-0.000 282 344 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 344(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
-0.000 282 344(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:

|-0.000 282 344| = 0.000 282 344


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: 0.000 282 344.

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) 0.000 282 344 × 2 = 0 + 0.000 564 688;
  • 2) 0.000 564 688 × 2 = 0 + 0.001 129 376;
  • 3) 0.001 129 376 × 2 = 0 + 0.002 258 752;
  • 4) 0.002 258 752 × 2 = 0 + 0.004 517 504;
  • 5) 0.004 517 504 × 2 = 0 + 0.009 035 008;
  • 6) 0.009 035 008 × 2 = 0 + 0.018 070 016;
  • 7) 0.018 070 016 × 2 = 0 + 0.036 140 032;
  • 8) 0.036 140 032 × 2 = 0 + 0.072 280 064;
  • 9) 0.072 280 064 × 2 = 0 + 0.144 560 128;
  • 10) 0.144 560 128 × 2 = 0 + 0.289 120 256;
  • 11) 0.289 120 256 × 2 = 0 + 0.578 240 512;
  • 12) 0.578 240 512 × 2 = 1 + 0.156 481 024;
  • 13) 0.156 481 024 × 2 = 0 + 0.312 962 048;
  • 14) 0.312 962 048 × 2 = 0 + 0.625 924 096;
  • 15) 0.625 924 096 × 2 = 1 + 0.251 848 192;
  • 16) 0.251 848 192 × 2 = 0 + 0.503 696 384;
  • 17) 0.503 696 384 × 2 = 1 + 0.007 392 768;
  • 18) 0.007 392 768 × 2 = 0 + 0.014 785 536;
  • 19) 0.014 785 536 × 2 = 0 + 0.029 571 072;
  • 20) 0.029 571 072 × 2 = 0 + 0.059 142 144;
  • 21) 0.059 142 144 × 2 = 0 + 0.118 284 288;
  • 22) 0.118 284 288 × 2 = 0 + 0.236 568 576;
  • 23) 0.236 568 576 × 2 = 0 + 0.473 137 152;
  • 24) 0.473 137 152 × 2 = 0 + 0.946 274 304;
  • 25) 0.946 274 304 × 2 = 1 + 0.892 548 608;
  • 26) 0.892 548 608 × 2 = 1 + 0.785 097 216;
  • 27) 0.785 097 216 × 2 = 1 + 0.570 194 432;
  • 28) 0.570 194 432 × 2 = 1 + 0.140 388 864;
  • 29) 0.140 388 864 × 2 = 0 + 0.280 777 728;
  • 30) 0.280 777 728 × 2 = 0 + 0.561 555 456;
  • 31) 0.561 555 456 × 2 = 1 + 0.123 110 912;
  • 32) 0.123 110 912 × 2 = 0 + 0.246 221 824;
  • 33) 0.246 221 824 × 2 = 0 + 0.492 443 648;
  • 34) 0.492 443 648 × 2 = 0 + 0.984 887 296;
  • 35) 0.984 887 296 × 2 = 1 + 0.969 774 592;
  • 36) 0.969 774 592 × 2 = 1 + 0.939 549 184;
  • 37) 0.939 549 184 × 2 = 1 + 0.879 098 368;
  • 38) 0.879 098 368 × 2 = 1 + 0.758 196 736;
  • 39) 0.758 196 736 × 2 = 1 + 0.516 393 472;
  • 40) 0.516 393 472 × 2 = 1 + 0.032 786 944;
  • 41) 0.032 786 944 × 2 = 0 + 0.065 573 888;
  • 42) 0.065 573 888 × 2 = 0 + 0.131 147 776;
  • 43) 0.131 147 776 × 2 = 0 + 0.262 295 552;
  • 44) 0.262 295 552 × 2 = 0 + 0.524 591 104;
  • 45) 0.524 591 104 × 2 = 1 + 0.049 182 208;
  • 46) 0.049 182 208 × 2 = 0 + 0.098 364 416;
  • 47) 0.098 364 416 × 2 = 0 + 0.196 728 832;
  • 48) 0.196 728 832 × 2 = 0 + 0.393 457 664;
  • 49) 0.393 457 664 × 2 = 0 + 0.786 915 328;
  • 50) 0.786 915 328 × 2 = 1 + 0.573 830 656;
  • 51) 0.573 830 656 × 2 = 1 + 0.147 661 312;
  • 52) 0.147 661 312 × 2 = 0 + 0.295 322 624;
  • 53) 0.295 322 624 × 2 = 0 + 0.590 645 248;
  • 54) 0.590 645 248 × 2 = 1 + 0.181 290 496;
  • 55) 0.181 290 496 × 2 = 0 + 0.362 580 992;
  • 56) 0.362 580 992 × 2 = 0 + 0.725 161 984;
  • 57) 0.725 161 984 × 2 = 1 + 0.450 323 968;
  • 58) 0.450 323 968 × 2 = 0 + 0.900 647 936;
  • 59) 0.900 647 936 × 2 = 1 + 0.801 295 872;
  • 60) 0.801 295 872 × 2 = 1 + 0.602 591 744;
  • 61) 0.602 591 744 × 2 = 1 + 0.205 183 488;
  • 62) 0.205 183 488 × 2 = 0 + 0.410 366 976;
  • 63) 0.410 366 976 × 2 = 0 + 0.820 733 952;
  • 64) 0.820 733 952 × 2 = 1 + 0.641 467 904;

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:


0.000 282 344(10) =

0.0000 0000 0001 0010 1000 0000 1111 0010 0011 1111 0000 1000 0110 0100 1011 1001(2)

6. Positive number before normalization:

0.000 282 344(10) =

0.0000 0000 0001 0010 1000 0000 1111 0010 0011 1111 0000 1000 0110 0100 1011 1001(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:


0.000 282 344(10) =

0.0000 0000 0001 0010 1000 0000 1111 0010 0011 1111 0000 1000 0110 0100 1011 1001(2) =

0.0000 0000 0001 0010 1000 0000 1111 0010 0011 1111 0000 1000 0110 0100 1011 1001(2) × 20 =

1.0010 1000 0000 1111 0010 0011 1111 0000 1000 0110 0100 1011 1001(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 0000 1111 0010 0011 1111 0000 1000 0110 0100 1011 1001


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 0000 1111 0010 0011 1111 0000 1000 0110 0100 1011 1001 =

0010 1000 0000 1111 0010 0011 1111 0000 1000 0110 0100 1011 1001


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 0000 1111 0010 0011 1111 0000 1000 0110 0100 1011 1001


Decimal number -0.000 282 344 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0000 1111 0010 0011 1111 0000 1000 0110 0100 1011 1001

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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