-0.016 738 891 601 562 497 32 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 497 32(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 32(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 32| = 0.016 738 891 601 562 497 32


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 32.

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 32 × 2 = 0 + 0.033 477 783 203 124 994 64;
  • 2) 0.033 477 783 203 124 994 64 × 2 = 0 + 0.066 955 566 406 249 989 28;
  • 3) 0.066 955 566 406 249 989 28 × 2 = 0 + 0.133 911 132 812 499 978 56;
  • 4) 0.133 911 132 812 499 978 56 × 2 = 0 + 0.267 822 265 624 999 957 12;
  • 5) 0.267 822 265 624 999 957 12 × 2 = 0 + 0.535 644 531 249 999 914 24;
  • 6) 0.535 644 531 249 999 914 24 × 2 = 1 + 0.071 289 062 499 999 828 48;
  • 7) 0.071 289 062 499 999 828 48 × 2 = 0 + 0.142 578 124 999 999 656 96;
  • 8) 0.142 578 124 999 999 656 96 × 2 = 0 + 0.285 156 249 999 999 313 92;
  • 9) 0.285 156 249 999 999 313 92 × 2 = 0 + 0.570 312 499 999 998 627 84;
  • 10) 0.570 312 499 999 998 627 84 × 2 = 1 + 0.140 624 999 999 997 255 68;
  • 11) 0.140 624 999 999 997 255 68 × 2 = 0 + 0.281 249 999 999 994 511 36;
  • 12) 0.281 249 999 999 994 511 36 × 2 = 0 + 0.562 499 999 999 989 022 72;
  • 13) 0.562 499 999 999 989 022 72 × 2 = 1 + 0.124 999 999 999 978 045 44;
  • 14) 0.124 999 999 999 978 045 44 × 2 = 0 + 0.249 999 999 999 956 090 88;
  • 15) 0.249 999 999 999 956 090 88 × 2 = 0 + 0.499 999 999 999 912 181 76;
  • 16) 0.499 999 999 999 912 181 76 × 2 = 0 + 0.999 999 999 999 824 363 52;
  • 17) 0.999 999 999 999 824 363 52 × 2 = 1 + 0.999 999 999 999 648 727 04;
  • 18) 0.999 999 999 999 648 727 04 × 2 = 1 + 0.999 999 999 999 297 454 08;
  • 19) 0.999 999 999 999 297 454 08 × 2 = 1 + 0.999 999 999 998 594 908 16;
  • 20) 0.999 999 999 998 594 908 16 × 2 = 1 + 0.999 999 999 997 189 816 32;
  • 21) 0.999 999 999 997 189 816 32 × 2 = 1 + 0.999 999 999 994 379 632 64;
  • 22) 0.999 999 999 994 379 632 64 × 2 = 1 + 0.999 999 999 988 759 265 28;
  • 23) 0.999 999 999 988 759 265 28 × 2 = 1 + 0.999 999 999 977 518 530 56;
  • 24) 0.999 999 999 977 518 530 56 × 2 = 1 + 0.999 999 999 955 037 061 12;
  • 25) 0.999 999 999 955 037 061 12 × 2 = 1 + 0.999 999 999 910 074 122 24;
  • 26) 0.999 999 999 910 074 122 24 × 2 = 1 + 0.999 999 999 820 148 244 48;
  • 27) 0.999 999 999 820 148 244 48 × 2 = 1 + 0.999 999 999 640 296 488 96;
  • 28) 0.999 999 999 640 296 488 96 × 2 = 1 + 0.999 999 999 280 592 977 92;
  • 29) 0.999 999 999 280 592 977 92 × 2 = 1 + 0.999 999 998 561 185 955 84;
  • 30) 0.999 999 998 561 185 955 84 × 2 = 1 + 0.999 999 997 122 371 911 68;
  • 31) 0.999 999 997 122 371 911 68 × 2 = 1 + 0.999 999 994 244 743 823 36;
  • 32) 0.999 999 994 244 743 823 36 × 2 = 1 + 0.999 999 988 489 487 646 72;
  • 33) 0.999 999 988 489 487 646 72 × 2 = 1 + 0.999 999 976 978 975 293 44;
  • 34) 0.999 999 976 978 975 293 44 × 2 = 1 + 0.999 999 953 957 950 586 88;
  • 35) 0.999 999 953 957 950 586 88 × 2 = 1 + 0.999 999 907 915 901 173 76;
  • 36) 0.999 999 907 915 901 173 76 × 2 = 1 + 0.999 999 815 831 802 347 52;
  • 37) 0.999 999 815 831 802 347 52 × 2 = 1 + 0.999 999 631 663 604 695 04;
  • 38) 0.999 999 631 663 604 695 04 × 2 = 1 + 0.999 999 263 327 209 390 08;
  • 39) 0.999 999 263 327 209 390 08 × 2 = 1 + 0.999 998 526 654 418 780 16;
  • 40) 0.999 998 526 654 418 780 16 × 2 = 1 + 0.999 997 053 308 837 560 32;
  • 41) 0.999 997 053 308 837 560 32 × 2 = 1 + 0.999 994 106 617 675 120 64;
  • 42) 0.999 994 106 617 675 120 64 × 2 = 1 + 0.999 988 213 235 350 241 28;
  • 43) 0.999 988 213 235 350 241 28 × 2 = 1 + 0.999 976 426 470 700 482 56;
  • 44) 0.999 976 426 470 700 482 56 × 2 = 1 + 0.999 952 852 941 400 965 12;
  • 45) 0.999 952 852 941 400 965 12 × 2 = 1 + 0.999 905 705 882 801 930 24;
  • 46) 0.999 905 705 882 801 930 24 × 2 = 1 + 0.999 811 411 765 603 860 48;
  • 47) 0.999 811 411 765 603 860 48 × 2 = 1 + 0.999 622 823 531 207 720 96;
  • 48) 0.999 622 823 531 207 720 96 × 2 = 1 + 0.999 245 647 062 415 441 92;
  • 49) 0.999 245 647 062 415 441 92 × 2 = 1 + 0.998 491 294 124 830 883 84;
  • 50) 0.998 491 294 124 830 883 84 × 2 = 1 + 0.996 982 588 249 661 767 68;
  • 51) 0.996 982 588 249 661 767 68 × 2 = 1 + 0.993 965 176 499 323 535 36;
  • 52) 0.993 965 176 499 323 535 36 × 2 = 1 + 0.987 930 352 998 647 070 72;
  • 53) 0.987 930 352 998 647 070 72 × 2 = 1 + 0.975 860 705 997 294 141 44;
  • 54) 0.975 860 705 997 294 141 44 × 2 = 1 + 0.951 721 411 994 588 282 88;
  • 55) 0.951 721 411 994 588 282 88 × 2 = 1 + 0.903 442 823 989 176 565 76;
  • 56) 0.903 442 823 989 176 565 76 × 2 = 1 + 0.806 885 647 978 353 131 52;
  • 57) 0.806 885 647 978 353 131 52 × 2 = 1 + 0.613 771 295 956 706 263 04;
  • 58) 0.613 771 295 956 706 263 04 × 2 = 1 + 0.227 542 591 913 412 526 08;

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 32(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 32(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 32(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 32 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