-0.000 282 002 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 002 92(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
-0.000 282 002 92(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:

|-0.000 282 002 92| = 0.000 282 002 92


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: 0.000 282 002 92.

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) 0.000 282 002 92 × 2 = 0 + 0.000 564 005 84;
  • 2) 0.000 564 005 84 × 2 = 0 + 0.001 128 011 68;
  • 3) 0.001 128 011 68 × 2 = 0 + 0.002 256 023 36;
  • 4) 0.002 256 023 36 × 2 = 0 + 0.004 512 046 72;
  • 5) 0.004 512 046 72 × 2 = 0 + 0.009 024 093 44;
  • 6) 0.009 024 093 44 × 2 = 0 + 0.018 048 186 88;
  • 7) 0.018 048 186 88 × 2 = 0 + 0.036 096 373 76;
  • 8) 0.036 096 373 76 × 2 = 0 + 0.072 192 747 52;
  • 9) 0.072 192 747 52 × 2 = 0 + 0.144 385 495 04;
  • 10) 0.144 385 495 04 × 2 = 0 + 0.288 770 990 08;
  • 11) 0.288 770 990 08 × 2 = 0 + 0.577 541 980 16;
  • 12) 0.577 541 980 16 × 2 = 1 + 0.155 083 960 32;
  • 13) 0.155 083 960 32 × 2 = 0 + 0.310 167 920 64;
  • 14) 0.310 167 920 64 × 2 = 0 + 0.620 335 841 28;
  • 15) 0.620 335 841 28 × 2 = 1 + 0.240 671 682 56;
  • 16) 0.240 671 682 56 × 2 = 0 + 0.481 343 365 12;
  • 17) 0.481 343 365 12 × 2 = 0 + 0.962 686 730 24;
  • 18) 0.962 686 730 24 × 2 = 1 + 0.925 373 460 48;
  • 19) 0.925 373 460 48 × 2 = 1 + 0.850 746 920 96;
  • 20) 0.850 746 920 96 × 2 = 1 + 0.701 493 841 92;
  • 21) 0.701 493 841 92 × 2 = 1 + 0.402 987 683 84;
  • 22) 0.402 987 683 84 × 2 = 0 + 0.805 975 367 68;
  • 23) 0.805 975 367 68 × 2 = 1 + 0.611 950 735 36;
  • 24) 0.611 950 735 36 × 2 = 1 + 0.223 901 470 72;
  • 25) 0.223 901 470 72 × 2 = 0 + 0.447 802 941 44;
  • 26) 0.447 802 941 44 × 2 = 0 + 0.895 605 882 88;
  • 27) 0.895 605 882 88 × 2 = 1 + 0.791 211 765 76;
  • 28) 0.791 211 765 76 × 2 = 1 + 0.582 423 531 52;
  • 29) 0.582 423 531 52 × 2 = 1 + 0.164 847 063 04;
  • 30) 0.164 847 063 04 × 2 = 0 + 0.329 694 126 08;
  • 31) 0.329 694 126 08 × 2 = 0 + 0.659 388 252 16;
  • 32) 0.659 388 252 16 × 2 = 1 + 0.318 776 504 32;
  • 33) 0.318 776 504 32 × 2 = 0 + 0.637 553 008 64;
  • 34) 0.637 553 008 64 × 2 = 1 + 0.275 106 017 28;
  • 35) 0.275 106 017 28 × 2 = 0 + 0.550 212 034 56;
  • 36) 0.550 212 034 56 × 2 = 1 + 0.100 424 069 12;
  • 37) 0.100 424 069 12 × 2 = 0 + 0.200 848 138 24;
  • 38) 0.200 848 138 24 × 2 = 0 + 0.401 696 276 48;
  • 39) 0.401 696 276 48 × 2 = 0 + 0.803 392 552 96;
  • 40) 0.803 392 552 96 × 2 = 1 + 0.606 785 105 92;
  • 41) 0.606 785 105 92 × 2 = 1 + 0.213 570 211 84;
  • 42) 0.213 570 211 84 × 2 = 0 + 0.427 140 423 68;
  • 43) 0.427 140 423 68 × 2 = 0 + 0.854 280 847 36;
  • 44) 0.854 280 847 36 × 2 = 1 + 0.708 561 694 72;
  • 45) 0.708 561 694 72 × 2 = 1 + 0.417 123 389 44;
  • 46) 0.417 123 389 44 × 2 = 0 + 0.834 246 778 88;
  • 47) 0.834 246 778 88 × 2 = 1 + 0.668 493 557 76;
  • 48) 0.668 493 557 76 × 2 = 1 + 0.336 987 115 52;
  • 49) 0.336 987 115 52 × 2 = 0 + 0.673 974 231 04;
  • 50) 0.673 974 231 04 × 2 = 1 + 0.347 948 462 08;
  • 51) 0.347 948 462 08 × 2 = 0 + 0.695 896 924 16;
  • 52) 0.695 896 924 16 × 2 = 1 + 0.391 793 848 32;
  • 53) 0.391 793 848 32 × 2 = 0 + 0.783 587 696 64;
  • 54) 0.783 587 696 64 × 2 = 1 + 0.567 175 393 28;
  • 55) 0.567 175 393 28 × 2 = 1 + 0.134 350 786 56;
  • 56) 0.134 350 786 56 × 2 = 0 + 0.268 701 573 12;
  • 57) 0.268 701 573 12 × 2 = 0 + 0.537 403 146 24;
  • 58) 0.537 403 146 24 × 2 = 1 + 0.074 806 292 48;
  • 59) 0.074 806 292 48 × 2 = 0 + 0.149 612 584 96;
  • 60) 0.149 612 584 96 × 2 = 0 + 0.299 225 169 92;
  • 61) 0.299 225 169 92 × 2 = 0 + 0.598 450 339 84;
  • 62) 0.598 450 339 84 × 2 = 1 + 0.196 900 679 68;
  • 63) 0.196 900 679 68 × 2 = 0 + 0.393 801 359 36;
  • 64) 0.393 801 359 36 × 2 = 0 + 0.787 602 718 72;

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:


0.000 282 002 92(10) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0101 0001 1001 1011 0101 0110 0100 0100(2)

6. Positive number before normalization:

0.000 282 002 92(10) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0101 0001 1001 1011 0101 0110 0100 0100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 002 92(10) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0101 0001 1001 1011 0101 0110 0100 0100(2) =

0.0000 0000 0001 0010 0111 1011 0011 1001 0101 0001 1001 1011 0101 0110 0100 0100(2) × 20 =

1.0010 0111 1011 0011 1001 0101 0001 1001 1011 0101 0110 0100 0100(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0011 1001 0101 0001 1001 1011 0101 0110 0100 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0011 1001 0101 0001 1001 1011 0101 0110 0100 0100 =

0010 0111 1011 0011 1001 0101 0001 1001 1011 0101 0110 0100 0100


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 0011

Mantissa (52 bits) =
0010 0111 1011 0011 1001 0101 0001 1001 1011 0101 0110 0100 0100


Decimal number -0.000 282 002 92 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1001 0101 0001 1001 1011 0101 0110 0100 0100

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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