-0.016 738 891 601 562 495 22 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 495 22(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
-0.016 738 891 601 562 495 22(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:

|-0.016 738 891 601 562 495 22| = 0.016 738 891 601 562 495 22


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: 0.016 738 891 601 562 495 22.

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) 0.016 738 891 601 562 495 22 × 2 = 0 + 0.033 477 783 203 124 990 44;
  • 2) 0.033 477 783 203 124 990 44 × 2 = 0 + 0.066 955 566 406 249 980 88;
  • 3) 0.066 955 566 406 249 980 88 × 2 = 0 + 0.133 911 132 812 499 961 76;
  • 4) 0.133 911 132 812 499 961 76 × 2 = 0 + 0.267 822 265 624 999 923 52;
  • 5) 0.267 822 265 624 999 923 52 × 2 = 0 + 0.535 644 531 249 999 847 04;
  • 6) 0.535 644 531 249 999 847 04 × 2 = 1 + 0.071 289 062 499 999 694 08;
  • 7) 0.071 289 062 499 999 694 08 × 2 = 0 + 0.142 578 124 999 999 388 16;
  • 8) 0.142 578 124 999 999 388 16 × 2 = 0 + 0.285 156 249 999 998 776 32;
  • 9) 0.285 156 249 999 998 776 32 × 2 = 0 + 0.570 312 499 999 997 552 64;
  • 10) 0.570 312 499 999 997 552 64 × 2 = 1 + 0.140 624 999 999 995 105 28;
  • 11) 0.140 624 999 999 995 105 28 × 2 = 0 + 0.281 249 999 999 990 210 56;
  • 12) 0.281 249 999 999 990 210 56 × 2 = 0 + 0.562 499 999 999 980 421 12;
  • 13) 0.562 499 999 999 980 421 12 × 2 = 1 + 0.124 999 999 999 960 842 24;
  • 14) 0.124 999 999 999 960 842 24 × 2 = 0 + 0.249 999 999 999 921 684 48;
  • 15) 0.249 999 999 999 921 684 48 × 2 = 0 + 0.499 999 999 999 843 368 96;
  • 16) 0.499 999 999 999 843 368 96 × 2 = 0 + 0.999 999 999 999 686 737 92;
  • 17) 0.999 999 999 999 686 737 92 × 2 = 1 + 0.999 999 999 999 373 475 84;
  • 18) 0.999 999 999 999 373 475 84 × 2 = 1 + 0.999 999 999 998 746 951 68;
  • 19) 0.999 999 999 998 746 951 68 × 2 = 1 + 0.999 999 999 997 493 903 36;
  • 20) 0.999 999 999 997 493 903 36 × 2 = 1 + 0.999 999 999 994 987 806 72;
  • 21) 0.999 999 999 994 987 806 72 × 2 = 1 + 0.999 999 999 989 975 613 44;
  • 22) 0.999 999 999 989 975 613 44 × 2 = 1 + 0.999 999 999 979 951 226 88;
  • 23) 0.999 999 999 979 951 226 88 × 2 = 1 + 0.999 999 999 959 902 453 76;
  • 24) 0.999 999 999 959 902 453 76 × 2 = 1 + 0.999 999 999 919 804 907 52;
  • 25) 0.999 999 999 919 804 907 52 × 2 = 1 + 0.999 999 999 839 609 815 04;
  • 26) 0.999 999 999 839 609 815 04 × 2 = 1 + 0.999 999 999 679 219 630 08;
  • 27) 0.999 999 999 679 219 630 08 × 2 = 1 + 0.999 999 999 358 439 260 16;
  • 28) 0.999 999 999 358 439 260 16 × 2 = 1 + 0.999 999 998 716 878 520 32;
  • 29) 0.999 999 998 716 878 520 32 × 2 = 1 + 0.999 999 997 433 757 040 64;
  • 30) 0.999 999 997 433 757 040 64 × 2 = 1 + 0.999 999 994 867 514 081 28;
  • 31) 0.999 999 994 867 514 081 28 × 2 = 1 + 0.999 999 989 735 028 162 56;
  • 32) 0.999 999 989 735 028 162 56 × 2 = 1 + 0.999 999 979 470 056 325 12;
  • 33) 0.999 999 979 470 056 325 12 × 2 = 1 + 0.999 999 958 940 112 650 24;
  • 34) 0.999 999 958 940 112 650 24 × 2 = 1 + 0.999 999 917 880 225 300 48;
  • 35) 0.999 999 917 880 225 300 48 × 2 = 1 + 0.999 999 835 760 450 600 96;
  • 36) 0.999 999 835 760 450 600 96 × 2 = 1 + 0.999 999 671 520 901 201 92;
  • 37) 0.999 999 671 520 901 201 92 × 2 = 1 + 0.999 999 343 041 802 403 84;
  • 38) 0.999 999 343 041 802 403 84 × 2 = 1 + 0.999 998 686 083 604 807 68;
  • 39) 0.999 998 686 083 604 807 68 × 2 = 1 + 0.999 997 372 167 209 615 36;
  • 40) 0.999 997 372 167 209 615 36 × 2 = 1 + 0.999 994 744 334 419 230 72;
  • 41) 0.999 994 744 334 419 230 72 × 2 = 1 + 0.999 989 488 668 838 461 44;
  • 42) 0.999 989 488 668 838 461 44 × 2 = 1 + 0.999 978 977 337 676 922 88;
  • 43) 0.999 978 977 337 676 922 88 × 2 = 1 + 0.999 957 954 675 353 845 76;
  • 44) 0.999 957 954 675 353 845 76 × 2 = 1 + 0.999 915 909 350 707 691 52;
  • 45) 0.999 915 909 350 707 691 52 × 2 = 1 + 0.999 831 818 701 415 383 04;
  • 46) 0.999 831 818 701 415 383 04 × 2 = 1 + 0.999 663 637 402 830 766 08;
  • 47) 0.999 663 637 402 830 766 08 × 2 = 1 + 0.999 327 274 805 661 532 16;
  • 48) 0.999 327 274 805 661 532 16 × 2 = 1 + 0.998 654 549 611 323 064 32;
  • 49) 0.998 654 549 611 323 064 32 × 2 = 1 + 0.997 309 099 222 646 128 64;
  • 50) 0.997 309 099 222 646 128 64 × 2 = 1 + 0.994 618 198 445 292 257 28;
  • 51) 0.994 618 198 445 292 257 28 × 2 = 1 + 0.989 236 396 890 584 514 56;
  • 52) 0.989 236 396 890 584 514 56 × 2 = 1 + 0.978 472 793 781 169 029 12;
  • 53) 0.978 472 793 781 169 029 12 × 2 = 1 + 0.956 945 587 562 338 058 24;
  • 54) 0.956 945 587 562 338 058 24 × 2 = 1 + 0.913 891 175 124 676 116 48;
  • 55) 0.913 891 175 124 676 116 48 × 2 = 1 + 0.827 782 350 249 352 232 96;
  • 56) 0.827 782 350 249 352 232 96 × 2 = 1 + 0.655 564 700 498 704 465 92;
  • 57) 0.655 564 700 498 704 465 92 × 2 = 1 + 0.311 129 400 997 408 931 84;
  • 58) 0.311 129 400 997 408 931 84 × 2 = 0 + 0.622 258 801 994 817 863 68;

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:


0.016 738 891 601 562 495 22(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 495 22(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:


0.016 738 891 601 562 495 22(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 -0.016 738 891 601 562 495 22 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