-0.000 282 005 925 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 925(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
-0.000 282 005 925(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:

|-0.000 282 005 925| = 0.000 282 005 925


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: 0.000 282 005 925.

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) 0.000 282 005 925 × 2 = 0 + 0.000 564 011 85;
  • 2) 0.000 564 011 85 × 2 = 0 + 0.001 128 023 7;
  • 3) 0.001 128 023 7 × 2 = 0 + 0.002 256 047 4;
  • 4) 0.002 256 047 4 × 2 = 0 + 0.004 512 094 8;
  • 5) 0.004 512 094 8 × 2 = 0 + 0.009 024 189 6;
  • 6) 0.009 024 189 6 × 2 = 0 + 0.018 048 379 2;
  • 7) 0.018 048 379 2 × 2 = 0 + 0.036 096 758 4;
  • 8) 0.036 096 758 4 × 2 = 0 + 0.072 193 516 8;
  • 9) 0.072 193 516 8 × 2 = 0 + 0.144 387 033 6;
  • 10) 0.144 387 033 6 × 2 = 0 + 0.288 774 067 2;
  • 11) 0.288 774 067 2 × 2 = 0 + 0.577 548 134 4;
  • 12) 0.577 548 134 4 × 2 = 1 + 0.155 096 268 8;
  • 13) 0.155 096 268 8 × 2 = 0 + 0.310 192 537 6;
  • 14) 0.310 192 537 6 × 2 = 0 + 0.620 385 075 2;
  • 15) 0.620 385 075 2 × 2 = 1 + 0.240 770 150 4;
  • 16) 0.240 770 150 4 × 2 = 0 + 0.481 540 300 8;
  • 17) 0.481 540 300 8 × 2 = 0 + 0.963 080 601 6;
  • 18) 0.963 080 601 6 × 2 = 1 + 0.926 161 203 2;
  • 19) 0.926 161 203 2 × 2 = 1 + 0.852 322 406 4;
  • 20) 0.852 322 406 4 × 2 = 1 + 0.704 644 812 8;
  • 21) 0.704 644 812 8 × 2 = 1 + 0.409 289 625 6;
  • 22) 0.409 289 625 6 × 2 = 0 + 0.818 579 251 2;
  • 23) 0.818 579 251 2 × 2 = 1 + 0.637 158 502 4;
  • 24) 0.637 158 502 4 × 2 = 1 + 0.274 317 004 8;
  • 25) 0.274 317 004 8 × 2 = 0 + 0.548 634 009 6;
  • 26) 0.548 634 009 6 × 2 = 1 + 0.097 268 019 2;
  • 27) 0.097 268 019 2 × 2 = 0 + 0.194 536 038 4;
  • 28) 0.194 536 038 4 × 2 = 0 + 0.389 072 076 8;
  • 29) 0.389 072 076 8 × 2 = 0 + 0.778 144 153 6;
  • 30) 0.778 144 153 6 × 2 = 1 + 0.556 288 307 2;
  • 31) 0.556 288 307 2 × 2 = 1 + 0.112 576 614 4;
  • 32) 0.112 576 614 4 × 2 = 0 + 0.225 153 228 8;
  • 33) 0.225 153 228 8 × 2 = 0 + 0.450 306 457 6;
  • 34) 0.450 306 457 6 × 2 = 0 + 0.900 612 915 2;
  • 35) 0.900 612 915 2 × 2 = 1 + 0.801 225 830 4;
  • 36) 0.801 225 830 4 × 2 = 1 + 0.602 451 660 8;
  • 37) 0.602 451 660 8 × 2 = 1 + 0.204 903 321 6;
  • 38) 0.204 903 321 6 × 2 = 0 + 0.409 806 643 2;
  • 39) 0.409 806 643 2 × 2 = 0 + 0.819 613 286 4;
  • 40) 0.819 613 286 4 × 2 = 1 + 0.639 226 572 8;
  • 41) 0.639 226 572 8 × 2 = 1 + 0.278 453 145 6;
  • 42) 0.278 453 145 6 × 2 = 0 + 0.556 906 291 2;
  • 43) 0.556 906 291 2 × 2 = 1 + 0.113 812 582 4;
  • 44) 0.113 812 582 4 × 2 = 0 + 0.227 625 164 8;
  • 45) 0.227 625 164 8 × 2 = 0 + 0.455 250 329 6;
  • 46) 0.455 250 329 6 × 2 = 0 + 0.910 500 659 2;
  • 47) 0.910 500 659 2 × 2 = 1 + 0.821 001 318 4;
  • 48) 0.821 001 318 4 × 2 = 1 + 0.642 002 636 8;
  • 49) 0.642 002 636 8 × 2 = 1 + 0.284 005 273 6;
  • 50) 0.284 005 273 6 × 2 = 0 + 0.568 010 547 2;
  • 51) 0.568 010 547 2 × 2 = 1 + 0.136 021 094 4;
  • 52) 0.136 021 094 4 × 2 = 0 + 0.272 042 188 8;
  • 53) 0.272 042 188 8 × 2 = 0 + 0.544 084 377 6;
  • 54) 0.544 084 377 6 × 2 = 1 + 0.088 168 755 2;
  • 55) 0.088 168 755 2 × 2 = 0 + 0.176 337 510 4;
  • 56) 0.176 337 510 4 × 2 = 0 + 0.352 675 020 8;
  • 57) 0.352 675 020 8 × 2 = 0 + 0.705 350 041 6;
  • 58) 0.705 350 041 6 × 2 = 1 + 0.410 700 083 2;
  • 59) 0.410 700 083 2 × 2 = 0 + 0.821 400 166 4;
  • 60) 0.821 400 166 4 × 2 = 1 + 0.642 800 332 8;
  • 61) 0.642 800 332 8 × 2 = 1 + 0.285 600 665 6;
  • 62) 0.285 600 665 6 × 2 = 0 + 0.571 201 331 2;
  • 63) 0.571 201 331 2 × 2 = 1 + 0.142 402 662 4;
  • 64) 0.142 402 662 4 × 2 = 0 + 0.284 805 324 8;

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:


0.000 282 005 925(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1001 1010 0011 1010 0100 0101 1010(2)

6. Positive number before normalization:

0.000 282 005 925(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1001 1010 0011 1010 0100 0101 1010(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:


0.000 282 005 925(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1001 1010 0011 1010 0100 0101 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1001 1010 0011 1010 0100 0101 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0011 1001 1010 0011 1010 0100 0101 1010(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 0100 0110 0011 1001 1010 0011 1010 0100 0101 1010


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 0100 0110 0011 1001 1010 0011 1010 0100 0101 1010 =

0010 0111 1011 0100 0110 0011 1001 1010 0011 1010 0100 0101 1010


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 0100 0110 0011 1001 1010 0011 1010 0100 0101 1010


Decimal number -0.000 282 005 925 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 1001 1010 0011 1010 0100 0101 1010

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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