-0.000 282 006 396 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 396(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
-0.000 282 006 396(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:

|-0.000 282 006 396| = 0.000 282 006 396


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: 0.000 282 006 396.

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) 0.000 282 006 396 × 2 = 0 + 0.000 564 012 792;
  • 2) 0.000 564 012 792 × 2 = 0 + 0.001 128 025 584;
  • 3) 0.001 128 025 584 × 2 = 0 + 0.002 256 051 168;
  • 4) 0.002 256 051 168 × 2 = 0 + 0.004 512 102 336;
  • 5) 0.004 512 102 336 × 2 = 0 + 0.009 024 204 672;
  • 6) 0.009 024 204 672 × 2 = 0 + 0.018 048 409 344;
  • 7) 0.018 048 409 344 × 2 = 0 + 0.036 096 818 688;
  • 8) 0.036 096 818 688 × 2 = 0 + 0.072 193 637 376;
  • 9) 0.072 193 637 376 × 2 = 0 + 0.144 387 274 752;
  • 10) 0.144 387 274 752 × 2 = 0 + 0.288 774 549 504;
  • 11) 0.288 774 549 504 × 2 = 0 + 0.577 549 099 008;
  • 12) 0.577 549 099 008 × 2 = 1 + 0.155 098 198 016;
  • 13) 0.155 098 198 016 × 2 = 0 + 0.310 196 396 032;
  • 14) 0.310 196 396 032 × 2 = 0 + 0.620 392 792 064;
  • 15) 0.620 392 792 064 × 2 = 1 + 0.240 785 584 128;
  • 16) 0.240 785 584 128 × 2 = 0 + 0.481 571 168 256;
  • 17) 0.481 571 168 256 × 2 = 0 + 0.963 142 336 512;
  • 18) 0.963 142 336 512 × 2 = 1 + 0.926 284 673 024;
  • 19) 0.926 284 673 024 × 2 = 1 + 0.852 569 346 048;
  • 20) 0.852 569 346 048 × 2 = 1 + 0.705 138 692 096;
  • 21) 0.705 138 692 096 × 2 = 1 + 0.410 277 384 192;
  • 22) 0.410 277 384 192 × 2 = 0 + 0.820 554 768 384;
  • 23) 0.820 554 768 384 × 2 = 1 + 0.641 109 536 768;
  • 24) 0.641 109 536 768 × 2 = 1 + 0.282 219 073 536;
  • 25) 0.282 219 073 536 × 2 = 0 + 0.564 438 147 072;
  • 26) 0.564 438 147 072 × 2 = 1 + 0.128 876 294 144;
  • 27) 0.128 876 294 144 × 2 = 0 + 0.257 752 588 288;
  • 28) 0.257 752 588 288 × 2 = 0 + 0.515 505 176 576;
  • 29) 0.515 505 176 576 × 2 = 1 + 0.031 010 353 152;
  • 30) 0.031 010 353 152 × 2 = 0 + 0.062 020 706 304;
  • 31) 0.062 020 706 304 × 2 = 0 + 0.124 041 412 608;
  • 32) 0.124 041 412 608 × 2 = 0 + 0.248 082 825 216;
  • 33) 0.248 082 825 216 × 2 = 0 + 0.496 165 650 432;
  • 34) 0.496 165 650 432 × 2 = 0 + 0.992 331 300 864;
  • 35) 0.992 331 300 864 × 2 = 1 + 0.984 662 601 728;
  • 36) 0.984 662 601 728 × 2 = 1 + 0.969 325 203 456;
  • 37) 0.969 325 203 456 × 2 = 1 + 0.938 650 406 912;
  • 38) 0.938 650 406 912 × 2 = 1 + 0.877 300 813 824;
  • 39) 0.877 300 813 824 × 2 = 1 + 0.754 601 627 648;
  • 40) 0.754 601 627 648 × 2 = 1 + 0.509 203 255 296;
  • 41) 0.509 203 255 296 × 2 = 1 + 0.018 406 510 592;
  • 42) 0.018 406 510 592 × 2 = 0 + 0.036 813 021 184;
  • 43) 0.036 813 021 184 × 2 = 0 + 0.073 626 042 368;
  • 44) 0.073 626 042 368 × 2 = 0 + 0.147 252 084 736;
  • 45) 0.147 252 084 736 × 2 = 0 + 0.294 504 169 472;
  • 46) 0.294 504 169 472 × 2 = 0 + 0.589 008 338 944;
  • 47) 0.589 008 338 944 × 2 = 1 + 0.178 016 677 888;
  • 48) 0.178 016 677 888 × 2 = 0 + 0.356 033 355 776;
  • 49) 0.356 033 355 776 × 2 = 0 + 0.712 066 711 552;
  • 50) 0.712 066 711 552 × 2 = 1 + 0.424 133 423 104;
  • 51) 0.424 133 423 104 × 2 = 0 + 0.848 266 846 208;
  • 52) 0.848 266 846 208 × 2 = 1 + 0.696 533 692 416;
  • 53) 0.696 533 692 416 × 2 = 1 + 0.393 067 384 832;
  • 54) 0.393 067 384 832 × 2 = 0 + 0.786 134 769 664;
  • 55) 0.786 134 769 664 × 2 = 1 + 0.572 269 539 328;
  • 56) 0.572 269 539 328 × 2 = 1 + 0.144 539 078 656;
  • 57) 0.144 539 078 656 × 2 = 0 + 0.289 078 157 312;
  • 58) 0.289 078 157 312 × 2 = 0 + 0.578 156 314 624;
  • 59) 0.578 156 314 624 × 2 = 1 + 0.156 312 629 248;
  • 60) 0.156 312 629 248 × 2 = 0 + 0.312 625 258 496;
  • 61) 0.312 625 258 496 × 2 = 0 + 0.625 250 516 992;
  • 62) 0.625 250 516 992 × 2 = 1 + 0.250 501 033 984;
  • 63) 0.250 501 033 984 × 2 = 0 + 0.501 002 067 968;
  • 64) 0.501 002 067 968 × 2 = 1 + 0.002 004 135 936;

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:


0.000 282 006 396(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0011 1111 1000 0010 0101 1011 0010 0101(2)

6. Positive number before normalization:

0.000 282 006 396(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0011 1111 1000 0010 0101 1011 0010 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:


0.000 282 006 396(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0011 1111 1000 0010 0101 1011 0010 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0011 1111 1000 0010 0101 1011 0010 0101(2) × 20 =

1.0010 0111 1011 0100 1000 0011 1111 1000 0010 0101 1011 0010 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 1011 0100 1000 0011 1111 1000 0010 0101 1011 0010 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 1011 0100 1000 0011 1111 1000 0010 0101 1011 0010 0101 =

0010 0111 1011 0100 1000 0011 1111 1000 0010 0101 1011 0010 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 1011 0100 1000 0011 1111 1000 0010 0101 1011 0010 0101


Decimal number -0.000 282 006 396 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0011 1111 1000 0010 0101 1011 0010 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