-0.000 282 006 064 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 064(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
-0.000 282 006 064(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:

|-0.000 282 006 064| = 0.000 282 006 064


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: 0.000 282 006 064.

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) 0.000 282 006 064 × 2 = 0 + 0.000 564 012 128;
  • 2) 0.000 564 012 128 × 2 = 0 + 0.001 128 024 256;
  • 3) 0.001 128 024 256 × 2 = 0 + 0.002 256 048 512;
  • 4) 0.002 256 048 512 × 2 = 0 + 0.004 512 097 024;
  • 5) 0.004 512 097 024 × 2 = 0 + 0.009 024 194 048;
  • 6) 0.009 024 194 048 × 2 = 0 + 0.018 048 388 096;
  • 7) 0.018 048 388 096 × 2 = 0 + 0.036 096 776 192;
  • 8) 0.036 096 776 192 × 2 = 0 + 0.072 193 552 384;
  • 9) 0.072 193 552 384 × 2 = 0 + 0.144 387 104 768;
  • 10) 0.144 387 104 768 × 2 = 0 + 0.288 774 209 536;
  • 11) 0.288 774 209 536 × 2 = 0 + 0.577 548 419 072;
  • 12) 0.577 548 419 072 × 2 = 1 + 0.155 096 838 144;
  • 13) 0.155 096 838 144 × 2 = 0 + 0.310 193 676 288;
  • 14) 0.310 193 676 288 × 2 = 0 + 0.620 387 352 576;
  • 15) 0.620 387 352 576 × 2 = 1 + 0.240 774 705 152;
  • 16) 0.240 774 705 152 × 2 = 0 + 0.481 549 410 304;
  • 17) 0.481 549 410 304 × 2 = 0 + 0.963 098 820 608;
  • 18) 0.963 098 820 608 × 2 = 1 + 0.926 197 641 216;
  • 19) 0.926 197 641 216 × 2 = 1 + 0.852 395 282 432;
  • 20) 0.852 395 282 432 × 2 = 1 + 0.704 790 564 864;
  • 21) 0.704 790 564 864 × 2 = 1 + 0.409 581 129 728;
  • 22) 0.409 581 129 728 × 2 = 0 + 0.819 162 259 456;
  • 23) 0.819 162 259 456 × 2 = 1 + 0.638 324 518 912;
  • 24) 0.638 324 518 912 × 2 = 1 + 0.276 649 037 824;
  • 25) 0.276 649 037 824 × 2 = 0 + 0.553 298 075 648;
  • 26) 0.553 298 075 648 × 2 = 1 + 0.106 596 151 296;
  • 27) 0.106 596 151 296 × 2 = 0 + 0.213 192 302 592;
  • 28) 0.213 192 302 592 × 2 = 0 + 0.426 384 605 184;
  • 29) 0.426 384 605 184 × 2 = 0 + 0.852 769 210 368;
  • 30) 0.852 769 210 368 × 2 = 1 + 0.705 538 420 736;
  • 31) 0.705 538 420 736 × 2 = 1 + 0.411 076 841 472;
  • 32) 0.411 076 841 472 × 2 = 0 + 0.822 153 682 944;
  • 33) 0.822 153 682 944 × 2 = 1 + 0.644 307 365 888;
  • 34) 0.644 307 365 888 × 2 = 1 + 0.288 614 731 776;
  • 35) 0.288 614 731 776 × 2 = 0 + 0.577 229 463 552;
  • 36) 0.577 229 463 552 × 2 = 1 + 0.154 458 927 104;
  • 37) 0.154 458 927 104 × 2 = 0 + 0.308 917 854 208;
  • 38) 0.308 917 854 208 × 2 = 0 + 0.617 835 708 416;
  • 39) 0.617 835 708 416 × 2 = 1 + 0.235 671 416 832;
  • 40) 0.235 671 416 832 × 2 = 0 + 0.471 342 833 664;
  • 41) 0.471 342 833 664 × 2 = 0 + 0.942 685 667 328;
  • 42) 0.942 685 667 328 × 2 = 1 + 0.885 371 334 656;
  • 43) 0.885 371 334 656 × 2 = 1 + 0.770 742 669 312;
  • 44) 0.770 742 669 312 × 2 = 1 + 0.541 485 338 624;
  • 45) 0.541 485 338 624 × 2 = 1 + 0.082 970 677 248;
  • 46) 0.082 970 677 248 × 2 = 0 + 0.165 941 354 496;
  • 47) 0.165 941 354 496 × 2 = 0 + 0.331 882 708 992;
  • 48) 0.331 882 708 992 × 2 = 0 + 0.663 765 417 984;
  • 49) 0.663 765 417 984 × 2 = 1 + 0.327 530 835 968;
  • 50) 0.327 530 835 968 × 2 = 0 + 0.655 061 671 936;
  • 51) 0.655 061 671 936 × 2 = 1 + 0.310 123 343 872;
  • 52) 0.310 123 343 872 × 2 = 0 + 0.620 246 687 744;
  • 53) 0.620 246 687 744 × 2 = 1 + 0.240 493 375 488;
  • 54) 0.240 493 375 488 × 2 = 0 + 0.480 986 750 976;
  • 55) 0.480 986 750 976 × 2 = 0 + 0.961 973 501 952;
  • 56) 0.961 973 501 952 × 2 = 1 + 0.923 947 003 904;
  • 57) 0.923 947 003 904 × 2 = 1 + 0.847 894 007 808;
  • 58) 0.847 894 007 808 × 2 = 1 + 0.695 788 015 616;
  • 59) 0.695 788 015 616 × 2 = 1 + 0.391 576 031 232;
  • 60) 0.391 576 031 232 × 2 = 0 + 0.783 152 062 464;
  • 61) 0.783 152 062 464 × 2 = 1 + 0.566 304 124 928;
  • 62) 0.566 304 124 928 × 2 = 1 + 0.132 608 249 856;
  • 63) 0.132 608 249 856 × 2 = 0 + 0.265 216 499 712;
  • 64) 0.265 216 499 712 × 2 = 0 + 0.530 432 999 424;

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:


0.000 282 006 064(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1101 0010 0111 1000 1010 1001 1110 1100(2)

6. Positive number before normalization:

0.000 282 006 064(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1101 0010 0111 1000 1010 1001 1110 1100(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:


0.000 282 006 064(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1101 0010 0111 1000 1010 1001 1110 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1101 0010 0111 1000 1010 1001 1110 1100(2) × 20 =

1.0010 0111 1011 0100 0110 1101 0010 0111 1000 1010 1001 1110 1100(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 1101 0010 0111 1000 1010 1001 1110 1100


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 1101 0010 0111 1000 1010 1001 1110 1100 =

0010 0111 1011 0100 0110 1101 0010 0111 1000 1010 1001 1110 1100


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 1101 0010 0111 1000 1010 1001 1110 1100


Decimal number -0.000 282 006 064 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1101 0010 0111 1000 1010 1001 1110 1100

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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