-0.000 282 005 902 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 902(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
-0.000 282 005 902(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:

|-0.000 282 005 902| = 0.000 282 005 902


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: 0.000 282 005 902.

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) 0.000 282 005 902 × 2 = 0 + 0.000 564 011 804;
  • 2) 0.000 564 011 804 × 2 = 0 + 0.001 128 023 608;
  • 3) 0.001 128 023 608 × 2 = 0 + 0.002 256 047 216;
  • 4) 0.002 256 047 216 × 2 = 0 + 0.004 512 094 432;
  • 5) 0.004 512 094 432 × 2 = 0 + 0.009 024 188 864;
  • 6) 0.009 024 188 864 × 2 = 0 + 0.018 048 377 728;
  • 7) 0.018 048 377 728 × 2 = 0 + 0.036 096 755 456;
  • 8) 0.036 096 755 456 × 2 = 0 + 0.072 193 510 912;
  • 9) 0.072 193 510 912 × 2 = 0 + 0.144 387 021 824;
  • 10) 0.144 387 021 824 × 2 = 0 + 0.288 774 043 648;
  • 11) 0.288 774 043 648 × 2 = 0 + 0.577 548 087 296;
  • 12) 0.577 548 087 296 × 2 = 1 + 0.155 096 174 592;
  • 13) 0.155 096 174 592 × 2 = 0 + 0.310 192 349 184;
  • 14) 0.310 192 349 184 × 2 = 0 + 0.620 384 698 368;
  • 15) 0.620 384 698 368 × 2 = 1 + 0.240 769 396 736;
  • 16) 0.240 769 396 736 × 2 = 0 + 0.481 538 793 472;
  • 17) 0.481 538 793 472 × 2 = 0 + 0.963 077 586 944;
  • 18) 0.963 077 586 944 × 2 = 1 + 0.926 155 173 888;
  • 19) 0.926 155 173 888 × 2 = 1 + 0.852 310 347 776;
  • 20) 0.852 310 347 776 × 2 = 1 + 0.704 620 695 552;
  • 21) 0.704 620 695 552 × 2 = 1 + 0.409 241 391 104;
  • 22) 0.409 241 391 104 × 2 = 0 + 0.818 482 782 208;
  • 23) 0.818 482 782 208 × 2 = 1 + 0.636 965 564 416;
  • 24) 0.636 965 564 416 × 2 = 1 + 0.273 931 128 832;
  • 25) 0.273 931 128 832 × 2 = 0 + 0.547 862 257 664;
  • 26) 0.547 862 257 664 × 2 = 1 + 0.095 724 515 328;
  • 27) 0.095 724 515 328 × 2 = 0 + 0.191 449 030 656;
  • 28) 0.191 449 030 656 × 2 = 0 + 0.382 898 061 312;
  • 29) 0.382 898 061 312 × 2 = 0 + 0.765 796 122 624;
  • 30) 0.765 796 122 624 × 2 = 1 + 0.531 592 245 248;
  • 31) 0.531 592 245 248 × 2 = 1 + 0.063 184 490 496;
  • 32) 0.063 184 490 496 × 2 = 0 + 0.126 368 980 992;
  • 33) 0.126 368 980 992 × 2 = 0 + 0.252 737 961 984;
  • 34) 0.252 737 961 984 × 2 = 0 + 0.505 475 923 968;
  • 35) 0.505 475 923 968 × 2 = 1 + 0.010 951 847 936;
  • 36) 0.010 951 847 936 × 2 = 0 + 0.021 903 695 872;
  • 37) 0.021 903 695 872 × 2 = 0 + 0.043 807 391 744;
  • 38) 0.043 807 391 744 × 2 = 0 + 0.087 614 783 488;
  • 39) 0.087 614 783 488 × 2 = 0 + 0.175 229 566 976;
  • 40) 0.175 229 566 976 × 2 = 0 + 0.350 459 133 952;
  • 41) 0.350 459 133 952 × 2 = 0 + 0.700 918 267 904;
  • 42) 0.700 918 267 904 × 2 = 1 + 0.401 836 535 808;
  • 43) 0.401 836 535 808 × 2 = 0 + 0.803 673 071 616;
  • 44) 0.803 673 071 616 × 2 = 1 + 0.607 346 143 232;
  • 45) 0.607 346 143 232 × 2 = 1 + 0.214 692 286 464;
  • 46) 0.214 692 286 464 × 2 = 0 + 0.429 384 572 928;
  • 47) 0.429 384 572 928 × 2 = 0 + 0.858 769 145 856;
  • 48) 0.858 769 145 856 × 2 = 1 + 0.717 538 291 712;
  • 49) 0.717 538 291 712 × 2 = 1 + 0.435 076 583 424;
  • 50) 0.435 076 583 424 × 2 = 0 + 0.870 153 166 848;
  • 51) 0.870 153 166 848 × 2 = 1 + 0.740 306 333 696;
  • 52) 0.740 306 333 696 × 2 = 1 + 0.480 612 667 392;
  • 53) 0.480 612 667 392 × 2 = 0 + 0.961 225 334 784;
  • 54) 0.961 225 334 784 × 2 = 1 + 0.922 450 669 568;
  • 55) 0.922 450 669 568 × 2 = 1 + 0.844 901 339 136;
  • 56) 0.844 901 339 136 × 2 = 1 + 0.689 802 678 272;
  • 57) 0.689 802 678 272 × 2 = 1 + 0.379 605 356 544;
  • 58) 0.379 605 356 544 × 2 = 0 + 0.759 210 713 088;
  • 59) 0.759 210 713 088 × 2 = 1 + 0.518 421 426 176;
  • 60) 0.518 421 426 176 × 2 = 1 + 0.036 842 852 352;
  • 61) 0.036 842 852 352 × 2 = 0 + 0.073 685 704 704;
  • 62) 0.073 685 704 704 × 2 = 0 + 0.147 371 409 408;
  • 63) 0.147 371 409 408 × 2 = 0 + 0.294 742 818 816;
  • 64) 0.294 742 818 816 × 2 = 0 + 0.589 485 637 632;

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:


0.000 282 005 902(10) =

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

6. Positive number before normalization:

0.000 282 005 902(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0000 0101 1001 1011 0111 1011 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:


0.000 282 005 902(10) =

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

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

1.0010 0111 1011 0100 0110 0010 0000 0101 1001 1011 0111 1011 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 0110 0010 0000 0101 1001 1011 0111 1011 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 0110 0010 0000 0101 1001 1011 0111 1011 0000 =

0010 0111 1011 0100 0110 0010 0000 0101 1001 1011 0111 1011 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 0110 0010 0000 0101 1001 1011 0111 1011 0000


Decimal number -0.000 282 005 902 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0000 0101 1001 1011 0111 1011 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