-0.016 738 891 601 562 494 97 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 494 97(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
-0.016 738 891 601 562 494 97(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:

|-0.016 738 891 601 562 494 97| = 0.016 738 891 601 562 494 97


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: 0.016 738 891 601 562 494 97.

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) 0.016 738 891 601 562 494 97 × 2 = 0 + 0.033 477 783 203 124 989 94;
  • 2) 0.033 477 783 203 124 989 94 × 2 = 0 + 0.066 955 566 406 249 979 88;
  • 3) 0.066 955 566 406 249 979 88 × 2 = 0 + 0.133 911 132 812 499 959 76;
  • 4) 0.133 911 132 812 499 959 76 × 2 = 0 + 0.267 822 265 624 999 919 52;
  • 5) 0.267 822 265 624 999 919 52 × 2 = 0 + 0.535 644 531 249 999 839 04;
  • 6) 0.535 644 531 249 999 839 04 × 2 = 1 + 0.071 289 062 499 999 678 08;
  • 7) 0.071 289 062 499 999 678 08 × 2 = 0 + 0.142 578 124 999 999 356 16;
  • 8) 0.142 578 124 999 999 356 16 × 2 = 0 + 0.285 156 249 999 998 712 32;
  • 9) 0.285 156 249 999 998 712 32 × 2 = 0 + 0.570 312 499 999 997 424 64;
  • 10) 0.570 312 499 999 997 424 64 × 2 = 1 + 0.140 624 999 999 994 849 28;
  • 11) 0.140 624 999 999 994 849 28 × 2 = 0 + 0.281 249 999 999 989 698 56;
  • 12) 0.281 249 999 999 989 698 56 × 2 = 0 + 0.562 499 999 999 979 397 12;
  • 13) 0.562 499 999 999 979 397 12 × 2 = 1 + 0.124 999 999 999 958 794 24;
  • 14) 0.124 999 999 999 958 794 24 × 2 = 0 + 0.249 999 999 999 917 588 48;
  • 15) 0.249 999 999 999 917 588 48 × 2 = 0 + 0.499 999 999 999 835 176 96;
  • 16) 0.499 999 999 999 835 176 96 × 2 = 0 + 0.999 999 999 999 670 353 92;
  • 17) 0.999 999 999 999 670 353 92 × 2 = 1 + 0.999 999 999 999 340 707 84;
  • 18) 0.999 999 999 999 340 707 84 × 2 = 1 + 0.999 999 999 998 681 415 68;
  • 19) 0.999 999 999 998 681 415 68 × 2 = 1 + 0.999 999 999 997 362 831 36;
  • 20) 0.999 999 999 997 362 831 36 × 2 = 1 + 0.999 999 999 994 725 662 72;
  • 21) 0.999 999 999 994 725 662 72 × 2 = 1 + 0.999 999 999 989 451 325 44;
  • 22) 0.999 999 999 989 451 325 44 × 2 = 1 + 0.999 999 999 978 902 650 88;
  • 23) 0.999 999 999 978 902 650 88 × 2 = 1 + 0.999 999 999 957 805 301 76;
  • 24) 0.999 999 999 957 805 301 76 × 2 = 1 + 0.999 999 999 915 610 603 52;
  • 25) 0.999 999 999 915 610 603 52 × 2 = 1 + 0.999 999 999 831 221 207 04;
  • 26) 0.999 999 999 831 221 207 04 × 2 = 1 + 0.999 999 999 662 442 414 08;
  • 27) 0.999 999 999 662 442 414 08 × 2 = 1 + 0.999 999 999 324 884 828 16;
  • 28) 0.999 999 999 324 884 828 16 × 2 = 1 + 0.999 999 998 649 769 656 32;
  • 29) 0.999 999 998 649 769 656 32 × 2 = 1 + 0.999 999 997 299 539 312 64;
  • 30) 0.999 999 997 299 539 312 64 × 2 = 1 + 0.999 999 994 599 078 625 28;
  • 31) 0.999 999 994 599 078 625 28 × 2 = 1 + 0.999 999 989 198 157 250 56;
  • 32) 0.999 999 989 198 157 250 56 × 2 = 1 + 0.999 999 978 396 314 501 12;
  • 33) 0.999 999 978 396 314 501 12 × 2 = 1 + 0.999 999 956 792 629 002 24;
  • 34) 0.999 999 956 792 629 002 24 × 2 = 1 + 0.999 999 913 585 258 004 48;
  • 35) 0.999 999 913 585 258 004 48 × 2 = 1 + 0.999 999 827 170 516 008 96;
  • 36) 0.999 999 827 170 516 008 96 × 2 = 1 + 0.999 999 654 341 032 017 92;
  • 37) 0.999 999 654 341 032 017 92 × 2 = 1 + 0.999 999 308 682 064 035 84;
  • 38) 0.999 999 308 682 064 035 84 × 2 = 1 + 0.999 998 617 364 128 071 68;
  • 39) 0.999 998 617 364 128 071 68 × 2 = 1 + 0.999 997 234 728 256 143 36;
  • 40) 0.999 997 234 728 256 143 36 × 2 = 1 + 0.999 994 469 456 512 286 72;
  • 41) 0.999 994 469 456 512 286 72 × 2 = 1 + 0.999 988 938 913 024 573 44;
  • 42) 0.999 988 938 913 024 573 44 × 2 = 1 + 0.999 977 877 826 049 146 88;
  • 43) 0.999 977 877 826 049 146 88 × 2 = 1 + 0.999 955 755 652 098 293 76;
  • 44) 0.999 955 755 652 098 293 76 × 2 = 1 + 0.999 911 511 304 196 587 52;
  • 45) 0.999 911 511 304 196 587 52 × 2 = 1 + 0.999 823 022 608 393 175 04;
  • 46) 0.999 823 022 608 393 175 04 × 2 = 1 + 0.999 646 045 216 786 350 08;
  • 47) 0.999 646 045 216 786 350 08 × 2 = 1 + 0.999 292 090 433 572 700 16;
  • 48) 0.999 292 090 433 572 700 16 × 2 = 1 + 0.998 584 180 867 145 400 32;
  • 49) 0.998 584 180 867 145 400 32 × 2 = 1 + 0.997 168 361 734 290 800 64;
  • 50) 0.997 168 361 734 290 800 64 × 2 = 1 + 0.994 336 723 468 581 601 28;
  • 51) 0.994 336 723 468 581 601 28 × 2 = 1 + 0.988 673 446 937 163 202 56;
  • 52) 0.988 673 446 937 163 202 56 × 2 = 1 + 0.977 346 893 874 326 405 12;
  • 53) 0.977 346 893 874 326 405 12 × 2 = 1 + 0.954 693 787 748 652 810 24;
  • 54) 0.954 693 787 748 652 810 24 × 2 = 1 + 0.909 387 575 497 305 620 48;
  • 55) 0.909 387 575 497 305 620 48 × 2 = 1 + 0.818 775 150 994 611 240 96;
  • 56) 0.818 775 150 994 611 240 96 × 2 = 1 + 0.637 550 301 989 222 481 92;
  • 57) 0.637 550 301 989 222 481 92 × 2 = 1 + 0.275 100 603 978 444 963 84;
  • 58) 0.275 100 603 978 444 963 84 × 2 = 0 + 0.550 201 207 956 889 927 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:


0.016 738 891 601 562 494 97(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 494 97(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:


0.016 738 891 601 562 494 97(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 -0.016 738 891 601 562 494 97 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