-0.000 282 006 146 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 146(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
-0.000 282 006 146(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:

|-0.000 282 006 146| = 0.000 282 006 146


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: 0.000 282 006 146.

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) 0.000 282 006 146 × 2 = 0 + 0.000 564 012 292;
  • 2) 0.000 564 012 292 × 2 = 0 + 0.001 128 024 584;
  • 3) 0.001 128 024 584 × 2 = 0 + 0.002 256 049 168;
  • 4) 0.002 256 049 168 × 2 = 0 + 0.004 512 098 336;
  • 5) 0.004 512 098 336 × 2 = 0 + 0.009 024 196 672;
  • 6) 0.009 024 196 672 × 2 = 0 + 0.018 048 393 344;
  • 7) 0.018 048 393 344 × 2 = 0 + 0.036 096 786 688;
  • 8) 0.036 096 786 688 × 2 = 0 + 0.072 193 573 376;
  • 9) 0.072 193 573 376 × 2 = 0 + 0.144 387 146 752;
  • 10) 0.144 387 146 752 × 2 = 0 + 0.288 774 293 504;
  • 11) 0.288 774 293 504 × 2 = 0 + 0.577 548 587 008;
  • 12) 0.577 548 587 008 × 2 = 1 + 0.155 097 174 016;
  • 13) 0.155 097 174 016 × 2 = 0 + 0.310 194 348 032;
  • 14) 0.310 194 348 032 × 2 = 0 + 0.620 388 696 064;
  • 15) 0.620 388 696 064 × 2 = 1 + 0.240 777 392 128;
  • 16) 0.240 777 392 128 × 2 = 0 + 0.481 554 784 256;
  • 17) 0.481 554 784 256 × 2 = 0 + 0.963 109 568 512;
  • 18) 0.963 109 568 512 × 2 = 1 + 0.926 219 137 024;
  • 19) 0.926 219 137 024 × 2 = 1 + 0.852 438 274 048;
  • 20) 0.852 438 274 048 × 2 = 1 + 0.704 876 548 096;
  • 21) 0.704 876 548 096 × 2 = 1 + 0.409 753 096 192;
  • 22) 0.409 753 096 192 × 2 = 0 + 0.819 506 192 384;
  • 23) 0.819 506 192 384 × 2 = 1 + 0.639 012 384 768;
  • 24) 0.639 012 384 768 × 2 = 1 + 0.278 024 769 536;
  • 25) 0.278 024 769 536 × 2 = 0 + 0.556 049 539 072;
  • 26) 0.556 049 539 072 × 2 = 1 + 0.112 099 078 144;
  • 27) 0.112 099 078 144 × 2 = 0 + 0.224 198 156 288;
  • 28) 0.224 198 156 288 × 2 = 0 + 0.448 396 312 576;
  • 29) 0.448 396 312 576 × 2 = 0 + 0.896 792 625 152;
  • 30) 0.896 792 625 152 × 2 = 1 + 0.793 585 250 304;
  • 31) 0.793 585 250 304 × 2 = 1 + 0.587 170 500 608;
  • 32) 0.587 170 500 608 × 2 = 1 + 0.174 341 001 216;
  • 33) 0.174 341 001 216 × 2 = 0 + 0.348 682 002 432;
  • 34) 0.348 682 002 432 × 2 = 0 + 0.697 364 004 864;
  • 35) 0.697 364 004 864 × 2 = 1 + 0.394 728 009 728;
  • 36) 0.394 728 009 728 × 2 = 0 + 0.789 456 019 456;
  • 37) 0.789 456 019 456 × 2 = 1 + 0.578 912 038 912;
  • 38) 0.578 912 038 912 × 2 = 1 + 0.157 824 077 824;
  • 39) 0.157 824 077 824 × 2 = 0 + 0.315 648 155 648;
  • 40) 0.315 648 155 648 × 2 = 0 + 0.631 296 311 296;
  • 41) 0.631 296 311 296 × 2 = 1 + 0.262 592 622 592;
  • 42) 0.262 592 622 592 × 2 = 0 + 0.525 185 245 184;
  • 43) 0.525 185 245 184 × 2 = 1 + 0.050 370 490 368;
  • 44) 0.050 370 490 368 × 2 = 0 + 0.100 740 980 736;
  • 45) 0.100 740 980 736 × 2 = 0 + 0.201 481 961 472;
  • 46) 0.201 481 961 472 × 2 = 0 + 0.402 963 922 944;
  • 47) 0.402 963 922 944 × 2 = 0 + 0.805 927 845 888;
  • 48) 0.805 927 845 888 × 2 = 1 + 0.611 855 691 776;
  • 49) 0.611 855 691 776 × 2 = 1 + 0.223 711 383 552;
  • 50) 0.223 711 383 552 × 2 = 0 + 0.447 422 767 104;
  • 51) 0.447 422 767 104 × 2 = 0 + 0.894 845 534 208;
  • 52) 0.894 845 534 208 × 2 = 1 + 0.789 691 068 416;
  • 53) 0.789 691 068 416 × 2 = 1 + 0.579 382 136 832;
  • 54) 0.579 382 136 832 × 2 = 1 + 0.158 764 273 664;
  • 55) 0.158 764 273 664 × 2 = 0 + 0.317 528 547 328;
  • 56) 0.317 528 547 328 × 2 = 0 + 0.635 057 094 656;
  • 57) 0.635 057 094 656 × 2 = 1 + 0.270 114 189 312;
  • 58) 0.270 114 189 312 × 2 = 0 + 0.540 228 378 624;
  • 59) 0.540 228 378 624 × 2 = 1 + 0.080 456 757 248;
  • 60) 0.080 456 757 248 × 2 = 0 + 0.160 913 514 496;
  • 61) 0.160 913 514 496 × 2 = 0 + 0.321 827 028 992;
  • 62) 0.321 827 028 992 × 2 = 0 + 0.643 654 057 984;
  • 63) 0.643 654 057 984 × 2 = 1 + 0.287 308 115 968;
  • 64) 0.287 308 115 968 × 2 = 0 + 0.574 616 231 936;

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:


0.000 282 006 146(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 1100 1010 0001 1001 1100 1010 0010(2)

6. Positive number before normalization:

0.000 282 006 146(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 1100 1010 0001 1001 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:


0.000 282 006 146(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 1100 1010 0001 1001 1100 1010 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0010 1100 1010 0001 1001 1100 1010 0010(2) × 20 =

1.0010 0111 1011 0100 0111 0010 1100 1010 0001 1001 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 0111 0010 1100 1010 0001 1001 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 0111 0010 1100 1010 0001 1001 1100 1010 0010 =

0010 0111 1011 0100 0111 0010 1100 1010 0001 1001 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 0111 0010 1100 1010 0001 1001 1100 1010 0010


Decimal number -0.000 282 006 146 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 0010 1100 1010 0001 1001 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