-0.066 298 274 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.066 298 274(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
-0.066 298 274(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:

|-0.066 298 274| = 0.066 298 274


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: 0.066 298 274.

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) 0.066 298 274 × 2 = 0 + 0.132 596 548;
  • 2) 0.132 596 548 × 2 = 0 + 0.265 193 096;
  • 3) 0.265 193 096 × 2 = 0 + 0.530 386 192;
  • 4) 0.530 386 192 × 2 = 1 + 0.060 772 384;
  • 5) 0.060 772 384 × 2 = 0 + 0.121 544 768;
  • 6) 0.121 544 768 × 2 = 0 + 0.243 089 536;
  • 7) 0.243 089 536 × 2 = 0 + 0.486 179 072;
  • 8) 0.486 179 072 × 2 = 0 + 0.972 358 144;
  • 9) 0.972 358 144 × 2 = 1 + 0.944 716 288;
  • 10) 0.944 716 288 × 2 = 1 + 0.889 432 576;
  • 11) 0.889 432 576 × 2 = 1 + 0.778 865 152;
  • 12) 0.778 865 152 × 2 = 1 + 0.557 730 304;
  • 13) 0.557 730 304 × 2 = 1 + 0.115 460 608;
  • 14) 0.115 460 608 × 2 = 0 + 0.230 921 216;
  • 15) 0.230 921 216 × 2 = 0 + 0.461 842 432;
  • 16) 0.461 842 432 × 2 = 0 + 0.923 684 864;
  • 17) 0.923 684 864 × 2 = 1 + 0.847 369 728;
  • 18) 0.847 369 728 × 2 = 1 + 0.694 739 456;
  • 19) 0.694 739 456 × 2 = 1 + 0.389 478 912;
  • 20) 0.389 478 912 × 2 = 0 + 0.778 957 824;
  • 21) 0.778 957 824 × 2 = 1 + 0.557 915 648;
  • 22) 0.557 915 648 × 2 = 1 + 0.115 831 296;
  • 23) 0.115 831 296 × 2 = 0 + 0.231 662 592;
  • 24) 0.231 662 592 × 2 = 0 + 0.463 325 184;
  • 25) 0.463 325 184 × 2 = 0 + 0.926 650 368;
  • 26) 0.926 650 368 × 2 = 1 + 0.853 300 736;
  • 27) 0.853 300 736 × 2 = 1 + 0.706 601 472;
  • 28) 0.706 601 472 × 2 = 1 + 0.413 202 944;
  • 29) 0.413 202 944 × 2 = 0 + 0.826 405 888;
  • 30) 0.826 405 888 × 2 = 1 + 0.652 811 776;
  • 31) 0.652 811 776 × 2 = 1 + 0.305 623 552;
  • 32) 0.305 623 552 × 2 = 0 + 0.611 247 104;
  • 33) 0.611 247 104 × 2 = 1 + 0.222 494 208;
  • 34) 0.222 494 208 × 2 = 0 + 0.444 988 416;
  • 35) 0.444 988 416 × 2 = 0 + 0.889 976 832;
  • 36) 0.889 976 832 × 2 = 1 + 0.779 953 664;
  • 37) 0.779 953 664 × 2 = 1 + 0.559 907 328;
  • 38) 0.559 907 328 × 2 = 1 + 0.119 814 656;
  • 39) 0.119 814 656 × 2 = 0 + 0.239 629 312;
  • 40) 0.239 629 312 × 2 = 0 + 0.479 258 624;
  • 41) 0.479 258 624 × 2 = 0 + 0.958 517 248;
  • 42) 0.958 517 248 × 2 = 1 + 0.917 034 496;
  • 43) 0.917 034 496 × 2 = 1 + 0.834 068 992;
  • 44) 0.834 068 992 × 2 = 1 + 0.668 137 984;
  • 45) 0.668 137 984 × 2 = 1 + 0.336 275 968;
  • 46) 0.336 275 968 × 2 = 0 + 0.672 551 936;
  • 47) 0.672 551 936 × 2 = 1 + 0.345 103 872;
  • 48) 0.345 103 872 × 2 = 0 + 0.690 207 744;
  • 49) 0.690 207 744 × 2 = 1 + 0.380 415 488;
  • 50) 0.380 415 488 × 2 = 0 + 0.760 830 976;
  • 51) 0.760 830 976 × 2 = 1 + 0.521 661 952;
  • 52) 0.521 661 952 × 2 = 1 + 0.043 323 904;
  • 53) 0.043 323 904 × 2 = 0 + 0.086 647 808;
  • 54) 0.086 647 808 × 2 = 0 + 0.173 295 616;
  • 55) 0.173 295 616 × 2 = 0 + 0.346 591 232;
  • 56) 0.346 591 232 × 2 = 0 + 0.693 182 464;

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:


0.066 298 274(10) =

0.0001 0000 1111 1000 1110 1100 0111 0110 1001 1100 0111 1010 1011 0000(2)

6. Positive number before normalization:

0.066 298 274(10) =

0.0001 0000 1111 1000 1110 1100 0111 0110 1001 1100 0111 1010 1011 0000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.066 298 274(10) =

0.0001 0000 1111 1000 1110 1100 0111 0110 1001 1100 0111 1010 1011 0000(2) =

0.0001 0000 1111 1000 1110 1100 0111 0110 1001 1100 0111 1010 1011 0000(2) × 20 =

1.0000 1111 1000 1110 1100 0111 0110 1001 1100 0111 1010 1011 0000(2) × 2-4


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):
1.0000 1111 1000 1110 1100 0111 0110 1001 1100 0111 1010 1011 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 0000 1111 1000 1110 1100 0111 0110 1001 1100 0111 1010 1011 0000 =

0000 1111 1000 1110 1100 0111 0110 1001 1100 0111 1010 1011 0000


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 1011

Mantissa (52 bits) =
0000 1111 1000 1110 1100 0111 0110 1001 1100 0111 1010 1011 0000


Decimal number -0.066 298 274 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0000 1111 1000 1110 1100 0111 0110 1001 1100 0111 1010 1011 0000

» Convert decimal -0.066 298 207 to 64 bit double precision IEEE 754 binary floating point representation standard

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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