-0.000 282 005 609 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 609(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
-0.000 282 005 609(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:

|-0.000 282 005 609| = 0.000 282 005 609


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: 0.000 282 005 609.

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) 0.000 282 005 609 × 2 = 0 + 0.000 564 011 218;
  • 2) 0.000 564 011 218 × 2 = 0 + 0.001 128 022 436;
  • 3) 0.001 128 022 436 × 2 = 0 + 0.002 256 044 872;
  • 4) 0.002 256 044 872 × 2 = 0 + 0.004 512 089 744;
  • 5) 0.004 512 089 744 × 2 = 0 + 0.009 024 179 488;
  • 6) 0.009 024 179 488 × 2 = 0 + 0.018 048 358 976;
  • 7) 0.018 048 358 976 × 2 = 0 + 0.036 096 717 952;
  • 8) 0.036 096 717 952 × 2 = 0 + 0.072 193 435 904;
  • 9) 0.072 193 435 904 × 2 = 0 + 0.144 386 871 808;
  • 10) 0.144 386 871 808 × 2 = 0 + 0.288 773 743 616;
  • 11) 0.288 773 743 616 × 2 = 0 + 0.577 547 487 232;
  • 12) 0.577 547 487 232 × 2 = 1 + 0.155 094 974 464;
  • 13) 0.155 094 974 464 × 2 = 0 + 0.310 189 948 928;
  • 14) 0.310 189 948 928 × 2 = 0 + 0.620 379 897 856;
  • 15) 0.620 379 897 856 × 2 = 1 + 0.240 759 795 712;
  • 16) 0.240 759 795 712 × 2 = 0 + 0.481 519 591 424;
  • 17) 0.481 519 591 424 × 2 = 0 + 0.963 039 182 848;
  • 18) 0.963 039 182 848 × 2 = 1 + 0.926 078 365 696;
  • 19) 0.926 078 365 696 × 2 = 1 + 0.852 156 731 392;
  • 20) 0.852 156 731 392 × 2 = 1 + 0.704 313 462 784;
  • 21) 0.704 313 462 784 × 2 = 1 + 0.408 626 925 568;
  • 22) 0.408 626 925 568 × 2 = 0 + 0.817 253 851 136;
  • 23) 0.817 253 851 136 × 2 = 1 + 0.634 507 702 272;
  • 24) 0.634 507 702 272 × 2 = 1 + 0.269 015 404 544;
  • 25) 0.269 015 404 544 × 2 = 0 + 0.538 030 809 088;
  • 26) 0.538 030 809 088 × 2 = 1 + 0.076 061 618 176;
  • 27) 0.076 061 618 176 × 2 = 0 + 0.152 123 236 352;
  • 28) 0.152 123 236 352 × 2 = 0 + 0.304 246 472 704;
  • 29) 0.304 246 472 704 × 2 = 0 + 0.608 492 945 408;
  • 30) 0.608 492 945 408 × 2 = 1 + 0.216 985 890 816;
  • 31) 0.216 985 890 816 × 2 = 0 + 0.433 971 781 632;
  • 32) 0.433 971 781 632 × 2 = 0 + 0.867 943 563 264;
  • 33) 0.867 943 563 264 × 2 = 1 + 0.735 887 126 528;
  • 34) 0.735 887 126 528 × 2 = 1 + 0.471 774 253 056;
  • 35) 0.471 774 253 056 × 2 = 0 + 0.943 548 506 112;
  • 36) 0.943 548 506 112 × 2 = 1 + 0.887 097 012 224;
  • 37) 0.887 097 012 224 × 2 = 1 + 0.774 194 024 448;
  • 38) 0.774 194 024 448 × 2 = 1 + 0.548 388 048 896;
  • 39) 0.548 388 048 896 × 2 = 1 + 0.096 776 097 792;
  • 40) 0.096 776 097 792 × 2 = 0 + 0.193 552 195 584;
  • 41) 0.193 552 195 584 × 2 = 0 + 0.387 104 391 168;
  • 42) 0.387 104 391 168 × 2 = 0 + 0.774 208 782 336;
  • 43) 0.774 208 782 336 × 2 = 1 + 0.548 417 564 672;
  • 44) 0.548 417 564 672 × 2 = 1 + 0.096 835 129 344;
  • 45) 0.096 835 129 344 × 2 = 0 + 0.193 670 258 688;
  • 46) 0.193 670 258 688 × 2 = 0 + 0.387 340 517 376;
  • 47) 0.387 340 517 376 × 2 = 0 + 0.774 681 034 752;
  • 48) 0.774 681 034 752 × 2 = 1 + 0.549 362 069 504;
  • 49) 0.549 362 069 504 × 2 = 1 + 0.098 724 139 008;
  • 50) 0.098 724 139 008 × 2 = 0 + 0.197 448 278 016;
  • 51) 0.197 448 278 016 × 2 = 0 + 0.394 896 556 032;
  • 52) 0.394 896 556 032 × 2 = 0 + 0.789 793 112 064;
  • 53) 0.789 793 112 064 × 2 = 1 + 0.579 586 224 128;
  • 54) 0.579 586 224 128 × 2 = 1 + 0.159 172 448 256;
  • 55) 0.159 172 448 256 × 2 = 0 + 0.318 344 896 512;
  • 56) 0.318 344 896 512 × 2 = 0 + 0.636 689 793 024;
  • 57) 0.636 689 793 024 × 2 = 1 + 0.273 379 586 048;
  • 58) 0.273 379 586 048 × 2 = 0 + 0.546 759 172 096;
  • 59) 0.546 759 172 096 × 2 = 1 + 0.093 518 344 192;
  • 60) 0.093 518 344 192 × 2 = 0 + 0.187 036 688 384;
  • 61) 0.187 036 688 384 × 2 = 0 + 0.374 073 376 768;
  • 62) 0.374 073 376 768 × 2 = 0 + 0.748 146 753 536;
  • 63) 0.748 146 753 536 × 2 = 1 + 0.496 293 507 072;
  • 64) 0.496 293 507 072 × 2 = 0 + 0.992 587 014 144;

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:


0.000 282 005 609(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1101 1110 0011 0001 1000 1100 1010 0010(2)

6. Positive number before normalization:

0.000 282 005 609(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1101 1110 0011 0001 1000 1100 1010 0010(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:


0.000 282 005 609(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1101 1110 0011 0001 1000 1100 1010 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1101 1110 0011 0001 1000 1100 1010 0010(2) × 20 =

1.0010 0111 1011 0100 0100 1101 1110 0011 0001 1000 1100 1010 0010(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 0100 1101 1110 0011 0001 1000 1100 1010 0010


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 0100 1101 1110 0011 0001 1000 1100 1010 0010 =

0010 0111 1011 0100 0100 1101 1110 0011 0001 1000 1100 1010 0010


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 0100 1101 1110 0011 0001 1000 1100 1010 0010


Decimal number -0.000 282 005 609 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0100 1101 1110 0011 0001 1000 1100 1010 0010

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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