-0.000 282 005 957 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 957(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
-0.000 282 005 957(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:

|-0.000 282 005 957| = 0.000 282 005 957


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: 0.000 282 005 957.

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) 0.000 282 005 957 × 2 = 0 + 0.000 564 011 914;
  • 2) 0.000 564 011 914 × 2 = 0 + 0.001 128 023 828;
  • 3) 0.001 128 023 828 × 2 = 0 + 0.002 256 047 656;
  • 4) 0.002 256 047 656 × 2 = 0 + 0.004 512 095 312;
  • 5) 0.004 512 095 312 × 2 = 0 + 0.009 024 190 624;
  • 6) 0.009 024 190 624 × 2 = 0 + 0.018 048 381 248;
  • 7) 0.018 048 381 248 × 2 = 0 + 0.036 096 762 496;
  • 8) 0.036 096 762 496 × 2 = 0 + 0.072 193 524 992;
  • 9) 0.072 193 524 992 × 2 = 0 + 0.144 387 049 984;
  • 10) 0.144 387 049 984 × 2 = 0 + 0.288 774 099 968;
  • 11) 0.288 774 099 968 × 2 = 0 + 0.577 548 199 936;
  • 12) 0.577 548 199 936 × 2 = 1 + 0.155 096 399 872;
  • 13) 0.155 096 399 872 × 2 = 0 + 0.310 192 799 744;
  • 14) 0.310 192 799 744 × 2 = 0 + 0.620 385 599 488;
  • 15) 0.620 385 599 488 × 2 = 1 + 0.240 771 198 976;
  • 16) 0.240 771 198 976 × 2 = 0 + 0.481 542 397 952;
  • 17) 0.481 542 397 952 × 2 = 0 + 0.963 084 795 904;
  • 18) 0.963 084 795 904 × 2 = 1 + 0.926 169 591 808;
  • 19) 0.926 169 591 808 × 2 = 1 + 0.852 339 183 616;
  • 20) 0.852 339 183 616 × 2 = 1 + 0.704 678 367 232;
  • 21) 0.704 678 367 232 × 2 = 1 + 0.409 356 734 464;
  • 22) 0.409 356 734 464 × 2 = 0 + 0.818 713 468 928;
  • 23) 0.818 713 468 928 × 2 = 1 + 0.637 426 937 856;
  • 24) 0.637 426 937 856 × 2 = 1 + 0.274 853 875 712;
  • 25) 0.274 853 875 712 × 2 = 0 + 0.549 707 751 424;
  • 26) 0.549 707 751 424 × 2 = 1 + 0.099 415 502 848;
  • 27) 0.099 415 502 848 × 2 = 0 + 0.198 831 005 696;
  • 28) 0.198 831 005 696 × 2 = 0 + 0.397 662 011 392;
  • 29) 0.397 662 011 392 × 2 = 0 + 0.795 324 022 784;
  • 30) 0.795 324 022 784 × 2 = 1 + 0.590 648 045 568;
  • 31) 0.590 648 045 568 × 2 = 1 + 0.181 296 091 136;
  • 32) 0.181 296 091 136 × 2 = 0 + 0.362 592 182 272;
  • 33) 0.362 592 182 272 × 2 = 0 + 0.725 184 364 544;
  • 34) 0.725 184 364 544 × 2 = 1 + 0.450 368 729 088;
  • 35) 0.450 368 729 088 × 2 = 0 + 0.900 737 458 176;
  • 36) 0.900 737 458 176 × 2 = 1 + 0.801 474 916 352;
  • 37) 0.801 474 916 352 × 2 = 1 + 0.602 949 832 704;
  • 38) 0.602 949 832 704 × 2 = 1 + 0.205 899 665 408;
  • 39) 0.205 899 665 408 × 2 = 0 + 0.411 799 330 816;
  • 40) 0.411 799 330 816 × 2 = 0 + 0.823 598 661 632;
  • 41) 0.823 598 661 632 × 2 = 1 + 0.647 197 323 264;
  • 42) 0.647 197 323 264 × 2 = 1 + 0.294 394 646 528;
  • 43) 0.294 394 646 528 × 2 = 0 + 0.588 789 293 056;
  • 44) 0.588 789 293 056 × 2 = 1 + 0.177 578 586 112;
  • 45) 0.177 578 586 112 × 2 = 0 + 0.355 157 172 224;
  • 46) 0.355 157 172 224 × 2 = 0 + 0.710 314 344 448;
  • 47) 0.710 314 344 448 × 2 = 1 + 0.420 628 688 896;
  • 48) 0.420 628 688 896 × 2 = 0 + 0.841 257 377 792;
  • 49) 0.841 257 377 792 × 2 = 1 + 0.682 514 755 584;
  • 50) 0.682 514 755 584 × 2 = 1 + 0.365 029 511 168;
  • 51) 0.365 029 511 168 × 2 = 0 + 0.730 059 022 336;
  • 52) 0.730 059 022 336 × 2 = 1 + 0.460 118 044 672;
  • 53) 0.460 118 044 672 × 2 = 0 + 0.920 236 089 344;
  • 54) 0.920 236 089 344 × 2 = 1 + 0.840 472 178 688;
  • 55) 0.840 472 178 688 × 2 = 1 + 0.680 944 357 376;
  • 56) 0.680 944 357 376 × 2 = 1 + 0.361 888 714 752;
  • 57) 0.361 888 714 752 × 2 = 0 + 0.723 777 429 504;
  • 58) 0.723 777 429 504 × 2 = 1 + 0.447 554 859 008;
  • 59) 0.447 554 859 008 × 2 = 0 + 0.895 109 718 016;
  • 60) 0.895 109 718 016 × 2 = 1 + 0.790 219 436 032;
  • 61) 0.790 219 436 032 × 2 = 1 + 0.580 438 872 064;
  • 62) 0.580 438 872 064 × 2 = 1 + 0.160 877 744 128;
  • 63) 0.160 877 744 128 × 2 = 0 + 0.321 755 488 256;
  • 64) 0.321 755 488 256 × 2 = 0 + 0.643 510 976 512;

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:


0.000 282 005 957(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 1101 0010 1101 0111 0101 1100(2)

6. Positive number before normalization:

0.000 282 005 957(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 1101 0010 1101 0111 0101 1100(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:


0.000 282 005 957(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 1101 0010 1101 0111 0101 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1100 1101 0010 1101 0111 0101 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0101 1100 1101 0010 1101 0111 0101 1100(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 0101 1100 1101 0010 1101 0111 0101 1100


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 0101 1100 1101 0010 1101 0111 0101 1100 =

0010 0111 1011 0100 0110 0101 1100 1101 0010 1101 0111 0101 1100


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 0101 1100 1101 0010 1101 0111 0101 1100


Decimal number -0.000 282 005 957 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0101 1100 1101 0010 1101 0111 0101 1100

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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