-0.016 738 891 601 562 482 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 482 3(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
-0.016 738 891 601 562 482 3(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:

|-0.016 738 891 601 562 482 3| = 0.016 738 891 601 562 482 3


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: 0.016 738 891 601 562 482 3.

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) 0.016 738 891 601 562 482 3 × 2 = 0 + 0.033 477 783 203 124 964 6;
  • 2) 0.033 477 783 203 124 964 6 × 2 = 0 + 0.066 955 566 406 249 929 2;
  • 3) 0.066 955 566 406 249 929 2 × 2 = 0 + 0.133 911 132 812 499 858 4;
  • 4) 0.133 911 132 812 499 858 4 × 2 = 0 + 0.267 822 265 624 999 716 8;
  • 5) 0.267 822 265 624 999 716 8 × 2 = 0 + 0.535 644 531 249 999 433 6;
  • 6) 0.535 644 531 249 999 433 6 × 2 = 1 + 0.071 289 062 499 998 867 2;
  • 7) 0.071 289 062 499 998 867 2 × 2 = 0 + 0.142 578 124 999 997 734 4;
  • 8) 0.142 578 124 999 997 734 4 × 2 = 0 + 0.285 156 249 999 995 468 8;
  • 9) 0.285 156 249 999 995 468 8 × 2 = 0 + 0.570 312 499 999 990 937 6;
  • 10) 0.570 312 499 999 990 937 6 × 2 = 1 + 0.140 624 999 999 981 875 2;
  • 11) 0.140 624 999 999 981 875 2 × 2 = 0 + 0.281 249 999 999 963 750 4;
  • 12) 0.281 249 999 999 963 750 4 × 2 = 0 + 0.562 499 999 999 927 500 8;
  • 13) 0.562 499 999 999 927 500 8 × 2 = 1 + 0.124 999 999 999 855 001 6;
  • 14) 0.124 999 999 999 855 001 6 × 2 = 0 + 0.249 999 999 999 710 003 2;
  • 15) 0.249 999 999 999 710 003 2 × 2 = 0 + 0.499 999 999 999 420 006 4;
  • 16) 0.499 999 999 999 420 006 4 × 2 = 0 + 0.999 999 999 998 840 012 8;
  • 17) 0.999 999 999 998 840 012 8 × 2 = 1 + 0.999 999 999 997 680 025 6;
  • 18) 0.999 999 999 997 680 025 6 × 2 = 1 + 0.999 999 999 995 360 051 2;
  • 19) 0.999 999 999 995 360 051 2 × 2 = 1 + 0.999 999 999 990 720 102 4;
  • 20) 0.999 999 999 990 720 102 4 × 2 = 1 + 0.999 999 999 981 440 204 8;
  • 21) 0.999 999 999 981 440 204 8 × 2 = 1 + 0.999 999 999 962 880 409 6;
  • 22) 0.999 999 999 962 880 409 6 × 2 = 1 + 0.999 999 999 925 760 819 2;
  • 23) 0.999 999 999 925 760 819 2 × 2 = 1 + 0.999 999 999 851 521 638 4;
  • 24) 0.999 999 999 851 521 638 4 × 2 = 1 + 0.999 999 999 703 043 276 8;
  • 25) 0.999 999 999 703 043 276 8 × 2 = 1 + 0.999 999 999 406 086 553 6;
  • 26) 0.999 999 999 406 086 553 6 × 2 = 1 + 0.999 999 998 812 173 107 2;
  • 27) 0.999 999 998 812 173 107 2 × 2 = 1 + 0.999 999 997 624 346 214 4;
  • 28) 0.999 999 997 624 346 214 4 × 2 = 1 + 0.999 999 995 248 692 428 8;
  • 29) 0.999 999 995 248 692 428 8 × 2 = 1 + 0.999 999 990 497 384 857 6;
  • 30) 0.999 999 990 497 384 857 6 × 2 = 1 + 0.999 999 980 994 769 715 2;
  • 31) 0.999 999 980 994 769 715 2 × 2 = 1 + 0.999 999 961 989 539 430 4;
  • 32) 0.999 999 961 989 539 430 4 × 2 = 1 + 0.999 999 923 979 078 860 8;
  • 33) 0.999 999 923 979 078 860 8 × 2 = 1 + 0.999 999 847 958 157 721 6;
  • 34) 0.999 999 847 958 157 721 6 × 2 = 1 + 0.999 999 695 916 315 443 2;
  • 35) 0.999 999 695 916 315 443 2 × 2 = 1 + 0.999 999 391 832 630 886 4;
  • 36) 0.999 999 391 832 630 886 4 × 2 = 1 + 0.999 998 783 665 261 772 8;
  • 37) 0.999 998 783 665 261 772 8 × 2 = 1 + 0.999 997 567 330 523 545 6;
  • 38) 0.999 997 567 330 523 545 6 × 2 = 1 + 0.999 995 134 661 047 091 2;
  • 39) 0.999 995 134 661 047 091 2 × 2 = 1 + 0.999 990 269 322 094 182 4;
  • 40) 0.999 990 269 322 094 182 4 × 2 = 1 + 0.999 980 538 644 188 364 8;
  • 41) 0.999 980 538 644 188 364 8 × 2 = 1 + 0.999 961 077 288 376 729 6;
  • 42) 0.999 961 077 288 376 729 6 × 2 = 1 + 0.999 922 154 576 753 459 2;
  • 43) 0.999 922 154 576 753 459 2 × 2 = 1 + 0.999 844 309 153 506 918 4;
  • 44) 0.999 844 309 153 506 918 4 × 2 = 1 + 0.999 688 618 307 013 836 8;
  • 45) 0.999 688 618 307 013 836 8 × 2 = 1 + 0.999 377 236 614 027 673 6;
  • 46) 0.999 377 236 614 027 673 6 × 2 = 1 + 0.998 754 473 228 055 347 2;
  • 47) 0.998 754 473 228 055 347 2 × 2 = 1 + 0.997 508 946 456 110 694 4;
  • 48) 0.997 508 946 456 110 694 4 × 2 = 1 + 0.995 017 892 912 221 388 8;
  • 49) 0.995 017 892 912 221 388 8 × 2 = 1 + 0.990 035 785 824 442 777 6;
  • 50) 0.990 035 785 824 442 777 6 × 2 = 1 + 0.980 071 571 648 885 555 2;
  • 51) 0.980 071 571 648 885 555 2 × 2 = 1 + 0.960 143 143 297 771 110 4;
  • 52) 0.960 143 143 297 771 110 4 × 2 = 1 + 0.920 286 286 595 542 220 8;
  • 53) 0.920 286 286 595 542 220 8 × 2 = 1 + 0.840 572 573 191 084 441 6;
  • 54) 0.840 572 573 191 084 441 6 × 2 = 1 + 0.681 145 146 382 168 883 2;
  • 55) 0.681 145 146 382 168 883 2 × 2 = 1 + 0.362 290 292 764 337 766 4;
  • 56) 0.362 290 292 764 337 766 4 × 2 = 0 + 0.724 580 585 528 675 532 8;
  • 57) 0.724 580 585 528 675 532 8 × 2 = 1 + 0.449 161 171 057 351 065 6;
  • 58) 0.449 161 171 057 351 065 6 × 2 = 0 + 0.898 322 342 114 702 131 2;

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:


0.016 738 891 601 562 482 3(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 482 3(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 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:


0.016 738 891 601 562 482 3(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010(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 1010


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 1010 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010


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 1010


Decimal number -0.016 738 891 601 562 482 3 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 1010

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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