-0.000 282 006 114 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 114(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
-0.000 282 006 114(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:

|-0.000 282 006 114| = 0.000 282 006 114


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: 0.000 282 006 114.

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) 0.000 282 006 114 × 2 = 0 + 0.000 564 012 228;
  • 2) 0.000 564 012 228 × 2 = 0 + 0.001 128 024 456;
  • 3) 0.001 128 024 456 × 2 = 0 + 0.002 256 048 912;
  • 4) 0.002 256 048 912 × 2 = 0 + 0.004 512 097 824;
  • 5) 0.004 512 097 824 × 2 = 0 + 0.009 024 195 648;
  • 6) 0.009 024 195 648 × 2 = 0 + 0.018 048 391 296;
  • 7) 0.018 048 391 296 × 2 = 0 + 0.036 096 782 592;
  • 8) 0.036 096 782 592 × 2 = 0 + 0.072 193 565 184;
  • 9) 0.072 193 565 184 × 2 = 0 + 0.144 387 130 368;
  • 10) 0.144 387 130 368 × 2 = 0 + 0.288 774 260 736;
  • 11) 0.288 774 260 736 × 2 = 0 + 0.577 548 521 472;
  • 12) 0.577 548 521 472 × 2 = 1 + 0.155 097 042 944;
  • 13) 0.155 097 042 944 × 2 = 0 + 0.310 194 085 888;
  • 14) 0.310 194 085 888 × 2 = 0 + 0.620 388 171 776;
  • 15) 0.620 388 171 776 × 2 = 1 + 0.240 776 343 552;
  • 16) 0.240 776 343 552 × 2 = 0 + 0.481 552 687 104;
  • 17) 0.481 552 687 104 × 2 = 0 + 0.963 105 374 208;
  • 18) 0.963 105 374 208 × 2 = 1 + 0.926 210 748 416;
  • 19) 0.926 210 748 416 × 2 = 1 + 0.852 421 496 832;
  • 20) 0.852 421 496 832 × 2 = 1 + 0.704 842 993 664;
  • 21) 0.704 842 993 664 × 2 = 1 + 0.409 685 987 328;
  • 22) 0.409 685 987 328 × 2 = 0 + 0.819 371 974 656;
  • 23) 0.819 371 974 656 × 2 = 1 + 0.638 743 949 312;
  • 24) 0.638 743 949 312 × 2 = 1 + 0.277 487 898 624;
  • 25) 0.277 487 898 624 × 2 = 0 + 0.554 975 797 248;
  • 26) 0.554 975 797 248 × 2 = 1 + 0.109 951 594 496;
  • 27) 0.109 951 594 496 × 2 = 0 + 0.219 903 188 992;
  • 28) 0.219 903 188 992 × 2 = 0 + 0.439 806 377 984;
  • 29) 0.439 806 377 984 × 2 = 0 + 0.879 612 755 968;
  • 30) 0.879 612 755 968 × 2 = 1 + 0.759 225 511 936;
  • 31) 0.759 225 511 936 × 2 = 1 + 0.518 451 023 872;
  • 32) 0.518 451 023 872 × 2 = 1 + 0.036 902 047 744;
  • 33) 0.036 902 047 744 × 2 = 0 + 0.073 804 095 488;
  • 34) 0.073 804 095 488 × 2 = 0 + 0.147 608 190 976;
  • 35) 0.147 608 190 976 × 2 = 0 + 0.295 216 381 952;
  • 36) 0.295 216 381 952 × 2 = 0 + 0.590 432 763 904;
  • 37) 0.590 432 763 904 × 2 = 1 + 0.180 865 527 808;
  • 38) 0.180 865 527 808 × 2 = 0 + 0.361 731 055 616;
  • 39) 0.361 731 055 616 × 2 = 0 + 0.723 462 111 232;
  • 40) 0.723 462 111 232 × 2 = 1 + 0.446 924 222 464;
  • 41) 0.446 924 222 464 × 2 = 0 + 0.893 848 444 928;
  • 42) 0.893 848 444 928 × 2 = 1 + 0.787 696 889 856;
  • 43) 0.787 696 889 856 × 2 = 1 + 0.575 393 779 712;
  • 44) 0.575 393 779 712 × 2 = 1 + 0.150 787 559 424;
  • 45) 0.150 787 559 424 × 2 = 0 + 0.301 575 118 848;
  • 46) 0.301 575 118 848 × 2 = 0 + 0.603 150 237 696;
  • 47) 0.603 150 237 696 × 2 = 1 + 0.206 300 475 392;
  • 48) 0.206 300 475 392 × 2 = 0 + 0.412 600 950 784;
  • 49) 0.412 600 950 784 × 2 = 0 + 0.825 201 901 568;
  • 50) 0.825 201 901 568 × 2 = 1 + 0.650 403 803 136;
  • 51) 0.650 403 803 136 × 2 = 1 + 0.300 807 606 272;
  • 52) 0.300 807 606 272 × 2 = 0 + 0.601 615 212 544;
  • 53) 0.601 615 212 544 × 2 = 1 + 0.203 230 425 088;
  • 54) 0.203 230 425 088 × 2 = 0 + 0.406 460 850 176;
  • 55) 0.406 460 850 176 × 2 = 0 + 0.812 921 700 352;
  • 56) 0.812 921 700 352 × 2 = 1 + 0.625 843 400 704;
  • 57) 0.625 843 400 704 × 2 = 1 + 0.251 686 801 408;
  • 58) 0.251 686 801 408 × 2 = 0 + 0.503 373 602 816;
  • 59) 0.503 373 602 816 × 2 = 1 + 0.006 747 205 632;
  • 60) 0.006 747 205 632 × 2 = 0 + 0.013 494 411 264;
  • 61) 0.013 494 411 264 × 2 = 0 + 0.026 988 822 528;
  • 62) 0.026 988 822 528 × 2 = 0 + 0.053 977 645 056;
  • 63) 0.053 977 645 056 × 2 = 0 + 0.107 955 290 112;
  • 64) 0.107 955 290 112 × 2 = 0 + 0.215 910 580 224;

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:


0.000 282 006 114(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0000 1001 0111 0010 0110 1001 1010 0000(2)

6. Positive number before normalization:

0.000 282 006 114(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0000 1001 0111 0010 0110 1001 1010 0000(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:


0.000 282 006 114(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0000 1001 0111 0010 0110 1001 1010 0000(2) =

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

1.0010 0111 1011 0100 0111 0000 1001 0111 0010 0110 1001 1010 0000(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 0000 1001 0111 0010 0110 1001 1010 0000


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 0000 1001 0111 0010 0110 1001 1010 0000 =

0010 0111 1011 0100 0111 0000 1001 0111 0010 0110 1001 1010 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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0111 0000 1001 0111 0010 0110 1001 1010 0000


Decimal number -0.000 282 006 114 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 0000 1001 0111 0010 0110 1001 1010 0000

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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