + 0.576 342 364 16;
  • 13) 0.576 342 364 16 × 2 = 1 + 0.152 684 728 32;
  • 14) 0.152 684 728 32 × 2 = 0 + 0.305 369 456 64;
  • 15) 0.305 369 456 64 × 2 = 0 + 0.610 738 913 28;
  • 16) 0.610 738 913 28 × 2 = 1 + 0.221 477 826 56;
  • 17) 0.221 477 826 56 × 2 = 0 + 0.442 955 653 12;
  • 18) 0.442 955 653 12 × 2 = 0 + 0.885 911 306 24;
  • 19) 0.885 911 306 24 × 2 = 1 + 0.771 822 612 48;
  • 20) 0.771 822 612 48 × 2 = 1 + 0.543 645 224 96;
  • 21) 0.543 645 224 96 × 2 = 1 + 0.087 290 449 92;
  • 22) 0.087 290 449 92 × 2 = 0 + 0.174 580 899 84;
  • 23) 0.174 580 899 84 × 2 = 0 + 0.349 161 799 68;
  • 24) 0.349 161 799 68 × 2 = 0 + 0.698 323 599 36;
  • 25) 0.698 323 599 36 × 2 = 1 + 0.396 647 198 72;
  • 26) 0.396 647 198 72 × 2 = 0 + 0.793 294 397 44;
  • 27) 0.793 294 397 44 × 2 = 1 + 0.586 588 794 88;
  • 28) 0.586 588 794 88 × 2 = 1 + 0.173 177 589 76;
  • 29) 0.173 177 589 76 × 2 = 0 + 0.346 355 179 52;
  • 30) 0.346 355 179 52 × 2 = 0 + 0.692 710 359 04;
  • 31) 0.692 710 359 04 × 2 = 1 + 0.385 420 718 08;
  • 32) 0.385 420 718 08 × 2 = 0 + 0.770 841 436 16;
  • 33) 0.770 841 436 16 × 2 = 1 + 0.541 682 872 32;
  • 34) 0.541 682 872 32 × 2 = 1 + 0.083 365 744 64;
  • 35) 0.083 365 744 64 × 2 = 0 + 0.166 731 489 28;
  • 36) 0.166 731 489 28 × 2 = 0 + 0.333 462 978 56;
  • 37) 0.333 462 978 56 × 2 = 0 + 0.666 925 957 12;
  • 38) 0.666 925 957 12 × 2 = 1 + 0.333 851 914 24;
  • 39) 0.333 851 914 24 × 2 = 0 + 0.667 703 828 48;
  • 40) 0.667 703 828 48 × 2 = 1 + 0.335 407 656 96;
  • 41) 0.335 407 656 96 × 2 = 0 + 0.670 815 313 92;
  • 42) 0.670 815 313 92 × 2 = 1 + 0.341 630 627 84;
  • 43) 0.341 630 627 84 × 2 = 0 + 0.683 261 255 68;
  • 44) 0.683 261 255 68 × 2 = 1 + 0.366 522 511 36;
  • 45) 0.366 522 511 36 × 2 = 0 + 0.733 045 022 72;
  • 46) 0.733 045 022 72 × 2 = 1 + 0.466 090 045 44;
  • 47) 0.466 090 045 44 × 2 = 0 + 0.932 180 090 88;
  • 48) 0.932 180 090 88 × 2 = 1 + 0.864 360 181 76;
  • 49) 0.864 360 181 76 × 2 = 1 + 0.728 720 363 52;
  • 50) 0.728 720 363 52 × 2 = 1 + 0.457 440 727 04;
  • 51) 0.457 440 727 04 × 2 = 0 + 0.914 881 454 08;
  • 52) 0.914 881 454 08 × 2 = 1 + 0.829 762 908 16;
  • 53) 0.829 762 908 16 × 2 = 1 + 0.659 525 816 32;

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:


0.833 880 942 96(10) =

0.1101 0101 0111 1001 0011 1000 1011 0010 1100 0101 0101 0101 1101 1(2)

6. Positive number before normalization:

1.833 880 942 96(10) =

1.1101 0101 0111 1001 0011 1000 1011 0010 1100 0101 0101 0101 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:


1.833 880 942 96(10) =

1.1101 0101 0111 1001 0011 1000 1011 0010 1100 0101 0101 0101 1101 1(2) =

1.1101 0101 0111 1001 0011 1000 1011 0010 1100 0101 0101 0101 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 0010 1100 0101 0101 0101 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 0010 1100 0101 0101 0101 1101 1 =

1101 0101 0111 1001 0011 1000 1011 0010 1100 0101 0101 0101 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 0010 1100 0101 0101 0101 1101


Decimal number -1.833 880 942 96 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1111 - 1101 0101 0111 1001 0011 1000 1011 0010 1100 0101 0101 0101 1101

» Convert decimal -1.833 880 943 65 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