-0.016 738 891 601 562 495 16 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 495 16(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 16(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 16| = 0.016 738 891 601 562 495 16


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 16.

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 16 × 2 = 0 + 0.033 477 783 203 124 990 32;
  • 2) 0.033 477 783 203 124 990 32 × 2 = 0 + 0.066 955 566 406 249 980 64;
  • 3) 0.066 955 566 406 249 980 64 × 2 = 0 + 0.133 911 132 812 499 961 28;
  • 4) 0.133 911 132 812 499 961 28 × 2 = 0 + 0.267 822 265 624 999 922 56;
  • 5) 0.267 822 265 624 999 922 56 × 2 = 0 + 0.535 644 531 249 999 845 12;
  • 6) 0.535 644 531 249 999 845 12 × 2 = 1 + 0.071 289 062 499 999 690 24;
  • 7) 0.071 289 062 499 999 690 24 × 2 = 0 + 0.142 578 124 999 999 380 48;
  • 8) 0.142 578 124 999 999 380 48 × 2 = 0 + 0.285 156 249 999 998 760 96;
  • 9) 0.285 156 249 999 998 760 96 × 2 = 0 + 0.570 312 499 999 997 521 92;
  • 10) 0.570 312 499 999 997 521 92 × 2 = 1 + 0.140 624 999 999 995 043 84;
  • 11) 0.140 624 999 999 995 043 84 × 2 = 0 + 0.281 249 999 999 990 087 68;
  • 12) 0.281 249 999 999 990 087 68 × 2 = 0 + 0.562 499 999 999 980 175 36;
  • 13) 0.562 499 999 999 980 175 36 × 2 = 1 + 0.124 999 999 999 960 350 72;
  • 14) 0.124 999 999 999 960 350 72 × 2 = 0 + 0.249 999 999 999 920 701 44;
  • 15) 0.249 999 999 999 920 701 44 × 2 = 0 + 0.499 999 999 999 841 402 88;
  • 16) 0.499 999 999 999 841 402 88 × 2 = 0 + 0.999 999 999 999 682 805 76;
  • 17) 0.999 999 999 999 682 805 76 × 2 = 1 + 0.999 999 999 999 365 611 52;
  • 18) 0.999 999 999 999 365 611 52 × 2 = 1 + 0.999 999 999 998 731 223 04;
  • 19) 0.999 999 999 998 731 223 04 × 2 = 1 + 0.999 999 999 997 462 446 08;
  • 20) 0.999 999 999 997 462 446 08 × 2 = 1 + 0.999 999 999 994 924 892 16;
  • 21) 0.999 999 999 994 924 892 16 × 2 = 1 + 0.999 999 999 989 849 784 32;
  • 22) 0.999 999 999 989 849 784 32 × 2 = 1 + 0.999 999 999 979 699 568 64;
  • 23) 0.999 999 999 979 699 568 64 × 2 = 1 + 0.999 999 999 959 399 137 28;
  • 24) 0.999 999 999 959 399 137 28 × 2 = 1 + 0.999 999 999 918 798 274 56;
  • 25) 0.999 999 999 918 798 274 56 × 2 = 1 + 0.999 999 999 837 596 549 12;
  • 26) 0.999 999 999 837 596 549 12 × 2 = 1 + 0.999 999 999 675 193 098 24;
  • 27) 0.999 999 999 675 193 098 24 × 2 = 1 + 0.999 999 999 350 386 196 48;
  • 28) 0.999 999 999 350 386 196 48 × 2 = 1 + 0.999 999 998 700 772 392 96;
  • 29) 0.999 999 998 700 772 392 96 × 2 = 1 + 0.999 999 997 401 544 785 92;
  • 30) 0.999 999 997 401 544 785 92 × 2 = 1 + 0.999 999 994 803 089 571 84;
  • 31) 0.999 999 994 803 089 571 84 × 2 = 1 + 0.999 999 989 606 179 143 68;
  • 32) 0.999 999 989 606 179 143 68 × 2 = 1 + 0.999 999 979 212 358 287 36;
  • 33) 0.999 999 979 212 358 287 36 × 2 = 1 + 0.999 999 958 424 716 574 72;
  • 34) 0.999 999 958 424 716 574 72 × 2 = 1 + 0.999 999 916 849 433 149 44;
  • 35) 0.999 999 916 849 433 149 44 × 2 = 1 + 0.999 999 833 698 866 298 88;
  • 36) 0.999 999 833 698 866 298 88 × 2 = 1 + 0.999 999 667 397 732 597 76;
  • 37) 0.999 999 667 397 732 597 76 × 2 = 1 + 0.999 999 334 795 465 195 52;
  • 38) 0.999 999 334 795 465 195 52 × 2 = 1 + 0.999 998 669 590 930 391 04;
  • 39) 0.999 998 669 590 930 391 04 × 2 = 1 + 0.999 997 339 181 860 782 08;
  • 40) 0.999 997 339 181 860 782 08 × 2 = 1 + 0.999 994 678 363 721 564 16;
  • 41) 0.999 994 678 363 721 564 16 × 2 = 1 + 0.999 989 356 727 443 128 32;
  • 42) 0.999 989 356 727 443 128 32 × 2 = 1 + 0.999 978 713 454 886 256 64;
  • 43) 0.999 978 713 454 886 256 64 × 2 = 1 + 0.999 957 426 909 772 513 28;
  • 44) 0.999 957 426 909 772 513 28 × 2 = 1 + 0.999 914 853 819 545 026 56;
  • 45) 0.999 914 853 819 545 026 56 × 2 = 1 + 0.999 829 707 639 090 053 12;
  • 46) 0.999 829 707 639 090 053 12 × 2 = 1 + 0.999 659 415 278 180 106 24;
  • 47) 0.999 659 415 278 180 106 24 × 2 = 1 + 0.999 318 830 556 360 212 48;
  • 48) 0.999 318 830 556 360 212 48 × 2 = 1 + 0.998 637 661 112 720 424 96;
  • 49) 0.998 637 661 112 720 424 96 × 2 = 1 + 0.997 275 322 225 440 849 92;
  • 50) 0.997 275 322 225 440 849 92 × 2 = 1 + 0.994 550 644 450 881 699 84;
  • 51) 0.994 550 644 450 881 699 84 × 2 = 1 + 0.989 101 288 901 763 399 68;
  • 52) 0.989 101 288 901 763 399 68 × 2 = 1 + 0.978 202 577 803 526 799 36;
  • 53) 0.978 202 577 803 526 799 36 × 2 = 1 + 0.956 405 155 607 053 598 72;
  • 54) 0.956 405 155 607 053 598 72 × 2 = 1 + 0.912 810 311 214 107 197 44;
  • 55) 0.912 810 311 214 107 197 44 × 2 = 1 + 0.825 620 622 428 214 394 88;
  • 56) 0.825 620 622 428 214 394 88 × 2 = 1 + 0.651 241 244 856 428 789 76;
  • 57) 0.651 241 244 856 428 789 76 × 2 = 1 + 0.302 482 489 712 857 579 52;
  • 58) 0.302 482 489 712 857 579 52 × 2 = 0 + 0.604 964 979 425 715 159 04;

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 16(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 16(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 16(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 16 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