-0.000 009 141 962 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 009 141 962(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
-0.000 009 141 962(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:

|-0.000 009 141 962| = 0.000 009 141 962


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: 0.000 009 141 962.

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) 0.000 009 141 962 × 2 = 0 + 0.000 018 283 924;
  • 2) 0.000 018 283 924 × 2 = 0 + 0.000 036 567 848;
  • 3) 0.000 036 567 848 × 2 = 0 + 0.000 073 135 696;
  • 4) 0.000 073 135 696 × 2 = 0 + 0.000 146 271 392;
  • 5) 0.000 146 271 392 × 2 = 0 + 0.000 292 542 784;
  • 6) 0.000 292 542 784 × 2 = 0 + 0.000 585 085 568;
  • 7) 0.000 585 085 568 × 2 = 0 + 0.001 170 171 136;
  • 8) 0.001 170 171 136 × 2 = 0 + 0.002 340 342 272;
  • 9) 0.002 340 342 272 × 2 = 0 + 0.004 680 684 544;
  • 10) 0.004 680 684 544 × 2 = 0 + 0.009 361 369 088;
  • 11) 0.009 361 369 088 × 2 = 0 + 0.018 722 738 176;
  • 12) 0.018 722 738 176 × 2 = 0 + 0.037 445 476 352;
  • 13) 0.037 445 476 352 × 2 = 0 + 0.074 890 952 704;
  • 14) 0.074 890 952 704 × 2 = 0 + 0.149 781 905 408;
  • 15) 0.149 781 905 408 × 2 = 0 + 0.299 563 810 816;
  • 16) 0.299 563 810 816 × 2 = 0 + 0.599 127 621 632;
  • 17) 0.599 127 621 632 × 2 = 1 + 0.198 255 243 264;
  • 18) 0.198 255 243 264 × 2 = 0 + 0.396 510 486 528;
  • 19) 0.396 510 486 528 × 2 = 0 + 0.793 020 973 056;
  • 20) 0.793 020 973 056 × 2 = 1 + 0.586 041 946 112;
  • 21) 0.586 041 946 112 × 2 = 1 + 0.172 083 892 224;
  • 22) 0.172 083 892 224 × 2 = 0 + 0.344 167 784 448;
  • 23) 0.344 167 784 448 × 2 = 0 + 0.688 335 568 896;
  • 24) 0.688 335 568 896 × 2 = 1 + 0.376 671 137 792;
  • 25) 0.376 671 137 792 × 2 = 0 + 0.753 342 275 584;
  • 26) 0.753 342 275 584 × 2 = 1 + 0.506 684 551 168;
  • 27) 0.506 684 551 168 × 2 = 1 + 0.013 369 102 336;
  • 28) 0.013 369 102 336 × 2 = 0 + 0.026 738 204 672;
  • 29) 0.026 738 204 672 × 2 = 0 + 0.053 476 409 344;
  • 30) 0.053 476 409 344 × 2 = 0 + 0.106 952 818 688;
  • 31) 0.106 952 818 688 × 2 = 0 + 0.213 905 637 376;
  • 32) 0.213 905 637 376 × 2 = 0 + 0.427 811 274 752;
  • 33) 0.427 811 274 752 × 2 = 0 + 0.855 622 549 504;
  • 34) 0.855 622 549 504 × 2 = 1 + 0.711 245 099 008;
  • 35) 0.711 245 099 008 × 2 = 1 + 0.422 490 198 016;
  • 36) 0.422 490 198 016 × 2 = 0 + 0.844 980 396 032;
  • 37) 0.844 980 396 032 × 2 = 1 + 0.689 960 792 064;
  • 38) 0.689 960 792 064 × 2 = 1 + 0.379 921 584 128;
  • 39) 0.379 921 584 128 × 2 = 0 + 0.759 843 168 256;
  • 40) 0.759 843 168 256 × 2 = 1 + 0.519 686 336 512;
  • 41) 0.519 686 336 512 × 2 = 1 + 0.039 372 673 024;
  • 42) 0.039 372 673 024 × 2 = 0 + 0.078 745 346 048;
  • 43) 0.078 745 346 048 × 2 = 0 + 0.157 490 692 096;
  • 44) 0.157 490 692 096 × 2 = 0 + 0.314 981 384 192;
  • 45) 0.314 981 384 192 × 2 = 0 + 0.629 962 768 384;
  • 46) 0.629 962 768 384 × 2 = 1 + 0.259 925 536 768;
  • 47) 0.259 925 536 768 × 2 = 0 + 0.519 851 073 536;
  • 48) 0.519 851 073 536 × 2 = 1 + 0.039 702 147 072;
  • 49) 0.039 702 147 072 × 2 = 0 + 0.079 404 294 144;
  • 50) 0.079 404 294 144 × 2 = 0 + 0.158 808 588 288;
  • 51) 0.158 808 588 288 × 2 = 0 + 0.317 617 176 576;
  • 52) 0.317 617 176 576 × 2 = 0 + 0.635 234 353 152;
  • 53) 0.635 234 353 152 × 2 = 1 + 0.270 468 706 304;
  • 54) 0.270 468 706 304 × 2 = 0 + 0.540 937 412 608;
  • 55) 0.540 937 412 608 × 2 = 1 + 0.081 874 825 216;
  • 56) 0.081 874 825 216 × 2 = 0 + 0.163 749 650 432;
  • 57) 0.163 749 650 432 × 2 = 0 + 0.327 499 300 864;
  • 58) 0.327 499 300 864 × 2 = 0 + 0.654 998 601 728;
  • 59) 0.654 998 601 728 × 2 = 1 + 0.309 997 203 456;
  • 60) 0.309 997 203 456 × 2 = 0 + 0.619 994 406 912;
  • 61) 0.619 994 406 912 × 2 = 1 + 0.239 988 813 824;
  • 62) 0.239 988 813 824 × 2 = 0 + 0.479 977 627 648;
  • 63) 0.479 977 627 648 × 2 = 0 + 0.959 955 255 296;
  • 64) 0.959 955 255 296 × 2 = 1 + 0.919 910 510 592;
  • 65) 0.919 910 510 592 × 2 = 1 + 0.839 821 021 184;
  • 66) 0.839 821 021 184 × 2 = 1 + 0.679 642 042 368;
  • 67) 0.679 642 042 368 × 2 = 1 + 0.359 284 084 736;
  • 68) 0.359 284 084 736 × 2 = 0 + 0.718 568 169 472;
  • 69) 0.718 568 169 472 × 2 = 1 + 0.437 136 338 944;

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:


0.000 009 141 962(10) =

0.0000 0000 0000 0000 1001 1001 0110 0000 0110 1101 1000 0101 0000 1010 0010 1001 1110 1(2)

6. Positive number before normalization:

0.000 009 141 962(10) =

0.0000 0000 0000 0000 1001 1001 0110 0000 0110 1101 1000 0101 0000 1010 0010 1001 1110 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 009 141 962(10) =

0.0000 0000 0000 0000 1001 1001 0110 0000 0110 1101 1000 0101 0000 1010 0010 1001 1110 1(2) =

0.0000 0000 0000 0000 1001 1001 0110 0000 0110 1101 1000 0101 0000 1010 0010 1001 1110 1(2) × 20 =

1.0011 0010 1100 0000 1101 1011 0000 1010 0001 0100 0101 0011 1101(2) × 2-17


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): -17

Mantissa (not normalized):
1.0011 0010 1100 0000 1101 1011 0000 1010 0001 0100 0101 0011 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(11-1) - 1 =

(-17 + 1 023)(10) =

1 006(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;
  • 1 006 ÷ 2 = 503 + 0;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 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) =

1006(10) =

011 1110 1110(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. 0011 0010 1100 0000 1101 1011 0000 1010 0001 0100 0101 0011 1101 =

0011 0010 1100 0000 1101 1011 0000 1010 0001 0100 0101 0011 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 1110 1110

Mantissa (52 bits) =
0011 0010 1100 0000 1101 1011 0000 1010 0001 0100 0101 0011 1101


Decimal number -0.000 009 141 962 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0011 0010 1100 0000 1101 1011 0000 1010 0001 0100 0101 0011 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