-0.000 282 005 747 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 747(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
-0.000 282 005 747(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:

|-0.000 282 005 747| = 0.000 282 005 747


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: 0.000 282 005 747.

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) 0.000 282 005 747 × 2 = 0 + 0.000 564 011 494;
  • 2) 0.000 564 011 494 × 2 = 0 + 0.001 128 022 988;
  • 3) 0.001 128 022 988 × 2 = 0 + 0.002 256 045 976;
  • 4) 0.002 256 045 976 × 2 = 0 + 0.004 512 091 952;
  • 5) 0.004 512 091 952 × 2 = 0 + 0.009 024 183 904;
  • 6) 0.009 024 183 904 × 2 = 0 + 0.018 048 367 808;
  • 7) 0.018 048 367 808 × 2 = 0 + 0.036 096 735 616;
  • 8) 0.036 096 735 616 × 2 = 0 + 0.072 193 471 232;
  • 9) 0.072 193 471 232 × 2 = 0 + 0.144 386 942 464;
  • 10) 0.144 386 942 464 × 2 = 0 + 0.288 773 884 928;
  • 11) 0.288 773 884 928 × 2 = 0 + 0.577 547 769 856;
  • 12) 0.577 547 769 856 × 2 = 1 + 0.155 095 539 712;
  • 13) 0.155 095 539 712 × 2 = 0 + 0.310 191 079 424;
  • 14) 0.310 191 079 424 × 2 = 0 + 0.620 382 158 848;
  • 15) 0.620 382 158 848 × 2 = 1 + 0.240 764 317 696;
  • 16) 0.240 764 317 696 × 2 = 0 + 0.481 528 635 392;
  • 17) 0.481 528 635 392 × 2 = 0 + 0.963 057 270 784;
  • 18) 0.963 057 270 784 × 2 = 1 + 0.926 114 541 568;
  • 19) 0.926 114 541 568 × 2 = 1 + 0.852 229 083 136;
  • 20) 0.852 229 083 136 × 2 = 1 + 0.704 458 166 272;
  • 21) 0.704 458 166 272 × 2 = 1 + 0.408 916 332 544;
  • 22) 0.408 916 332 544 × 2 = 0 + 0.817 832 665 088;
  • 23) 0.817 832 665 088 × 2 = 1 + 0.635 665 330 176;
  • 24) 0.635 665 330 176 × 2 = 1 + 0.271 330 660 352;
  • 25) 0.271 330 660 352 × 2 = 0 + 0.542 661 320 704;
  • 26) 0.542 661 320 704 × 2 = 1 + 0.085 322 641 408;
  • 27) 0.085 322 641 408 × 2 = 0 + 0.170 645 282 816;
  • 28) 0.170 645 282 816 × 2 = 0 + 0.341 290 565 632;
  • 29) 0.341 290 565 632 × 2 = 0 + 0.682 581 131 264;
  • 30) 0.682 581 131 264 × 2 = 1 + 0.365 162 262 528;
  • 31) 0.365 162 262 528 × 2 = 0 + 0.730 324 525 056;
  • 32) 0.730 324 525 056 × 2 = 1 + 0.460 649 050 112;
  • 33) 0.460 649 050 112 × 2 = 0 + 0.921 298 100 224;
  • 34) 0.921 298 100 224 × 2 = 1 + 0.842 596 200 448;
  • 35) 0.842 596 200 448 × 2 = 1 + 0.685 192 400 896;
  • 36) 0.685 192 400 896 × 2 = 1 + 0.370 384 801 792;
  • 37) 0.370 384 801 792 × 2 = 0 + 0.740 769 603 584;
  • 38) 0.740 769 603 584 × 2 = 1 + 0.481 539 207 168;
  • 39) 0.481 539 207 168 × 2 = 0 + 0.963 078 414 336;
  • 40) 0.963 078 414 336 × 2 = 1 + 0.926 156 828 672;
  • 41) 0.926 156 828 672 × 2 = 1 + 0.852 313 657 344;
  • 42) 0.852 313 657 344 × 2 = 1 + 0.704 627 314 688;
  • 43) 0.704 627 314 688 × 2 = 1 + 0.409 254 629 376;
  • 44) 0.409 254 629 376 × 2 = 0 + 0.818 509 258 752;
  • 45) 0.818 509 258 752 × 2 = 1 + 0.637 018 517 504;
  • 46) 0.637 018 517 504 × 2 = 1 + 0.274 037 035 008;
  • 47) 0.274 037 035 008 × 2 = 0 + 0.548 074 070 016;
  • 48) 0.548 074 070 016 × 2 = 1 + 0.096 148 140 032;
  • 49) 0.096 148 140 032 × 2 = 0 + 0.192 296 280 064;
  • 50) 0.192 296 280 064 × 2 = 0 + 0.384 592 560 128;
  • 51) 0.384 592 560 128 × 2 = 0 + 0.769 185 120 256;
  • 52) 0.769 185 120 256 × 2 = 1 + 0.538 370 240 512;
  • 53) 0.538 370 240 512 × 2 = 1 + 0.076 740 481 024;
  • 54) 0.076 740 481 024 × 2 = 0 + 0.153 480 962 048;
  • 55) 0.153 480 962 048 × 2 = 0 + 0.306 961 924 096;
  • 56) 0.306 961 924 096 × 2 = 0 + 0.613 923 848 192;
  • 57) 0.613 923 848 192 × 2 = 1 + 0.227 847 696 384;
  • 58) 0.227 847 696 384 × 2 = 0 + 0.455 695 392 768;
  • 59) 0.455 695 392 768 × 2 = 0 + 0.911 390 785 536;
  • 60) 0.911 390 785 536 × 2 = 1 + 0.822 781 571 072;
  • 61) 0.822 781 571 072 × 2 = 1 + 0.645 563 142 144;
  • 62) 0.645 563 142 144 × 2 = 1 + 0.291 126 284 288;
  • 63) 0.291 126 284 288 × 2 = 0 + 0.582 252 568 576;
  • 64) 0.582 252 568 576 × 2 = 1 + 0.164 505 137 152;

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:


0.000 282 005 747(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0111 0101 1110 1101 0001 1000 1001 1101(2)

6. Positive number before normalization:

0.000 282 005 747(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0111 0101 1110 1101 0001 1000 1001 1101(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:


0.000 282 005 747(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0111 0101 1110 1101 0001 1000 1001 1101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0111 0101 1110 1101 0001 1000 1001 1101(2) × 20 =

1.0010 0111 1011 0100 0101 0111 0101 1110 1101 0001 1000 1001 1101(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 0101 0111 0101 1110 1101 0001 1000 1001 1101


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 0101 0111 0101 1110 1101 0001 1000 1001 1101 =

0010 0111 1011 0100 0101 0111 0101 1110 1101 0001 1000 1001 1101


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 0101 0111 0101 1110 1101 0001 1000 1001 1101


Decimal number -0.000 282 005 747 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0111 0101 1110 1101 0001 1000 1001 1101

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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