-0.000 282 005 717 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 717(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
-0.000 282 005 717(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:

|-0.000 282 005 717| = 0.000 282 005 717


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: 0.000 282 005 717.

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) 0.000 282 005 717 × 2 = 0 + 0.000 564 011 434;
  • 2) 0.000 564 011 434 × 2 = 0 + 0.001 128 022 868;
  • 3) 0.001 128 022 868 × 2 = 0 + 0.002 256 045 736;
  • 4) 0.002 256 045 736 × 2 = 0 + 0.004 512 091 472;
  • 5) 0.004 512 091 472 × 2 = 0 + 0.009 024 182 944;
  • 6) 0.009 024 182 944 × 2 = 0 + 0.018 048 365 888;
  • 7) 0.018 048 365 888 × 2 = 0 + 0.036 096 731 776;
  • 8) 0.036 096 731 776 × 2 = 0 + 0.072 193 463 552;
  • 9) 0.072 193 463 552 × 2 = 0 + 0.144 386 927 104;
  • 10) 0.144 386 927 104 × 2 = 0 + 0.288 773 854 208;
  • 11) 0.288 773 854 208 × 2 = 0 + 0.577 547 708 416;
  • 12) 0.577 547 708 416 × 2 = 1 + 0.155 095 416 832;
  • 13) 0.155 095 416 832 × 2 = 0 + 0.310 190 833 664;
  • 14) 0.310 190 833 664 × 2 = 0 + 0.620 381 667 328;
  • 15) 0.620 381 667 328 × 2 = 1 + 0.240 763 334 656;
  • 16) 0.240 763 334 656 × 2 = 0 + 0.481 526 669 312;
  • 17) 0.481 526 669 312 × 2 = 0 + 0.963 053 338 624;
  • 18) 0.963 053 338 624 × 2 = 1 + 0.926 106 677 248;
  • 19) 0.926 106 677 248 × 2 = 1 + 0.852 213 354 496;
  • 20) 0.852 213 354 496 × 2 = 1 + 0.704 426 708 992;
  • 21) 0.704 426 708 992 × 2 = 1 + 0.408 853 417 984;
  • 22) 0.408 853 417 984 × 2 = 0 + 0.817 706 835 968;
  • 23) 0.817 706 835 968 × 2 = 1 + 0.635 413 671 936;
  • 24) 0.635 413 671 936 × 2 = 1 + 0.270 827 343 872;
  • 25) 0.270 827 343 872 × 2 = 0 + 0.541 654 687 744;
  • 26) 0.541 654 687 744 × 2 = 1 + 0.083 309 375 488;
  • 27) 0.083 309 375 488 × 2 = 0 + 0.166 618 750 976;
  • 28) 0.166 618 750 976 × 2 = 0 + 0.333 237 501 952;
  • 29) 0.333 237 501 952 × 2 = 0 + 0.666 475 003 904;
  • 30) 0.666 475 003 904 × 2 = 1 + 0.332 950 007 808;
  • 31) 0.332 950 007 808 × 2 = 0 + 0.665 900 015 616;
  • 32) 0.665 900 015 616 × 2 = 1 + 0.331 800 031 232;
  • 33) 0.331 800 031 232 × 2 = 0 + 0.663 600 062 464;
  • 34) 0.663 600 062 464 × 2 = 1 + 0.327 200 124 928;
  • 35) 0.327 200 124 928 × 2 = 0 + 0.654 400 249 856;
  • 36) 0.654 400 249 856 × 2 = 1 + 0.308 800 499 712;
  • 37) 0.308 800 499 712 × 2 = 0 + 0.617 600 999 424;
  • 38) 0.617 600 999 424 × 2 = 1 + 0.235 201 998 848;
  • 39) 0.235 201 998 848 × 2 = 0 + 0.470 403 997 696;
  • 40) 0.470 403 997 696 × 2 = 0 + 0.940 807 995 392;
  • 41) 0.940 807 995 392 × 2 = 1 + 0.881 615 990 784;
  • 42) 0.881 615 990 784 × 2 = 1 + 0.763 231 981 568;
  • 43) 0.763 231 981 568 × 2 = 1 + 0.526 463 963 136;
  • 44) 0.526 463 963 136 × 2 = 1 + 0.052 927 926 272;
  • 45) 0.052 927 926 272 × 2 = 0 + 0.105 855 852 544;
  • 46) 0.105 855 852 544 × 2 = 0 + 0.211 711 705 088;
  • 47) 0.211 711 705 088 × 2 = 0 + 0.423 423 410 176;
  • 48) 0.423 423 410 176 × 2 = 0 + 0.846 846 820 352;
  • 49) 0.846 846 820 352 × 2 = 1 + 0.693 693 640 704;
  • 50) 0.693 693 640 704 × 2 = 1 + 0.387 387 281 408;
  • 51) 0.387 387 281 408 × 2 = 0 + 0.774 774 562 816;
  • 52) 0.774 774 562 816 × 2 = 1 + 0.549 549 125 632;
  • 53) 0.549 549 125 632 × 2 = 1 + 0.099 098 251 264;
  • 54) 0.099 098 251 264 × 2 = 0 + 0.198 196 502 528;
  • 55) 0.198 196 502 528 × 2 = 0 + 0.396 393 005 056;
  • 56) 0.396 393 005 056 × 2 = 0 + 0.792 786 010 112;
  • 57) 0.792 786 010 112 × 2 = 1 + 0.585 572 020 224;
  • 58) 0.585 572 020 224 × 2 = 1 + 0.171 144 040 448;
  • 59) 0.171 144 040 448 × 2 = 0 + 0.342 288 080 896;
  • 60) 0.342 288 080 896 × 2 = 0 + 0.684 576 161 792;
  • 61) 0.684 576 161 792 × 2 = 1 + 0.369 152 323 584;
  • 62) 0.369 152 323 584 × 2 = 0 + 0.738 304 647 168;
  • 63) 0.738 304 647 168 × 2 = 1 + 0.476 609 294 336;
  • 64) 0.476 609 294 336 × 2 = 0 + 0.953 218 588 672;

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:


0.000 282 005 717(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 0100 1111 0000 1101 1000 1100 1010(2)

6. Positive number before normalization:

0.000 282 005 717(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 0100 1111 0000 1101 1000 1100 1010(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:


0.000 282 005 717(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 0100 1111 0000 1101 1000 1100 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 0100 1111 0000 1101 1000 1100 1010(2) × 20 =

1.0010 0111 1011 0100 0101 0101 0100 1111 0000 1101 1000 1100 1010(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 0101 0101 0100 1111 0000 1101 1000 1100 1010


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 0101 0101 0100 1111 0000 1101 1000 1100 1010 =

0010 0111 1011 0100 0101 0101 0100 1111 0000 1101 1000 1100 1010


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 0101 0101 0100 1111 0000 1101 1000 1100 1010


Decimal number -0.000 282 005 717 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0101 0100 1111 0000 1101 1000 1100 1010

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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