-0.000 282 672 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 672(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
-0.000 282 672(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:

|-0.000 282 672| = 0.000 282 672


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: 0.000 282 672.

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) 0.000 282 672 × 2 = 0 + 0.000 565 344;
  • 2) 0.000 565 344 × 2 = 0 + 0.001 130 688;
  • 3) 0.001 130 688 × 2 = 0 + 0.002 261 376;
  • 4) 0.002 261 376 × 2 = 0 + 0.004 522 752;
  • 5) 0.004 522 752 × 2 = 0 + 0.009 045 504;
  • 6) 0.009 045 504 × 2 = 0 + 0.018 091 008;
  • 7) 0.018 091 008 × 2 = 0 + 0.036 182 016;
  • 8) 0.036 182 016 × 2 = 0 + 0.072 364 032;
  • 9) 0.072 364 032 × 2 = 0 + 0.144 728 064;
  • 10) 0.144 728 064 × 2 = 0 + 0.289 456 128;
  • 11) 0.289 456 128 × 2 = 0 + 0.578 912 256;
  • 12) 0.578 912 256 × 2 = 1 + 0.157 824 512;
  • 13) 0.157 824 512 × 2 = 0 + 0.315 649 024;
  • 14) 0.315 649 024 × 2 = 0 + 0.631 298 048;
  • 15) 0.631 298 048 × 2 = 1 + 0.262 596 096;
  • 16) 0.262 596 096 × 2 = 0 + 0.525 192 192;
  • 17) 0.525 192 192 × 2 = 1 + 0.050 384 384;
  • 18) 0.050 384 384 × 2 = 0 + 0.100 768 768;
  • 19) 0.100 768 768 × 2 = 0 + 0.201 537 536;
  • 20) 0.201 537 536 × 2 = 0 + 0.403 075 072;
  • 21) 0.403 075 072 × 2 = 0 + 0.806 150 144;
  • 22) 0.806 150 144 × 2 = 1 + 0.612 300 288;
  • 23) 0.612 300 288 × 2 = 1 + 0.224 600 576;
  • 24) 0.224 600 576 × 2 = 0 + 0.449 201 152;
  • 25) 0.449 201 152 × 2 = 0 + 0.898 402 304;
  • 26) 0.898 402 304 × 2 = 1 + 0.796 804 608;
  • 27) 0.796 804 608 × 2 = 1 + 0.593 609 216;
  • 28) 0.593 609 216 × 2 = 1 + 0.187 218 432;
  • 29) 0.187 218 432 × 2 = 0 + 0.374 436 864;
  • 30) 0.374 436 864 × 2 = 0 + 0.748 873 728;
  • 31) 0.748 873 728 × 2 = 1 + 0.497 747 456;
  • 32) 0.497 747 456 × 2 = 0 + 0.995 494 912;
  • 33) 0.995 494 912 × 2 = 1 + 0.990 989 824;
  • 34) 0.990 989 824 × 2 = 1 + 0.981 979 648;
  • 35) 0.981 979 648 × 2 = 1 + 0.963 959 296;
  • 36) 0.963 959 296 × 2 = 1 + 0.927 918 592;
  • 37) 0.927 918 592 × 2 = 1 + 0.855 837 184;
  • 38) 0.855 837 184 × 2 = 1 + 0.711 674 368;
  • 39) 0.711 674 368 × 2 = 1 + 0.423 348 736;
  • 40) 0.423 348 736 × 2 = 0 + 0.846 697 472;
  • 41) 0.846 697 472 × 2 = 1 + 0.693 394 944;
  • 42) 0.693 394 944 × 2 = 1 + 0.386 789 888;
  • 43) 0.386 789 888 × 2 = 0 + 0.773 579 776;
  • 44) 0.773 579 776 × 2 = 1 + 0.547 159 552;
  • 45) 0.547 159 552 × 2 = 1 + 0.094 319 104;
  • 46) 0.094 319 104 × 2 = 0 + 0.188 638 208;
  • 47) 0.188 638 208 × 2 = 0 + 0.377 276 416;
  • 48) 0.377 276 416 × 2 = 0 + 0.754 552 832;
  • 49) 0.754 552 832 × 2 = 1 + 0.509 105 664;
  • 50) 0.509 105 664 × 2 = 1 + 0.018 211 328;
  • 51) 0.018 211 328 × 2 = 0 + 0.036 422 656;
  • 52) 0.036 422 656 × 2 = 0 + 0.072 845 312;
  • 53) 0.072 845 312 × 2 = 0 + 0.145 690 624;
  • 54) 0.145 690 624 × 2 = 0 + 0.291 381 248;
  • 55) 0.291 381 248 × 2 = 0 + 0.582 762 496;
  • 56) 0.582 762 496 × 2 = 1 + 0.165 524 992;
  • 57) 0.165 524 992 × 2 = 0 + 0.331 049 984;
  • 58) 0.331 049 984 × 2 = 0 + 0.662 099 968;
  • 59) 0.662 099 968 × 2 = 1 + 0.324 199 936;
  • 60) 0.324 199 936 × 2 = 0 + 0.648 399 872;
  • 61) 0.648 399 872 × 2 = 1 + 0.296 799 744;
  • 62) 0.296 799 744 × 2 = 0 + 0.593 599 488;
  • 63) 0.593 599 488 × 2 = 1 + 0.187 198 976;
  • 64) 0.187 198 976 × 2 = 0 + 0.374 397 952;

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:


0.000 282 672(10) =

0.0000 0000 0001 0010 1000 0110 0111 0010 1111 1110 1101 1000 1100 0001 0010 1010(2)

6. Positive number before normalization:

0.000 282 672(10) =

0.0000 0000 0001 0010 1000 0110 0111 0010 1111 1110 1101 1000 1100 0001 0010 1010(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:


0.000 282 672(10) =

0.0000 0000 0001 0010 1000 0110 0111 0010 1111 1110 1101 1000 1100 0001 0010 1010(2) =

0.0000 0000 0001 0010 1000 0110 0111 0010 1111 1110 1101 1000 1100 0001 0010 1010(2) × 20 =

1.0010 1000 0110 0111 0010 1111 1110 1101 1000 1100 0001 0010 1010(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 1000 0110 0111 0010 1111 1110 1101 1000 1100 0001 0010 1010


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 1000 0110 0111 0010 1111 1110 1101 1000 1100 0001 0010 1010 =

0010 1000 0110 0111 0010 1111 1110 1101 1000 1100 0001 0010 1010


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 1000 0110 0111 0010 1111 1110 1101 1000 1100 0001 0010 1010


Decimal number -0.000 282 672 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0110 0111 0010 1111 1110 1101 1000 1100 0001 0010 1010

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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