-0.016 738 891 601 562 496 151 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 151(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
-0.016 738 891 601 562 496 151(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:

|-0.016 738 891 601 562 496 151| = 0.016 738 891 601 562 496 151


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: 0.016 738 891 601 562 496 151.

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) 0.016 738 891 601 562 496 151 × 2 = 0 + 0.033 477 783 203 124 992 302;
  • 2) 0.033 477 783 203 124 992 302 × 2 = 0 + 0.066 955 566 406 249 984 604;
  • 3) 0.066 955 566 406 249 984 604 × 2 = 0 + 0.133 911 132 812 499 969 208;
  • 4) 0.133 911 132 812 499 969 208 × 2 = 0 + 0.267 822 265 624 999 938 416;
  • 5) 0.267 822 265 624 999 938 416 × 2 = 0 + 0.535 644 531 249 999 876 832;
  • 6) 0.535 644 531 249 999 876 832 × 2 = 1 + 0.071 289 062 499 999 753 664;
  • 7) 0.071 289 062 499 999 753 664 × 2 = 0 + 0.142 578 124 999 999 507 328;
  • 8) 0.142 578 124 999 999 507 328 × 2 = 0 + 0.285 156 249 999 999 014 656;
  • 9) 0.285 156 249 999 999 014 656 × 2 = 0 + 0.570 312 499 999 998 029 312;
  • 10) 0.570 312 499 999 998 029 312 × 2 = 1 + 0.140 624 999 999 996 058 624;
  • 11) 0.140 624 999 999 996 058 624 × 2 = 0 + 0.281 249 999 999 992 117 248;
  • 12) 0.281 249 999 999 992 117 248 × 2 = 0 + 0.562 499 999 999 984 234 496;
  • 13) 0.562 499 999 999 984 234 496 × 2 = 1 + 0.124 999 999 999 968 468 992;
  • 14) 0.124 999 999 999 968 468 992 × 2 = 0 + 0.249 999 999 999 936 937 984;
  • 15) 0.249 999 999 999 936 937 984 × 2 = 0 + 0.499 999 999 999 873 875 968;
  • 16) 0.499 999 999 999 873 875 968 × 2 = 0 + 0.999 999 999 999 747 751 936;
  • 17) 0.999 999 999 999 747 751 936 × 2 = 1 + 0.999 999 999 999 495 503 872;
  • 18) 0.999 999 999 999 495 503 872 × 2 = 1 + 0.999 999 999 998 991 007 744;
  • 19) 0.999 999 999 998 991 007 744 × 2 = 1 + 0.999 999 999 997 982 015 488;
  • 20) 0.999 999 999 997 982 015 488 × 2 = 1 + 0.999 999 999 995 964 030 976;
  • 21) 0.999 999 999 995 964 030 976 × 2 = 1 + 0.999 999 999 991 928 061 952;
  • 22) 0.999 999 999 991 928 061 952 × 2 = 1 + 0.999 999 999 983 856 123 904;
  • 23) 0.999 999 999 983 856 123 904 × 2 = 1 + 0.999 999 999 967 712 247 808;
  • 24) 0.999 999 999 967 712 247 808 × 2 = 1 + 0.999 999 999 935 424 495 616;
  • 25) 0.999 999 999 935 424 495 616 × 2 = 1 + 0.999 999 999 870 848 991 232;
  • 26) 0.999 999 999 870 848 991 232 × 2 = 1 + 0.999 999 999 741 697 982 464;
  • 27) 0.999 999 999 741 697 982 464 × 2 = 1 + 0.999 999 999 483 395 964 928;
  • 28) 0.999 999 999 483 395 964 928 × 2 = 1 + 0.999 999 998 966 791 929 856;
  • 29) 0.999 999 998 966 791 929 856 × 2 = 1 + 0.999 999 997 933 583 859 712;
  • 30) 0.999 999 997 933 583 859 712 × 2 = 1 + 0.999 999 995 867 167 719 424;
  • 31) 0.999 999 995 867 167 719 424 × 2 = 1 + 0.999 999 991 734 335 438 848;
  • 32) 0.999 999 991 734 335 438 848 × 2 = 1 + 0.999 999 983 468 670 877 696;
  • 33) 0.999 999 983 468 670 877 696 × 2 = 1 + 0.999 999 966 937 341 755 392;
  • 34) 0.999 999 966 937 341 755 392 × 2 = 1 + 0.999 999 933 874 683 510 784;
  • 35) 0.999 999 933 874 683 510 784 × 2 = 1 + 0.999 999 867 749 367 021 568;
  • 36) 0.999 999 867 749 367 021 568 × 2 = 1 + 0.999 999 735 498 734 043 136;
  • 37) 0.999 999 735 498 734 043 136 × 2 = 1 + 0.999 999 470 997 468 086 272;
  • 38) 0.999 999 470 997 468 086 272 × 2 = 1 + 0.999 998 941 994 936 172 544;
  • 39) 0.999 998 941 994 936 172 544 × 2 = 1 + 0.999 997 883 989 872 345 088;
  • 40) 0.999 997 883 989 872 345 088 × 2 = 1 + 0.999 995 767 979 744 690 176;
  • 41) 0.999 995 767 979 744 690 176 × 2 = 1 + 0.999 991 535 959 489 380 352;
  • 42) 0.999 991 535 959 489 380 352 × 2 = 1 + 0.999 983 071 918 978 760 704;
  • 43) 0.999 983 071 918 978 760 704 × 2 = 1 + 0.999 966 143 837 957 521 408;
  • 44) 0.999 966 143 837 957 521 408 × 2 = 1 + 0.999 932 287 675 915 042 816;
  • 45) 0.999 932 287 675 915 042 816 × 2 = 1 + 0.999 864 575 351 830 085 632;
  • 46) 0.999 864 575 351 830 085 632 × 2 = 1 + 0.999 729 150 703 660 171 264;
  • 47) 0.999 729 150 703 660 171 264 × 2 = 1 + 0.999 458 301 407 320 342 528;
  • 48) 0.999 458 301 407 320 342 528 × 2 = 1 + 0.998 916 602 814 640 685 056;
  • 49) 0.998 916 602 814 640 685 056 × 2 = 1 + 0.997 833 205 629 281 370 112;
  • 50) 0.997 833 205 629 281 370 112 × 2 = 1 + 0.995 666 411 258 562 740 224;
  • 51) 0.995 666 411 258 562 740 224 × 2 = 1 + 0.991 332 822 517 125 480 448;
  • 52) 0.991 332 822 517 125 480 448 × 2 = 1 + 0.982 665 645 034 250 960 896;
  • 53) 0.982 665 645 034 250 960 896 × 2 = 1 + 0.965 331 290 068 501 921 792;
  • 54) 0.965 331 290 068 501 921 792 × 2 = 1 + 0.930 662 580 137 003 843 584;
  • 55) 0.930 662 580 137 003 843 584 × 2 = 1 + 0.861 325 160 274 007 687 168;
  • 56) 0.861 325 160 274 007 687 168 × 2 = 1 + 0.722 650 320 548 015 374 336;
  • 57) 0.722 650 320 548 015 374 336 × 2 = 1 + 0.445 300 641 096 030 748 672;
  • 58) 0.445 300 641 096 030 748 672 × 2 = 0 + 0.890 601 282 192 061 497 344;

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:


0.016 738 891 601 562 496 151(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 151(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 151(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1017(10) =

011 1111 1001(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. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 151 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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