-0.000 282 005 984 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 984(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
-0.000 282 005 984(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:

|-0.000 282 005 984| = 0.000 282 005 984


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: 0.000 282 005 984.

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) 0.000 282 005 984 × 2 = 0 + 0.000 564 011 968;
  • 2) 0.000 564 011 968 × 2 = 0 + 0.001 128 023 936;
  • 3) 0.001 128 023 936 × 2 = 0 + 0.002 256 047 872;
  • 4) 0.002 256 047 872 × 2 = 0 + 0.004 512 095 744;
  • 5) 0.004 512 095 744 × 2 = 0 + 0.009 024 191 488;
  • 6) 0.009 024 191 488 × 2 = 0 + 0.018 048 382 976;
  • 7) 0.018 048 382 976 × 2 = 0 + 0.036 096 765 952;
  • 8) 0.036 096 765 952 × 2 = 0 + 0.072 193 531 904;
  • 9) 0.072 193 531 904 × 2 = 0 + 0.144 387 063 808;
  • 10) 0.144 387 063 808 × 2 = 0 + 0.288 774 127 616;
  • 11) 0.288 774 127 616 × 2 = 0 + 0.577 548 255 232;
  • 12) 0.577 548 255 232 × 2 = 1 + 0.155 096 510 464;
  • 13) 0.155 096 510 464 × 2 = 0 + 0.310 193 020 928;
  • 14) 0.310 193 020 928 × 2 = 0 + 0.620 386 041 856;
  • 15) 0.620 386 041 856 × 2 = 1 + 0.240 772 083 712;
  • 16) 0.240 772 083 712 × 2 = 0 + 0.481 544 167 424;
  • 17) 0.481 544 167 424 × 2 = 0 + 0.963 088 334 848;
  • 18) 0.963 088 334 848 × 2 = 1 + 0.926 176 669 696;
  • 19) 0.926 176 669 696 × 2 = 1 + 0.852 353 339 392;
  • 20) 0.852 353 339 392 × 2 = 1 + 0.704 706 678 784;
  • 21) 0.704 706 678 784 × 2 = 1 + 0.409 413 357 568;
  • 22) 0.409 413 357 568 × 2 = 0 + 0.818 826 715 136;
  • 23) 0.818 826 715 136 × 2 = 1 + 0.637 653 430 272;
  • 24) 0.637 653 430 272 × 2 = 1 + 0.275 306 860 544;
  • 25) 0.275 306 860 544 × 2 = 0 + 0.550 613 721 088;
  • 26) 0.550 613 721 088 × 2 = 1 + 0.101 227 442 176;
  • 27) 0.101 227 442 176 × 2 = 0 + 0.202 454 884 352;
  • 28) 0.202 454 884 352 × 2 = 0 + 0.404 909 768 704;
  • 29) 0.404 909 768 704 × 2 = 0 + 0.809 819 537 408;
  • 30) 0.809 819 537 408 × 2 = 1 + 0.619 639 074 816;
  • 31) 0.619 639 074 816 × 2 = 1 + 0.239 278 149 632;
  • 32) 0.239 278 149 632 × 2 = 0 + 0.478 556 299 264;
  • 33) 0.478 556 299 264 × 2 = 0 + 0.957 112 598 528;
  • 34) 0.957 112 598 528 × 2 = 1 + 0.914 225 197 056;
  • 35) 0.914 225 197 056 × 2 = 1 + 0.828 450 394 112;
  • 36) 0.828 450 394 112 × 2 = 1 + 0.656 900 788 224;
  • 37) 0.656 900 788 224 × 2 = 1 + 0.313 801 576 448;
  • 38) 0.313 801 576 448 × 2 = 0 + 0.627 603 152 896;
  • 39) 0.627 603 152 896 × 2 = 1 + 0.255 206 305 792;
  • 40) 0.255 206 305 792 × 2 = 0 + 0.510 412 611 584;
  • 41) 0.510 412 611 584 × 2 = 1 + 0.020 825 223 168;
  • 42) 0.020 825 223 168 × 2 = 0 + 0.041 650 446 336;
  • 43) 0.041 650 446 336 × 2 = 0 + 0.083 300 892 672;
  • 44) 0.083 300 892 672 × 2 = 0 + 0.166 601 785 344;
  • 45) 0.166 601 785 344 × 2 = 0 + 0.333 203 570 688;
  • 46) 0.333 203 570 688 × 2 = 0 + 0.666 407 141 376;
  • 47) 0.666 407 141 376 × 2 = 1 + 0.332 814 282 752;
  • 48) 0.332 814 282 752 × 2 = 0 + 0.665 628 565 504;
  • 49) 0.665 628 565 504 × 2 = 1 + 0.331 257 131 008;
  • 50) 0.331 257 131 008 × 2 = 0 + 0.662 514 262 016;
  • 51) 0.662 514 262 016 × 2 = 1 + 0.325 028 524 032;
  • 52) 0.325 028 524 032 × 2 = 0 + 0.650 057 048 064;
  • 53) 0.650 057 048 064 × 2 = 1 + 0.300 114 096 128;
  • 54) 0.300 114 096 128 × 2 = 0 + 0.600 228 192 256;
  • 55) 0.600 228 192 256 × 2 = 1 + 0.200 456 384 512;
  • 56) 0.200 456 384 512 × 2 = 0 + 0.400 912 769 024;
  • 57) 0.400 912 769 024 × 2 = 0 + 0.801 825 538 048;
  • 58) 0.801 825 538 048 × 2 = 1 + 0.603 651 076 096;
  • 59) 0.603 651 076 096 × 2 = 1 + 0.207 302 152 192;
  • 60) 0.207 302 152 192 × 2 = 0 + 0.414 604 304 384;
  • 61) 0.414 604 304 384 × 2 = 0 + 0.829 208 608 768;
  • 62) 0.829 208 608 768 × 2 = 1 + 0.658 417 217 536;
  • 63) 0.658 417 217 536 × 2 = 1 + 0.316 834 435 072;
  • 64) 0.316 834 435 072 × 2 = 0 + 0.633 668 870 144;

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:


0.000 282 005 984(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0111 1010 1000 0010 1010 1010 0110 0110(2)

6. Positive number before normalization:

0.000 282 005 984(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0111 1010 1000 0010 1010 1010 0110 0110(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:


0.000 282 005 984(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0111 1010 1000 0010 1010 1010 0110 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0111 1010 1000 0010 1010 1010 0110 0110(2) × 20 =

1.0010 0111 1011 0100 0110 0111 1010 1000 0010 1010 1010 0110 0110(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 0111 1010 1000 0010 1010 1010 0110 0110


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 0111 1010 1000 0010 1010 1010 0110 0110 =

0010 0111 1011 0100 0110 0111 1010 1000 0010 1010 1010 0110 0110


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 0111 1010 1000 0010 1010 1010 0110 0110


Decimal number -0.000 282 005 984 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0111 1010 1000 0010 1010 1010 0110 0110

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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