-0.000 282 098 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 098(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
-0.000 282 098(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:

|-0.000 282 098| = 0.000 282 098


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: 0.000 282 098.

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) 0.000 282 098 × 2 = 0 + 0.000 564 196;
  • 2) 0.000 564 196 × 2 = 0 + 0.001 128 392;
  • 3) 0.001 128 392 × 2 = 0 + 0.002 256 784;
  • 4) 0.002 256 784 × 2 = 0 + 0.004 513 568;
  • 5) 0.004 513 568 × 2 = 0 + 0.009 027 136;
  • 6) 0.009 027 136 × 2 = 0 + 0.018 054 272;
  • 7) 0.018 054 272 × 2 = 0 + 0.036 108 544;
  • 8) 0.036 108 544 × 2 = 0 + 0.072 217 088;
  • 9) 0.072 217 088 × 2 = 0 + 0.144 434 176;
  • 10) 0.144 434 176 × 2 = 0 + 0.288 868 352;
  • 11) 0.288 868 352 × 2 = 0 + 0.577 736 704;
  • 12) 0.577 736 704 × 2 = 1 + 0.155 473 408;
  • 13) 0.155 473 408 × 2 = 0 + 0.310 946 816;
  • 14) 0.310 946 816 × 2 = 0 + 0.621 893 632;
  • 15) 0.621 893 632 × 2 = 1 + 0.243 787 264;
  • 16) 0.243 787 264 × 2 = 0 + 0.487 574 528;
  • 17) 0.487 574 528 × 2 = 0 + 0.975 149 056;
  • 18) 0.975 149 056 × 2 = 1 + 0.950 298 112;
  • 19) 0.950 298 112 × 2 = 1 + 0.900 596 224;
  • 20) 0.900 596 224 × 2 = 1 + 0.801 192 448;
  • 21) 0.801 192 448 × 2 = 1 + 0.602 384 896;
  • 22) 0.602 384 896 × 2 = 1 + 0.204 769 792;
  • 23) 0.204 769 792 × 2 = 0 + 0.409 539 584;
  • 24) 0.409 539 584 × 2 = 0 + 0.819 079 168;
  • 25) 0.819 079 168 × 2 = 1 + 0.638 158 336;
  • 26) 0.638 158 336 × 2 = 1 + 0.276 316 672;
  • 27) 0.276 316 672 × 2 = 0 + 0.552 633 344;
  • 28) 0.552 633 344 × 2 = 1 + 0.105 266 688;
  • 29) 0.105 266 688 × 2 = 0 + 0.210 533 376;
  • 30) 0.210 533 376 × 2 = 0 + 0.421 066 752;
  • 31) 0.421 066 752 × 2 = 0 + 0.842 133 504;
  • 32) 0.842 133 504 × 2 = 1 + 0.684 267 008;
  • 33) 0.684 267 008 × 2 = 1 + 0.368 534 016;
  • 34) 0.368 534 016 × 2 = 0 + 0.737 068 032;
  • 35) 0.737 068 032 × 2 = 1 + 0.474 136 064;
  • 36) 0.474 136 064 × 2 = 0 + 0.948 272 128;
  • 37) 0.948 272 128 × 2 = 1 + 0.896 544 256;
  • 38) 0.896 544 256 × 2 = 1 + 0.793 088 512;
  • 39) 0.793 088 512 × 2 = 1 + 0.586 177 024;
  • 40) 0.586 177 024 × 2 = 1 + 0.172 354 048;
  • 41) 0.172 354 048 × 2 = 0 + 0.344 708 096;
  • 42) 0.344 708 096 × 2 = 0 + 0.689 416 192;
  • 43) 0.689 416 192 × 2 = 1 + 0.378 832 384;
  • 44) 0.378 832 384 × 2 = 0 + 0.757 664 768;
  • 45) 0.757 664 768 × 2 = 1 + 0.515 329 536;
  • 46) 0.515 329 536 × 2 = 1 + 0.030 659 072;
  • 47) 0.030 659 072 × 2 = 0 + 0.061 318 144;
  • 48) 0.061 318 144 × 2 = 0 + 0.122 636 288;
  • 49) 0.122 636 288 × 2 = 0 + 0.245 272 576;
  • 50) 0.245 272 576 × 2 = 0 + 0.490 545 152;
  • 51) 0.490 545 152 × 2 = 0 + 0.981 090 304;
  • 52) 0.981 090 304 × 2 = 1 + 0.962 180 608;
  • 53) 0.962 180 608 × 2 = 1 + 0.924 361 216;
  • 54) 0.924 361 216 × 2 = 1 + 0.848 722 432;
  • 55) 0.848 722 432 × 2 = 1 + 0.697 444 864;
  • 56) 0.697 444 864 × 2 = 1 + 0.394 889 728;
  • 57) 0.394 889 728 × 2 = 0 + 0.789 779 456;
  • 58) 0.789 779 456 × 2 = 1 + 0.579 558 912;
  • 59) 0.579 558 912 × 2 = 1 + 0.159 117 824;
  • 60) 0.159 117 824 × 2 = 0 + 0.318 235 648;
  • 61) 0.318 235 648 × 2 = 0 + 0.636 471 296;
  • 62) 0.636 471 296 × 2 = 1 + 0.272 942 592;
  • 63) 0.272 942 592 × 2 = 0 + 0.545 885 184;
  • 64) 0.545 885 184 × 2 = 1 + 0.091 770 368;

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:


0.000 282 098(10) =

0.0000 0000 0001 0010 0111 1100 1101 0001 1010 1111 0010 1100 0001 1111 0110 0101(2)

6. Positive number before normalization:

0.000 282 098(10) =

0.0000 0000 0001 0010 0111 1100 1101 0001 1010 1111 0010 1100 0001 1111 0110 0101(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:


0.000 282 098(10) =

0.0000 0000 0001 0010 0111 1100 1101 0001 1010 1111 0010 1100 0001 1111 0110 0101(2) =

0.0000 0000 0001 0010 0111 1100 1101 0001 1010 1111 0010 1100 0001 1111 0110 0101(2) × 20 =

1.0010 0111 1100 1101 0001 1010 1111 0010 1100 0001 1111 0110 0101(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 1100 1101 0001 1010 1111 0010 1100 0001 1111 0110 0101


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 1100 1101 0001 1010 1111 0010 1100 0001 1111 0110 0101 =

0010 0111 1100 1101 0001 1010 1111 0010 1100 0001 1111 0110 0101


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 1100 1101 0001 1010 1111 0010 1100 0001 1111 0110 0101


Decimal number -0.000 282 098 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1100 1101 0001 1010 1111 0010 1100 0001 1111 0110 0101

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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