-0.000 282 006 239 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 239(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
-0.000 282 006 239(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:

|-0.000 282 006 239| = 0.000 282 006 239


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: 0.000 282 006 239.

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) 0.000 282 006 239 × 2 = 0 + 0.000 564 012 478;
  • 2) 0.000 564 012 478 × 2 = 0 + 0.001 128 024 956;
  • 3) 0.001 128 024 956 × 2 = 0 + 0.002 256 049 912;
  • 4) 0.002 256 049 912 × 2 = 0 + 0.004 512 099 824;
  • 5) 0.004 512 099 824 × 2 = 0 + 0.009 024 199 648;
  • 6) 0.009 024 199 648 × 2 = 0 + 0.018 048 399 296;
  • 7) 0.018 048 399 296 × 2 = 0 + 0.036 096 798 592;
  • 8) 0.036 096 798 592 × 2 = 0 + 0.072 193 597 184;
  • 9) 0.072 193 597 184 × 2 = 0 + 0.144 387 194 368;
  • 10) 0.144 387 194 368 × 2 = 0 + 0.288 774 388 736;
  • 11) 0.288 774 388 736 × 2 = 0 + 0.577 548 777 472;
  • 12) 0.577 548 777 472 × 2 = 1 + 0.155 097 554 944;
  • 13) 0.155 097 554 944 × 2 = 0 + 0.310 195 109 888;
  • 14) 0.310 195 109 888 × 2 = 0 + 0.620 390 219 776;
  • 15) 0.620 390 219 776 × 2 = 1 + 0.240 780 439 552;
  • 16) 0.240 780 439 552 × 2 = 0 + 0.481 560 879 104;
  • 17) 0.481 560 879 104 × 2 = 0 + 0.963 121 758 208;
  • 18) 0.963 121 758 208 × 2 = 1 + 0.926 243 516 416;
  • 19) 0.926 243 516 416 × 2 = 1 + 0.852 487 032 832;
  • 20) 0.852 487 032 832 × 2 = 1 + 0.704 974 065 664;
  • 21) 0.704 974 065 664 × 2 = 1 + 0.409 948 131 328;
  • 22) 0.409 948 131 328 × 2 = 0 + 0.819 896 262 656;
  • 23) 0.819 896 262 656 × 2 = 1 + 0.639 792 525 312;
  • 24) 0.639 792 525 312 × 2 = 1 + 0.279 585 050 624;
  • 25) 0.279 585 050 624 × 2 = 0 + 0.559 170 101 248;
  • 26) 0.559 170 101 248 × 2 = 1 + 0.118 340 202 496;
  • 27) 0.118 340 202 496 × 2 = 0 + 0.236 680 404 992;
  • 28) 0.236 680 404 992 × 2 = 0 + 0.473 360 809 984;
  • 29) 0.473 360 809 984 × 2 = 0 + 0.946 721 619 968;
  • 30) 0.946 721 619 968 × 2 = 1 + 0.893 443 239 936;
  • 31) 0.893 443 239 936 × 2 = 1 + 0.786 886 479 872;
  • 32) 0.786 886 479 872 × 2 = 1 + 0.573 772 959 744;
  • 33) 0.573 772 959 744 × 2 = 1 + 0.147 545 919 488;
  • 34) 0.147 545 919 488 × 2 = 0 + 0.295 091 838 976;
  • 35) 0.295 091 838 976 × 2 = 0 + 0.590 183 677 952;
  • 36) 0.590 183 677 952 × 2 = 1 + 0.180 367 355 904;
  • 37) 0.180 367 355 904 × 2 = 0 + 0.360 734 711 808;
  • 38) 0.360 734 711 808 × 2 = 0 + 0.721 469 423 616;
  • 39) 0.721 469 423 616 × 2 = 1 + 0.442 938 847 232;
  • 40) 0.442 938 847 232 × 2 = 0 + 0.885 877 694 464;
  • 41) 0.885 877 694 464 × 2 = 1 + 0.771 755 388 928;
  • 42) 0.771 755 388 928 × 2 = 1 + 0.543 510 777 856;
  • 43) 0.543 510 777 856 × 2 = 1 + 0.087 021 555 712;
  • 44) 0.087 021 555 712 × 2 = 0 + 0.174 043 111 424;
  • 45) 0.174 043 111 424 × 2 = 0 + 0.348 086 222 848;
  • 46) 0.348 086 222 848 × 2 = 0 + 0.696 172 445 696;
  • 47) 0.696 172 445 696 × 2 = 1 + 0.392 344 891 392;
  • 48) 0.392 344 891 392 × 2 = 0 + 0.784 689 782 784;
  • 49) 0.784 689 782 784 × 2 = 1 + 0.569 379 565 568;
  • 50) 0.569 379 565 568 × 2 = 1 + 0.138 759 131 136;
  • 51) 0.138 759 131 136 × 2 = 0 + 0.277 518 262 272;
  • 52) 0.277 518 262 272 × 2 = 0 + 0.555 036 524 544;
  • 53) 0.555 036 524 544 × 2 = 1 + 0.110 073 049 088;
  • 54) 0.110 073 049 088 × 2 = 0 + 0.220 146 098 176;
  • 55) 0.220 146 098 176 × 2 = 0 + 0.440 292 196 352;
  • 56) 0.440 292 196 352 × 2 = 0 + 0.880 584 392 704;
  • 57) 0.880 584 392 704 × 2 = 1 + 0.761 168 785 408;
  • 58) 0.761 168 785 408 × 2 = 1 + 0.522 337 570 816;
  • 59) 0.522 337 570 816 × 2 = 1 + 0.044 675 141 632;
  • 60) 0.044 675 141 632 × 2 = 0 + 0.089 350 283 264;
  • 61) 0.089 350 283 264 × 2 = 0 + 0.178 700 566 528;
  • 62) 0.178 700 566 528 × 2 = 0 + 0.357 401 133 056;
  • 63) 0.357 401 133 056 × 2 = 0 + 0.714 802 266 112;
  • 64) 0.714 802 266 112 × 2 = 1 + 0.429 604 532 224;

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:


0.000 282 006 239(10) =

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

6. Positive number before normalization:

0.000 282 006 239(10) =

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


0.000 282 006 239(10) =

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

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

1.0010 0111 1011 0100 0111 1001 0010 1110 0010 1100 1000 1110 0001(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 0111 1001 0010 1110 0010 1100 1000 1110 0001


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 0111 1001 0010 1110 0010 1100 1000 1110 0001 =

0010 0111 1011 0100 0111 1001 0010 1110 0010 1100 1000 1110 0001


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 0111 1001 0010 1110 0010 1100 1000 1110 0001


Decimal number -0.000 282 006 239 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1001 0010 1110 0010 1100 1000 1110 0001

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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