-0.000 282 338 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 338(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
-0.000 282 338(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:

|-0.000 282 338| = 0.000 282 338


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: 0.000 282 338.

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) 0.000 282 338 × 2 = 0 + 0.000 564 676;
  • 2) 0.000 564 676 × 2 = 0 + 0.001 129 352;
  • 3) 0.001 129 352 × 2 = 0 + 0.002 258 704;
  • 4) 0.002 258 704 × 2 = 0 + 0.004 517 408;
  • 5) 0.004 517 408 × 2 = 0 + 0.009 034 816;
  • 6) 0.009 034 816 × 2 = 0 + 0.018 069 632;
  • 7) 0.018 069 632 × 2 = 0 + 0.036 139 264;
  • 8) 0.036 139 264 × 2 = 0 + 0.072 278 528;
  • 9) 0.072 278 528 × 2 = 0 + 0.144 557 056;
  • 10) 0.144 557 056 × 2 = 0 + 0.289 114 112;
  • 11) 0.289 114 112 × 2 = 0 + 0.578 228 224;
  • 12) 0.578 228 224 × 2 = 1 + 0.156 456 448;
  • 13) 0.156 456 448 × 2 = 0 + 0.312 912 896;
  • 14) 0.312 912 896 × 2 = 0 + 0.625 825 792;
  • 15) 0.625 825 792 × 2 = 1 + 0.251 651 584;
  • 16) 0.251 651 584 × 2 = 0 + 0.503 303 168;
  • 17) 0.503 303 168 × 2 = 1 + 0.006 606 336;
  • 18) 0.006 606 336 × 2 = 0 + 0.013 212 672;
  • 19) 0.013 212 672 × 2 = 0 + 0.026 425 344;
  • 20) 0.026 425 344 × 2 = 0 + 0.052 850 688;
  • 21) 0.052 850 688 × 2 = 0 + 0.105 701 376;
  • 22) 0.105 701 376 × 2 = 0 + 0.211 402 752;
  • 23) 0.211 402 752 × 2 = 0 + 0.422 805 504;
  • 24) 0.422 805 504 × 2 = 0 + 0.845 611 008;
  • 25) 0.845 611 008 × 2 = 1 + 0.691 222 016;
  • 26) 0.691 222 016 × 2 = 1 + 0.382 444 032;
  • 27) 0.382 444 032 × 2 = 0 + 0.764 888 064;
  • 28) 0.764 888 064 × 2 = 1 + 0.529 776 128;
  • 29) 0.529 776 128 × 2 = 1 + 0.059 552 256;
  • 30) 0.059 552 256 × 2 = 0 + 0.119 104 512;
  • 31) 0.119 104 512 × 2 = 0 + 0.238 209 024;
  • 32) 0.238 209 024 × 2 = 0 + 0.476 418 048;
  • 33) 0.476 418 048 × 2 = 0 + 0.952 836 096;
  • 34) 0.952 836 096 × 2 = 1 + 0.905 672 192;
  • 35) 0.905 672 192 × 2 = 1 + 0.811 344 384;
  • 36) 0.811 344 384 × 2 = 1 + 0.622 688 768;
  • 37) 0.622 688 768 × 2 = 1 + 0.245 377 536;
  • 38) 0.245 377 536 × 2 = 0 + 0.490 755 072;
  • 39) 0.490 755 072 × 2 = 0 + 0.981 510 144;
  • 40) 0.981 510 144 × 2 = 1 + 0.963 020 288;
  • 41) 0.963 020 288 × 2 = 1 + 0.926 040 576;
  • 42) 0.926 040 576 × 2 = 1 + 0.852 081 152;
  • 43) 0.852 081 152 × 2 = 1 + 0.704 162 304;
  • 44) 0.704 162 304 × 2 = 1 + 0.408 324 608;
  • 45) 0.408 324 608 × 2 = 0 + 0.816 649 216;
  • 46) 0.816 649 216 × 2 = 1 + 0.633 298 432;
  • 47) 0.633 298 432 × 2 = 1 + 0.266 596 864;
  • 48) 0.266 596 864 × 2 = 0 + 0.533 193 728;
  • 49) 0.533 193 728 × 2 = 1 + 0.066 387 456;
  • 50) 0.066 387 456 × 2 = 0 + 0.132 774 912;
  • 51) 0.132 774 912 × 2 = 0 + 0.265 549 824;
  • 52) 0.265 549 824 × 2 = 0 + 0.531 099 648;
  • 53) 0.531 099 648 × 2 = 1 + 0.062 199 296;
  • 54) 0.062 199 296 × 2 = 0 + 0.124 398 592;
  • 55) 0.124 398 592 × 2 = 0 + 0.248 797 184;
  • 56) 0.248 797 184 × 2 = 0 + 0.497 594 368;
  • 57) 0.497 594 368 × 2 = 0 + 0.995 188 736;
  • 58) 0.995 188 736 × 2 = 1 + 0.990 377 472;
  • 59) 0.990 377 472 × 2 = 1 + 0.980 754 944;
  • 60) 0.980 754 944 × 2 = 1 + 0.961 509 888;
  • 61) 0.961 509 888 × 2 = 1 + 0.923 019 776;
  • 62) 0.923 019 776 × 2 = 1 + 0.846 039 552;
  • 63) 0.846 039 552 × 2 = 1 + 0.692 079 104;
  • 64) 0.692 079 104 × 2 = 1 + 0.384 158 208;

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:


0.000 282 338(10) =

0.0000 0000 0001 0010 1000 0000 1101 1000 0111 1001 1111 0110 1000 1000 0111 1111(2)

6. Positive number before normalization:

0.000 282 338(10) =

0.0000 0000 0001 0010 1000 0000 1101 1000 0111 1001 1111 0110 1000 1000 0111 1111(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:


0.000 282 338(10) =

0.0000 0000 0001 0010 1000 0000 1101 1000 0111 1001 1111 0110 1000 1000 0111 1111(2) =

0.0000 0000 0001 0010 1000 0000 1101 1000 0111 1001 1111 0110 1000 1000 0111 1111(2) × 20 =

1.0010 1000 0000 1101 1000 0111 1001 1111 0110 1000 1000 0111 1111(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 0000 1101 1000 0111 1001 1111 0110 1000 1000 0111 1111


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 0000 1101 1000 0111 1001 1111 0110 1000 1000 0111 1111 =

0010 1000 0000 1101 1000 0111 1001 1111 0110 1000 1000 0111 1111


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 0000 1101 1000 0111 1001 1111 0110 1000 1000 0111 1111


Decimal number -0.000 282 338 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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