-0.016 738 891 601 562 497 123 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 497 123(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
-0.016 738 891 601 562 497 123(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:

|-0.016 738 891 601 562 497 123| = 0.016 738 891 601 562 497 123


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: 0.016 738 891 601 562 497 123.

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) 0.016 738 891 601 562 497 123 × 2 = 0 + 0.033 477 783 203 124 994 246;
  • 2) 0.033 477 783 203 124 994 246 × 2 = 0 + 0.066 955 566 406 249 988 492;
  • 3) 0.066 955 566 406 249 988 492 × 2 = 0 + 0.133 911 132 812 499 976 984;
  • 4) 0.133 911 132 812 499 976 984 × 2 = 0 + 0.267 822 265 624 999 953 968;
  • 5) 0.267 822 265 624 999 953 968 × 2 = 0 + 0.535 644 531 249 999 907 936;
  • 6) 0.535 644 531 249 999 907 936 × 2 = 1 + 0.071 289 062 499 999 815 872;
  • 7) 0.071 289 062 499 999 815 872 × 2 = 0 + 0.142 578 124 999 999 631 744;
  • 8) 0.142 578 124 999 999 631 744 × 2 = 0 + 0.285 156 249 999 999 263 488;
  • 9) 0.285 156 249 999 999 263 488 × 2 = 0 + 0.570 312 499 999 998 526 976;
  • 10) 0.570 312 499 999 998 526 976 × 2 = 1 + 0.140 624 999 999 997 053 952;
  • 11) 0.140 624 999 999 997 053 952 × 2 = 0 + 0.281 249 999 999 994 107 904;
  • 12) 0.281 249 999 999 994 107 904 × 2 = 0 + 0.562 499 999 999 988 215 808;
  • 13) 0.562 499 999 999 988 215 808 × 2 = 1 + 0.124 999 999 999 976 431 616;
  • 14) 0.124 999 999 999 976 431 616 × 2 = 0 + 0.249 999 999 999 952 863 232;
  • 15) 0.249 999 999 999 952 863 232 × 2 = 0 + 0.499 999 999 999 905 726 464;
  • 16) 0.499 999 999 999 905 726 464 × 2 = 0 + 0.999 999 999 999 811 452 928;
  • 17) 0.999 999 999 999 811 452 928 × 2 = 1 + 0.999 999 999 999 622 905 856;
  • 18) 0.999 999 999 999 622 905 856 × 2 = 1 + 0.999 999 999 999 245 811 712;
  • 19) 0.999 999 999 999 245 811 712 × 2 = 1 + 0.999 999 999 998 491 623 424;
  • 20) 0.999 999 999 998 491 623 424 × 2 = 1 + 0.999 999 999 996 983 246 848;
  • 21) 0.999 999 999 996 983 246 848 × 2 = 1 + 0.999 999 999 993 966 493 696;
  • 22) 0.999 999 999 993 966 493 696 × 2 = 1 + 0.999 999 999 987 932 987 392;
  • 23) 0.999 999 999 987 932 987 392 × 2 = 1 + 0.999 999 999 975 865 974 784;
  • 24) 0.999 999 999 975 865 974 784 × 2 = 1 + 0.999 999 999 951 731 949 568;
  • 25) 0.999 999 999 951 731 949 568 × 2 = 1 + 0.999 999 999 903 463 899 136;
  • 26) 0.999 999 999 903 463 899 136 × 2 = 1 + 0.999 999 999 806 927 798 272;
  • 27) 0.999 999 999 806 927 798 272 × 2 = 1 + 0.999 999 999 613 855 596 544;
  • 28) 0.999 999 999 613 855 596 544 × 2 = 1 + 0.999 999 999 227 711 193 088;
  • 29) 0.999 999 999 227 711 193 088 × 2 = 1 + 0.999 999 998 455 422 386 176;
  • 30) 0.999 999 998 455 422 386 176 × 2 = 1 + 0.999 999 996 910 844 772 352;
  • 31) 0.999 999 996 910 844 772 352 × 2 = 1 + 0.999 999 993 821 689 544 704;
  • 32) 0.999 999 993 821 689 544 704 × 2 = 1 + 0.999 999 987 643 379 089 408;
  • 33) 0.999 999 987 643 379 089 408 × 2 = 1 + 0.999 999 975 286 758 178 816;
  • 34) 0.999 999 975 286 758 178 816 × 2 = 1 + 0.999 999 950 573 516 357 632;
  • 35) 0.999 999 950 573 516 357 632 × 2 = 1 + 0.999 999 901 147 032 715 264;
  • 36) 0.999 999 901 147 032 715 264 × 2 = 1 + 0.999 999 802 294 065 430 528;
  • 37) 0.999 999 802 294 065 430 528 × 2 = 1 + 0.999 999 604 588 130 861 056;
  • 38) 0.999 999 604 588 130 861 056 × 2 = 1 + 0.999 999 209 176 261 722 112;
  • 39) 0.999 999 209 176 261 722 112 × 2 = 1 + 0.999 998 418 352 523 444 224;
  • 40) 0.999 998 418 352 523 444 224 × 2 = 1 + 0.999 996 836 705 046 888 448;
  • 41) 0.999 996 836 705 046 888 448 × 2 = 1 + 0.999 993 673 410 093 776 896;
  • 42) 0.999 993 673 410 093 776 896 × 2 = 1 + 0.999 987 346 820 187 553 792;
  • 43) 0.999 987 346 820 187 553 792 × 2 = 1 + 0.999 974 693 640 375 107 584;
  • 44) 0.999 974 693 640 375 107 584 × 2 = 1 + 0.999 949 387 280 750 215 168;
  • 45) 0.999 949 387 280 750 215 168 × 2 = 1 + 0.999 898 774 561 500 430 336;
  • 46) 0.999 898 774 561 500 430 336 × 2 = 1 + 0.999 797 549 123 000 860 672;
  • 47) 0.999 797 549 123 000 860 672 × 2 = 1 + 0.999 595 098 246 001 721 344;
  • 48) 0.999 595 098 246 001 721 344 × 2 = 1 + 0.999 190 196 492 003 442 688;
  • 49) 0.999 190 196 492 003 442 688 × 2 = 1 + 0.998 380 392 984 006 885 376;
  • 50) 0.998 380 392 984 006 885 376 × 2 = 1 + 0.996 760 785 968 013 770 752;
  • 51) 0.996 760 785 968 013 770 752 × 2 = 1 + 0.993 521 571 936 027 541 504;
  • 52) 0.993 521 571 936 027 541 504 × 2 = 1 + 0.987 043 143 872 055 083 008;
  • 53) 0.987 043 143 872 055 083 008 × 2 = 1 + 0.974 086 287 744 110 166 016;
  • 54) 0.974 086 287 744 110 166 016 × 2 = 1 + 0.948 172 575 488 220 332 032;
  • 55) 0.948 172 575 488 220 332 032 × 2 = 1 + 0.896 345 150 976 440 664 064;
  • 56) 0.896 345 150 976 440 664 064 × 2 = 1 + 0.792 690 301 952 881 328 128;
  • 57) 0.792 690 301 952 881 328 128 × 2 = 1 + 0.585 380 603 905 762 656 256;
  • 58) 0.585 380 603 905 762 656 256 × 2 = 1 + 0.170 761 207 811 525 312 512;

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:


0.016 738 891 601 562 497 123(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 497 123(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(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:


0.016 738 891 601 562 497 123(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(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 1111


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 1111 =

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


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 1111


Decimal number -0.016 738 891 601 562 497 123 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 1111

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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