-0.000 282 468 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 468(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
-0.000 282 468(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:

|-0.000 282 468| = 0.000 282 468


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: 0.000 282 468.

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) 0.000 282 468 × 2 = 0 + 0.000 564 936;
  • 2) 0.000 564 936 × 2 = 0 + 0.001 129 872;
  • 3) 0.001 129 872 × 2 = 0 + 0.002 259 744;
  • 4) 0.002 259 744 × 2 = 0 + 0.004 519 488;
  • 5) 0.004 519 488 × 2 = 0 + 0.009 038 976;
  • 6) 0.009 038 976 × 2 = 0 + 0.018 077 952;
  • 7) 0.018 077 952 × 2 = 0 + 0.036 155 904;
  • 8) 0.036 155 904 × 2 = 0 + 0.072 311 808;
  • 9) 0.072 311 808 × 2 = 0 + 0.144 623 616;
  • 10) 0.144 623 616 × 2 = 0 + 0.289 247 232;
  • 11) 0.289 247 232 × 2 = 0 + 0.578 494 464;
  • 12) 0.578 494 464 × 2 = 1 + 0.156 988 928;
  • 13) 0.156 988 928 × 2 = 0 + 0.313 977 856;
  • 14) 0.313 977 856 × 2 = 0 + 0.627 955 712;
  • 15) 0.627 955 712 × 2 = 1 + 0.255 911 424;
  • 16) 0.255 911 424 × 2 = 0 + 0.511 822 848;
  • 17) 0.511 822 848 × 2 = 1 + 0.023 645 696;
  • 18) 0.023 645 696 × 2 = 0 + 0.047 291 392;
  • 19) 0.047 291 392 × 2 = 0 + 0.094 582 784;
  • 20) 0.094 582 784 × 2 = 0 + 0.189 165 568;
  • 21) 0.189 165 568 × 2 = 0 + 0.378 331 136;
  • 22) 0.378 331 136 × 2 = 0 + 0.756 662 272;
  • 23) 0.756 662 272 × 2 = 1 + 0.513 324 544;
  • 24) 0.513 324 544 × 2 = 1 + 0.026 649 088;
  • 25) 0.026 649 088 × 2 = 0 + 0.053 298 176;
  • 26) 0.053 298 176 × 2 = 0 + 0.106 596 352;
  • 27) 0.106 596 352 × 2 = 0 + 0.213 192 704;
  • 28) 0.213 192 704 × 2 = 0 + 0.426 385 408;
  • 29) 0.426 385 408 × 2 = 0 + 0.852 770 816;
  • 30) 0.852 770 816 × 2 = 1 + 0.705 541 632;
  • 31) 0.705 541 632 × 2 = 1 + 0.411 083 264;
  • 32) 0.411 083 264 × 2 = 0 + 0.822 166 528;
  • 33) 0.822 166 528 × 2 = 1 + 0.644 333 056;
  • 34) 0.644 333 056 × 2 = 1 + 0.288 666 112;
  • 35) 0.288 666 112 × 2 = 0 + 0.577 332 224;
  • 36) 0.577 332 224 × 2 = 1 + 0.154 664 448;
  • 37) 0.154 664 448 × 2 = 0 + 0.309 328 896;
  • 38) 0.309 328 896 × 2 = 0 + 0.618 657 792;
  • 39) 0.618 657 792 × 2 = 1 + 0.237 315 584;
  • 40) 0.237 315 584 × 2 = 0 + 0.474 631 168;
  • 41) 0.474 631 168 × 2 = 0 + 0.949 262 336;
  • 42) 0.949 262 336 × 2 = 1 + 0.898 524 672;
  • 43) 0.898 524 672 × 2 = 1 + 0.797 049 344;
  • 44) 0.797 049 344 × 2 = 1 + 0.594 098 688;
  • 45) 0.594 098 688 × 2 = 1 + 0.188 197 376;
  • 46) 0.188 197 376 × 2 = 0 + 0.376 394 752;
  • 47) 0.376 394 752 × 2 = 0 + 0.752 789 504;
  • 48) 0.752 789 504 × 2 = 1 + 0.505 579 008;
  • 49) 0.505 579 008 × 2 = 1 + 0.011 158 016;
  • 50) 0.011 158 016 × 2 = 0 + 0.022 316 032;
  • 51) 0.022 316 032 × 2 = 0 + 0.044 632 064;
  • 52) 0.044 632 064 × 2 = 0 + 0.089 264 128;
  • 53) 0.089 264 128 × 2 = 0 + 0.178 528 256;
  • 54) 0.178 528 256 × 2 = 0 + 0.357 056 512;
  • 55) 0.357 056 512 × 2 = 0 + 0.714 113 024;
  • 56) 0.714 113 024 × 2 = 1 + 0.428 226 048;
  • 57) 0.428 226 048 × 2 = 0 + 0.856 452 096;
  • 58) 0.856 452 096 × 2 = 1 + 0.712 904 192;
  • 59) 0.712 904 192 × 2 = 1 + 0.425 808 384;
  • 60) 0.425 808 384 × 2 = 0 + 0.851 616 768;
  • 61) 0.851 616 768 × 2 = 1 + 0.703 233 536;
  • 62) 0.703 233 536 × 2 = 1 + 0.406 467 072;
  • 63) 0.406 467 072 × 2 = 0 + 0.812 934 144;
  • 64) 0.812 934 144 × 2 = 1 + 0.625 868 288;

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:


0.000 282 468(10) =

0.0000 0000 0001 0010 1000 0011 0000 0110 1101 0010 0111 1001 1000 0001 0110 1101(2)

6. Positive number before normalization:

0.000 282 468(10) =

0.0000 0000 0001 0010 1000 0011 0000 0110 1101 0010 0111 1001 1000 0001 0110 1101(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:


0.000 282 468(10) =

0.0000 0000 0001 0010 1000 0011 0000 0110 1101 0010 0111 1001 1000 0001 0110 1101(2) =

0.0000 0000 0001 0010 1000 0011 0000 0110 1101 0010 0111 1001 1000 0001 0110 1101(2) × 20 =

1.0010 1000 0011 0000 0110 1101 0010 0111 1001 1000 0001 0110 1101(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 1000 0011 0000 0110 1101 0010 0111 1001 1000 0001 0110 1101


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 1000 0011 0000 0110 1101 0010 0111 1001 1000 0001 0110 1101 =

0010 1000 0011 0000 0110 1101 0010 0111 1001 1000 0001 0110 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 0011

Mantissa (52 bits) =
0010 1000 0011 0000 0110 1101 0010 0111 1001 1000 0001 0110 1101


Decimal number -0.000 282 468 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0011 0000 0110 1101 0010 0111 1001 1000 0001 0110 1101

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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