-0.000 282 006 393 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 393(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
-0.000 282 006 393(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:

|-0.000 282 006 393| = 0.000 282 006 393


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: 0.000 282 006 393.

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) 0.000 282 006 393 × 2 = 0 + 0.000 564 012 786;
  • 2) 0.000 564 012 786 × 2 = 0 + 0.001 128 025 572;
  • 3) 0.001 128 025 572 × 2 = 0 + 0.002 256 051 144;
  • 4) 0.002 256 051 144 × 2 = 0 + 0.004 512 102 288;
  • 5) 0.004 512 102 288 × 2 = 0 + 0.009 024 204 576;
  • 6) 0.009 024 204 576 × 2 = 0 + 0.018 048 409 152;
  • 7) 0.018 048 409 152 × 2 = 0 + 0.036 096 818 304;
  • 8) 0.036 096 818 304 × 2 = 0 + 0.072 193 636 608;
  • 9) 0.072 193 636 608 × 2 = 0 + 0.144 387 273 216;
  • 10) 0.144 387 273 216 × 2 = 0 + 0.288 774 546 432;
  • 11) 0.288 774 546 432 × 2 = 0 + 0.577 549 092 864;
  • 12) 0.577 549 092 864 × 2 = 1 + 0.155 098 185 728;
  • 13) 0.155 098 185 728 × 2 = 0 + 0.310 196 371 456;
  • 14) 0.310 196 371 456 × 2 = 0 + 0.620 392 742 912;
  • 15) 0.620 392 742 912 × 2 = 1 + 0.240 785 485 824;
  • 16) 0.240 785 485 824 × 2 = 0 + 0.481 570 971 648;
  • 17) 0.481 570 971 648 × 2 = 0 + 0.963 141 943 296;
  • 18) 0.963 141 943 296 × 2 = 1 + 0.926 283 886 592;
  • 19) 0.926 283 886 592 × 2 = 1 + 0.852 567 773 184;
  • 20) 0.852 567 773 184 × 2 = 1 + 0.705 135 546 368;
  • 21) 0.705 135 546 368 × 2 = 1 + 0.410 271 092 736;
  • 22) 0.410 271 092 736 × 2 = 0 + 0.820 542 185 472;
  • 23) 0.820 542 185 472 × 2 = 1 + 0.641 084 370 944;
  • 24) 0.641 084 370 944 × 2 = 1 + 0.282 168 741 888;
  • 25) 0.282 168 741 888 × 2 = 0 + 0.564 337 483 776;
  • 26) 0.564 337 483 776 × 2 = 1 + 0.128 674 967 552;
  • 27) 0.128 674 967 552 × 2 = 0 + 0.257 349 935 104;
  • 28) 0.257 349 935 104 × 2 = 0 + 0.514 699 870 208;
  • 29) 0.514 699 870 208 × 2 = 1 + 0.029 399 740 416;
  • 30) 0.029 399 740 416 × 2 = 0 + 0.058 799 480 832;
  • 31) 0.058 799 480 832 × 2 = 0 + 0.117 598 961 664;
  • 32) 0.117 598 961 664 × 2 = 0 + 0.235 197 923 328;
  • 33) 0.235 197 923 328 × 2 = 0 + 0.470 395 846 656;
  • 34) 0.470 395 846 656 × 2 = 0 + 0.940 791 693 312;
  • 35) 0.940 791 693 312 × 2 = 1 + 0.881 583 386 624;
  • 36) 0.881 583 386 624 × 2 = 1 + 0.763 166 773 248;
  • 37) 0.763 166 773 248 × 2 = 1 + 0.526 333 546 496;
  • 38) 0.526 333 546 496 × 2 = 1 + 0.052 667 092 992;
  • 39) 0.052 667 092 992 × 2 = 0 + 0.105 334 185 984;
  • 40) 0.105 334 185 984 × 2 = 0 + 0.210 668 371 968;
  • 41) 0.210 668 371 968 × 2 = 0 + 0.421 336 743 936;
  • 42) 0.421 336 743 936 × 2 = 0 + 0.842 673 487 872;
  • 43) 0.842 673 487 872 × 2 = 1 + 0.685 346 975 744;
  • 44) 0.685 346 975 744 × 2 = 1 + 0.370 693 951 488;
  • 45) 0.370 693 951 488 × 2 = 0 + 0.741 387 902 976;
  • 46) 0.741 387 902 976 × 2 = 1 + 0.482 775 805 952;
  • 47) 0.482 775 805 952 × 2 = 0 + 0.965 551 611 904;
  • 48) 0.965 551 611 904 × 2 = 1 + 0.931 103 223 808;
  • 49) 0.931 103 223 808 × 2 = 1 + 0.862 206 447 616;
  • 50) 0.862 206 447 616 × 2 = 1 + 0.724 412 895 232;
  • 51) 0.724 412 895 232 × 2 = 1 + 0.448 825 790 464;
  • 52) 0.448 825 790 464 × 2 = 0 + 0.897 651 580 928;
  • 53) 0.897 651 580 928 × 2 = 1 + 0.795 303 161 856;
  • 54) 0.795 303 161 856 × 2 = 1 + 0.590 606 323 712;
  • 55) 0.590 606 323 712 × 2 = 1 + 0.181 212 647 424;
  • 56) 0.181 212 647 424 × 2 = 0 + 0.362 425 294 848;
  • 57) 0.362 425 294 848 × 2 = 0 + 0.724 850 589 696;
  • 58) 0.724 850 589 696 × 2 = 1 + 0.449 701 179 392;
  • 59) 0.449 701 179 392 × 2 = 0 + 0.899 402 358 784;
  • 60) 0.899 402 358 784 × 2 = 1 + 0.798 804 717 568;
  • 61) 0.798 804 717 568 × 2 = 1 + 0.597 609 435 136;
  • 62) 0.597 609 435 136 × 2 = 1 + 0.195 218 870 272;
  • 63) 0.195 218 870 272 × 2 = 0 + 0.390 437 740 544;
  • 64) 0.390 437 740 544 × 2 = 0 + 0.780 875 481 088;

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:


0.000 282 006 393(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0011 1100 0011 0101 1110 1110 0101 1100(2)

6. Positive number before normalization:

0.000 282 006 393(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0011 1100 0011 0101 1110 1110 0101 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:


0.000 282 006 393(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0011 1100 0011 0101 1110 1110 0101 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0011 1100 0011 0101 1110 1110 0101 1100(2) × 20 =

1.0010 0111 1011 0100 1000 0011 1100 0011 0101 1110 1110 0101 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 1000 0011 1100 0011 0101 1110 1110 0101 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 1000 0011 1100 0011 0101 1110 1110 0101 1100 =

0010 0111 1011 0100 1000 0011 1100 0011 0101 1110 1110 0101 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 1000 0011 1100 0011 0101 1110 1110 0101 1100


Decimal number -0.000 282 006 393 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0011 1100 0011 0101 1110 1110 0101 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