-0.000 282 005 864 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 864(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
-0.000 282 005 864(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:

|-0.000 282 005 864| = 0.000 282 005 864


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: 0.000 282 005 864.

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) 0.000 282 005 864 × 2 = 0 + 0.000 564 011 728;
  • 2) 0.000 564 011 728 × 2 = 0 + 0.001 128 023 456;
  • 3) 0.001 128 023 456 × 2 = 0 + 0.002 256 046 912;
  • 4) 0.002 256 046 912 × 2 = 0 + 0.004 512 093 824;
  • 5) 0.004 512 093 824 × 2 = 0 + 0.009 024 187 648;
  • 6) 0.009 024 187 648 × 2 = 0 + 0.018 048 375 296;
  • 7) 0.018 048 375 296 × 2 = 0 + 0.036 096 750 592;
  • 8) 0.036 096 750 592 × 2 = 0 + 0.072 193 501 184;
  • 9) 0.072 193 501 184 × 2 = 0 + 0.144 387 002 368;
  • 10) 0.144 387 002 368 × 2 = 0 + 0.288 774 004 736;
  • 11) 0.288 774 004 736 × 2 = 0 + 0.577 548 009 472;
  • 12) 0.577 548 009 472 × 2 = 1 + 0.155 096 018 944;
  • 13) 0.155 096 018 944 × 2 = 0 + 0.310 192 037 888;
  • 14) 0.310 192 037 888 × 2 = 0 + 0.620 384 075 776;
  • 15) 0.620 384 075 776 × 2 = 1 + 0.240 768 151 552;
  • 16) 0.240 768 151 552 × 2 = 0 + 0.481 536 303 104;
  • 17) 0.481 536 303 104 × 2 = 0 + 0.963 072 606 208;
  • 18) 0.963 072 606 208 × 2 = 1 + 0.926 145 212 416;
  • 19) 0.926 145 212 416 × 2 = 1 + 0.852 290 424 832;
  • 20) 0.852 290 424 832 × 2 = 1 + 0.704 580 849 664;
  • 21) 0.704 580 849 664 × 2 = 1 + 0.409 161 699 328;
  • 22) 0.409 161 699 328 × 2 = 0 + 0.818 323 398 656;
  • 23) 0.818 323 398 656 × 2 = 1 + 0.636 646 797 312;
  • 24) 0.636 646 797 312 × 2 = 1 + 0.273 293 594 624;
  • 25) 0.273 293 594 624 × 2 = 0 + 0.546 587 189 248;
  • 26) 0.546 587 189 248 × 2 = 1 + 0.093 174 378 496;
  • 27) 0.093 174 378 496 × 2 = 0 + 0.186 348 756 992;
  • 28) 0.186 348 756 992 × 2 = 0 + 0.372 697 513 984;
  • 29) 0.372 697 513 984 × 2 = 0 + 0.745 395 027 968;
  • 30) 0.745 395 027 968 × 2 = 1 + 0.490 790 055 936;
  • 31) 0.490 790 055 936 × 2 = 0 + 0.981 580 111 872;
  • 32) 0.981 580 111 872 × 2 = 1 + 0.963 160 223 744;
  • 33) 0.963 160 223 744 × 2 = 1 + 0.926 320 447 488;
  • 34) 0.926 320 447 488 × 2 = 1 + 0.852 640 894 976;
  • 35) 0.852 640 894 976 × 2 = 1 + 0.705 281 789 952;
  • 36) 0.705 281 789 952 × 2 = 1 + 0.410 563 579 904;
  • 37) 0.410 563 579 904 × 2 = 0 + 0.821 127 159 808;
  • 38) 0.821 127 159 808 × 2 = 1 + 0.642 254 319 616;
  • 39) 0.642 254 319 616 × 2 = 1 + 0.284 508 639 232;
  • 40) 0.284 508 639 232 × 2 = 0 + 0.569 017 278 464;
  • 41) 0.569 017 278 464 × 2 = 1 + 0.138 034 556 928;
  • 42) 0.138 034 556 928 × 2 = 0 + 0.276 069 113 856;
  • 43) 0.276 069 113 856 × 2 = 0 + 0.552 138 227 712;
  • 44) 0.552 138 227 712 × 2 = 1 + 0.104 276 455 424;
  • 45) 0.104 276 455 424 × 2 = 0 + 0.208 552 910 848;
  • 46) 0.208 552 910 848 × 2 = 0 + 0.417 105 821 696;
  • 47) 0.417 105 821 696 × 2 = 0 + 0.834 211 643 392;
  • 48) 0.834 211 643 392 × 2 = 1 + 0.668 423 286 784;
  • 49) 0.668 423 286 784 × 2 = 1 + 0.336 846 573 568;
  • 50) 0.336 846 573 568 × 2 = 0 + 0.673 693 147 136;
  • 51) 0.673 693 147 136 × 2 = 1 + 0.347 386 294 272;
  • 52) 0.347 386 294 272 × 2 = 0 + 0.694 772 588 544;
  • 53) 0.694 772 588 544 × 2 = 1 + 0.389 545 177 088;
  • 54) 0.389 545 177 088 × 2 = 0 + 0.779 090 354 176;
  • 55) 0.779 090 354 176 × 2 = 1 + 0.558 180 708 352;
  • 56) 0.558 180 708 352 × 2 = 1 + 0.116 361 416 704;
  • 57) 0.116 361 416 704 × 2 = 0 + 0.232 722 833 408;
  • 58) 0.232 722 833 408 × 2 = 0 + 0.465 445 666 816;
  • 59) 0.465 445 666 816 × 2 = 0 + 0.930 891 333 632;
  • 60) 0.930 891 333 632 × 2 = 1 + 0.861 782 667 264;
  • 61) 0.861 782 667 264 × 2 = 1 + 0.723 565 334 528;
  • 62) 0.723 565 334 528 × 2 = 1 + 0.447 130 669 056;
  • 63) 0.447 130 669 056 × 2 = 0 + 0.894 261 338 112;
  • 64) 0.894 261 338 112 × 2 = 1 + 0.788 522 676 224;

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:


0.000 282 005 864(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0110 1001 0001 1010 1011 0001 1101(2)

6. Positive number before normalization:

0.000 282 005 864(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0110 1001 0001 1010 1011 0001 1101(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:


0.000 282 005 864(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0110 1001 0001 1010 1011 0001 1101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 0110 1001 0001 1010 1011 0001 1101(2) × 20 =

1.0010 0111 1011 0100 0101 1111 0110 1001 0001 1010 1011 0001 1101(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 1111 0110 1001 0001 1010 1011 0001 1101


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 1111 0110 1001 0001 1010 1011 0001 1101 =

0010 0111 1011 0100 0101 1111 0110 1001 0001 1010 1011 0001 1101


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 1111 0110 1001 0001 1010 1011 0001 1101


Decimal number -0.000 282 005 864 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 0110 1001 0001 1010 1011 0001 1101

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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