-0.000 001 324 439 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 324 439(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
-0.000 001 324 439(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:

|-0.000 001 324 439| = 0.000 001 324 439


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: 0.000 001 324 439.

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) 0.000 001 324 439 × 2 = 0 + 0.000 002 648 878;
  • 2) 0.000 002 648 878 × 2 = 0 + 0.000 005 297 756;
  • 3) 0.000 005 297 756 × 2 = 0 + 0.000 010 595 512;
  • 4) 0.000 010 595 512 × 2 = 0 + 0.000 021 191 024;
  • 5) 0.000 021 191 024 × 2 = 0 + 0.000 042 382 048;
  • 6) 0.000 042 382 048 × 2 = 0 + 0.000 084 764 096;
  • 7) 0.000 084 764 096 × 2 = 0 + 0.000 169 528 192;
  • 8) 0.000 169 528 192 × 2 = 0 + 0.000 339 056 384;
  • 9) 0.000 339 056 384 × 2 = 0 + 0.000 678 112 768;
  • 10) 0.000 678 112 768 × 2 = 0 + 0.001 356 225 536;
  • 11) 0.001 356 225 536 × 2 = 0 + 0.002 712 451 072;
  • 12) 0.002 712 451 072 × 2 = 0 + 0.005 424 902 144;
  • 13) 0.005 424 902 144 × 2 = 0 + 0.010 849 804 288;
  • 14) 0.010 849 804 288 × 2 = 0 + 0.021 699 608 576;
  • 15) 0.021 699 608 576 × 2 = 0 + 0.043 399 217 152;
  • 16) 0.043 399 217 152 × 2 = 0 + 0.086 798 434 304;
  • 17) 0.086 798 434 304 × 2 = 0 + 0.173 596 868 608;
  • 18) 0.173 596 868 608 × 2 = 0 + 0.347 193 737 216;
  • 19) 0.347 193 737 216 × 2 = 0 + 0.694 387 474 432;
  • 20) 0.694 387 474 432 × 2 = 1 + 0.388 774 948 864;
  • 21) 0.388 774 948 864 × 2 = 0 + 0.777 549 897 728;
  • 22) 0.777 549 897 728 × 2 = 1 + 0.555 099 795 456;
  • 23) 0.555 099 795 456 × 2 = 1 + 0.110 199 590 912;
  • 24) 0.110 199 590 912 × 2 = 0 + 0.220 399 181 824;
  • 25) 0.220 399 181 824 × 2 = 0 + 0.440 798 363 648;
  • 26) 0.440 798 363 648 × 2 = 0 + 0.881 596 727 296;
  • 27) 0.881 596 727 296 × 2 = 1 + 0.763 193 454 592;
  • 28) 0.763 193 454 592 × 2 = 1 + 0.526 386 909 184;
  • 29) 0.526 386 909 184 × 2 = 1 + 0.052 773 818 368;
  • 30) 0.052 773 818 368 × 2 = 0 + 0.105 547 636 736;
  • 31) 0.105 547 636 736 × 2 = 0 + 0.211 095 273 472;
  • 32) 0.211 095 273 472 × 2 = 0 + 0.422 190 546 944;
  • 33) 0.422 190 546 944 × 2 = 0 + 0.844 381 093 888;
  • 34) 0.844 381 093 888 × 2 = 1 + 0.688 762 187 776;
  • 35) 0.688 762 187 776 × 2 = 1 + 0.377 524 375 552;
  • 36) 0.377 524 375 552 × 2 = 0 + 0.755 048 751 104;
  • 37) 0.755 048 751 104 × 2 = 1 + 0.510 097 502 208;
  • 38) 0.510 097 502 208 × 2 = 1 + 0.020 195 004 416;
  • 39) 0.020 195 004 416 × 2 = 0 + 0.040 390 008 832;
  • 40) 0.040 390 008 832 × 2 = 0 + 0.080 780 017 664;
  • 41) 0.080 780 017 664 × 2 = 0 + 0.161 560 035 328;
  • 42) 0.161 560 035 328 × 2 = 0 + 0.323 120 070 656;
  • 43) 0.323 120 070 656 × 2 = 0 + 0.646 240 141 312;
  • 44) 0.646 240 141 312 × 2 = 1 + 0.292 480 282 624;
  • 45) 0.292 480 282 624 × 2 = 0 + 0.584 960 565 248;
  • 46) 0.584 960 565 248 × 2 = 1 + 0.169 921 130 496;
  • 47) 0.169 921 130 496 × 2 = 0 + 0.339 842 260 992;
  • 48) 0.339 842 260 992 × 2 = 0 + 0.679 684 521 984;
  • 49) 0.679 684 521 984 × 2 = 1 + 0.359 369 043 968;
  • 50) 0.359 369 043 968 × 2 = 0 + 0.718 738 087 936;
  • 51) 0.718 738 087 936 × 2 = 1 + 0.437 476 175 872;
  • 52) 0.437 476 175 872 × 2 = 0 + 0.874 952 351 744;
  • 53) 0.874 952 351 744 × 2 = 1 + 0.749 904 703 488;
  • 54) 0.749 904 703 488 × 2 = 1 + 0.499 809 406 976;
  • 55) 0.499 809 406 976 × 2 = 0 + 0.999 618 813 952;
  • 56) 0.999 618 813 952 × 2 = 1 + 0.999 237 627 904;
  • 57) 0.999 237 627 904 × 2 = 1 + 0.998 475 255 808;
  • 58) 0.998 475 255 808 × 2 = 1 + 0.996 950 511 616;
  • 59) 0.996 950 511 616 × 2 = 1 + 0.993 901 023 232;
  • 60) 0.993 901 023 232 × 2 = 1 + 0.987 802 046 464;
  • 61) 0.987 802 046 464 × 2 = 1 + 0.975 604 092 928;
  • 62) 0.975 604 092 928 × 2 = 1 + 0.951 208 185 856;
  • 63) 0.951 208 185 856 × 2 = 1 + 0.902 416 371 712;
  • 64) 0.902 416 371 712 × 2 = 1 + 0.804 832 743 424;
  • 65) 0.804 832 743 424 × 2 = 1 + 0.609 665 486 848;
  • 66) 0.609 665 486 848 × 2 = 1 + 0.219 330 973 696;
  • 67) 0.219 330 973 696 × 2 = 0 + 0.438 661 947 392;
  • 68) 0.438 661 947 392 × 2 = 0 + 0.877 323 894 784;
  • 69) 0.877 323 894 784 × 2 = 1 + 0.754 647 789 568;
  • 70) 0.754 647 789 568 × 2 = 1 + 0.509 295 579 136;
  • 71) 0.509 295 579 136 × 2 = 1 + 0.018 591 158 272;
  • 72) 0.018 591 158 272 × 2 = 0 + 0.037 182 316 544;

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:


0.000 001 324 439(10) =

0.0000 0000 0000 0000 0001 0110 0011 1000 0110 1100 0001 0100 1010 1101 1111 1111 1100 1110(2)

6. Positive number before normalization:

0.000 001 324 439(10) =

0.0000 0000 0000 0000 0001 0110 0011 1000 0110 1100 0001 0100 1010 1101 1111 1111 1100 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 324 439(10) =

0.0000 0000 0000 0000 0001 0110 0011 1000 0110 1100 0001 0100 1010 1101 1111 1111 1100 1110(2) =

0.0000 0000 0000 0000 0001 0110 0011 1000 0110 1100 0001 0100 1010 1101 1111 1111 1100 1110(2) × 20 =

1.0110 0011 1000 0110 1100 0001 0100 1010 1101 1111 1111 1100 1110(2) × 2-20


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): -20

Mantissa (not normalized):
1.0110 0011 1000 0110 1100 0001 0100 1010 1101 1111 1111 1100 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-20 + 2(11-1) - 1 =

(-20 + 1 023)(10) =

1 003(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;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1003(10) =

011 1110 1011(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. 0110 0011 1000 0110 1100 0001 0100 1010 1101 1111 1111 1100 1110 =

0110 0011 1000 0110 1100 0001 0100 1010 1101 1111 1111 1100 1110


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 1110 1011

Mantissa (52 bits) =
0110 0011 1000 0110 1100 0001 0100 1010 1101 1111 1111 1100 1110


Decimal number -0.000 001 324 439 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1011 - 0110 0011 1000 0110 1100 0001 0100 1010 1101 1111 1111 1100 1110

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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