-0.000 282 005 914 537 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 537(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
-0.000 282 005 914 537(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:

|-0.000 282 005 914 537| = 0.000 282 005 914 537


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: 0.000 282 005 914 537.

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) 0.000 282 005 914 537 × 2 = 0 + 0.000 564 011 829 074;
  • 2) 0.000 564 011 829 074 × 2 = 0 + 0.001 128 023 658 148;
  • 3) 0.001 128 023 658 148 × 2 = 0 + 0.002 256 047 316 296;
  • 4) 0.002 256 047 316 296 × 2 = 0 + 0.004 512 094 632 592;
  • 5) 0.004 512 094 632 592 × 2 = 0 + 0.009 024 189 265 184;
  • 6) 0.009 024 189 265 184 × 2 = 0 + 0.018 048 378 530 368;
  • 7) 0.018 048 378 530 368 × 2 = 0 + 0.036 096 757 060 736;
  • 8) 0.036 096 757 060 736 × 2 = 0 + 0.072 193 514 121 472;
  • 9) 0.072 193 514 121 472 × 2 = 0 + 0.144 387 028 242 944;
  • 10) 0.144 387 028 242 944 × 2 = 0 + 0.288 774 056 485 888;
  • 11) 0.288 774 056 485 888 × 2 = 0 + 0.577 548 112 971 776;
  • 12) 0.577 548 112 971 776 × 2 = 1 + 0.155 096 225 943 552;
  • 13) 0.155 096 225 943 552 × 2 = 0 + 0.310 192 451 887 104;
  • 14) 0.310 192 451 887 104 × 2 = 0 + 0.620 384 903 774 208;
  • 15) 0.620 384 903 774 208 × 2 = 1 + 0.240 769 807 548 416;
  • 16) 0.240 769 807 548 416 × 2 = 0 + 0.481 539 615 096 832;
  • 17) 0.481 539 615 096 832 × 2 = 0 + 0.963 079 230 193 664;
  • 18) 0.963 079 230 193 664 × 2 = 1 + 0.926 158 460 387 328;
  • 19) 0.926 158 460 387 328 × 2 = 1 + 0.852 316 920 774 656;
  • 20) 0.852 316 920 774 656 × 2 = 1 + 0.704 633 841 549 312;
  • 21) 0.704 633 841 549 312 × 2 = 1 + 0.409 267 683 098 624;
  • 22) 0.409 267 683 098 624 × 2 = 0 + 0.818 535 366 197 248;
  • 23) 0.818 535 366 197 248 × 2 = 1 + 0.637 070 732 394 496;
  • 24) 0.637 070 732 394 496 × 2 = 1 + 0.274 141 464 788 992;
  • 25) 0.274 141 464 788 992 × 2 = 0 + 0.548 282 929 577 984;
  • 26) 0.548 282 929 577 984 × 2 = 1 + 0.096 565 859 155 968;
  • 27) 0.096 565 859 155 968 × 2 = 0 + 0.193 131 718 311 936;
  • 28) 0.193 131 718 311 936 × 2 = 0 + 0.386 263 436 623 872;
  • 29) 0.386 263 436 623 872 × 2 = 0 + 0.772 526 873 247 744;
  • 30) 0.772 526 873 247 744 × 2 = 1 + 0.545 053 746 495 488;
  • 31) 0.545 053 746 495 488 × 2 = 1 + 0.090 107 492 990 976;
  • 32) 0.090 107 492 990 976 × 2 = 0 + 0.180 214 985 981 952;
  • 33) 0.180 214 985 981 952 × 2 = 0 + 0.360 429 971 963 904;
  • 34) 0.360 429 971 963 904 × 2 = 0 + 0.720 859 943 927 808;
  • 35) 0.720 859 943 927 808 × 2 = 1 + 0.441 719 887 855 616;
  • 36) 0.441 719 887 855 616 × 2 = 0 + 0.883 439 775 711 232;
  • 37) 0.883 439 775 711 232 × 2 = 1 + 0.766 879 551 422 464;
  • 38) 0.766 879 551 422 464 × 2 = 1 + 0.533 759 102 844 928;
  • 39) 0.533 759 102 844 928 × 2 = 1 + 0.067 518 205 689 856;
  • 40) 0.067 518 205 689 856 × 2 = 0 + 0.135 036 411 379 712;
  • 41) 0.135 036 411 379 712 × 2 = 0 + 0.270 072 822 759 424;
  • 42) 0.270 072 822 759 424 × 2 = 0 + 0.540 145 645 518 848;
  • 43) 0.540 145 645 518 848 × 2 = 1 + 0.080 291 291 037 696;
  • 44) 0.080 291 291 037 696 × 2 = 0 + 0.160 582 582 075 392;
  • 45) 0.160 582 582 075 392 × 2 = 0 + 0.321 165 164 150 784;
  • 46) 0.321 165 164 150 784 × 2 = 0 + 0.642 330 328 301 568;
  • 47) 0.642 330 328 301 568 × 2 = 1 + 0.284 660 656 603 136;
  • 48) 0.284 660 656 603 136 × 2 = 0 + 0.569 321 313 206 272;
  • 49) 0.569 321 313 206 272 × 2 = 1 + 0.138 642 626 412 544;
  • 50) 0.138 642 626 412 544 × 2 = 0 + 0.277 285 252 825 088;
  • 51) 0.277 285 252 825 088 × 2 = 0 + 0.554 570 505 650 176;
  • 52) 0.554 570 505 650 176 × 2 = 1 + 0.109 141 011 300 352;
  • 53) 0.109 141 011 300 352 × 2 = 0 + 0.218 282 022 600 704;
  • 54) 0.218 282 022 600 704 × 2 = 0 + 0.436 564 045 201 408;
  • 55) 0.436 564 045 201 408 × 2 = 0 + 0.873 128 090 402 816;
  • 56) 0.873 128 090 402 816 × 2 = 1 + 0.746 256 180 805 632;
  • 57) 0.746 256 180 805 632 × 2 = 1 + 0.492 512 361 611 264;
  • 58) 0.492 512 361 611 264 × 2 = 0 + 0.985 024 723 222 528;
  • 59) 0.985 024 723 222 528 × 2 = 1 + 0.970 049 446 445 056;
  • 60) 0.970 049 446 445 056 × 2 = 1 + 0.940 098 892 890 112;
  • 61) 0.940 098 892 890 112 × 2 = 1 + 0.880 197 785 780 224;
  • 62) 0.880 197 785 780 224 × 2 = 1 + 0.760 395 571 560 448;
  • 63) 0.760 395 571 560 448 × 2 = 1 + 0.520 791 143 120 896;
  • 64) 0.520 791 143 120 896 × 2 = 1 + 0.041 582 286 241 792;

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:


0.000 282 005 914 537(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0010 1001 0001 1011 1111(2)

6. Positive number before normalization:

0.000 282 005 914 537(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0010 1001 0001 1011 1111(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:


0.000 282 005 914 537(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0010 1001 0001 1011 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0010 1001 0001 1011 1111(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0010 0010 1001 0001 1011 1111(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 0010 1110 0010 0010 1001 0001 1011 1111


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 0010 1110 0010 0010 1001 0001 1011 1111 =

0010 0111 1011 0100 0110 0010 1110 0010 0010 1001 0001 1011 1111


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 0010 1110 0010 0010 1001 0001 1011 1111


Decimal number -0.000 282 005 914 537 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0010 0010 1001 0001 1011 1111

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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