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

Convert decimal -0.016 738 891 601 562 495 64(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 64(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 64| = 0.016 738 891 601 562 495 64


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 64.

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 64 × 2 = 0 + 0.033 477 783 203 124 991 28;
  • 2) 0.033 477 783 203 124 991 28 × 2 = 0 + 0.066 955 566 406 249 982 56;
  • 3) 0.066 955 566 406 249 982 56 × 2 = 0 + 0.133 911 132 812 499 965 12;
  • 4) 0.133 911 132 812 499 965 12 × 2 = 0 + 0.267 822 265 624 999 930 24;
  • 5) 0.267 822 265 624 999 930 24 × 2 = 0 + 0.535 644 531 249 999 860 48;
  • 6) 0.535 644 531 249 999 860 48 × 2 = 1 + 0.071 289 062 499 999 720 96;
  • 7) 0.071 289 062 499 999 720 96 × 2 = 0 + 0.142 578 124 999 999 441 92;
  • 8) 0.142 578 124 999 999 441 92 × 2 = 0 + 0.285 156 249 999 998 883 84;
  • 9) 0.285 156 249 999 998 883 84 × 2 = 0 + 0.570 312 499 999 997 767 68;
  • 10) 0.570 312 499 999 997 767 68 × 2 = 1 + 0.140 624 999 999 995 535 36;
  • 11) 0.140 624 999 999 995 535 36 × 2 = 0 + 0.281 249 999 999 991 070 72;
  • 12) 0.281 249 999 999 991 070 72 × 2 = 0 + 0.562 499 999 999 982 141 44;
  • 13) 0.562 499 999 999 982 141 44 × 2 = 1 + 0.124 999 999 999 964 282 88;
  • 14) 0.124 999 999 999 964 282 88 × 2 = 0 + 0.249 999 999 999 928 565 76;
  • 15) 0.249 999 999 999 928 565 76 × 2 = 0 + 0.499 999 999 999 857 131 52;
  • 16) 0.499 999 999 999 857 131 52 × 2 = 0 + 0.999 999 999 999 714 263 04;
  • 17) 0.999 999 999 999 714 263 04 × 2 = 1 + 0.999 999 999 999 428 526 08;
  • 18) 0.999 999 999 999 428 526 08 × 2 = 1 + 0.999 999 999 998 857 052 16;
  • 19) 0.999 999 999 998 857 052 16 × 2 = 1 + 0.999 999 999 997 714 104 32;
  • 20) 0.999 999 999 997 714 104 32 × 2 = 1 + 0.999 999 999 995 428 208 64;
  • 21) 0.999 999 999 995 428 208 64 × 2 = 1 + 0.999 999 999 990 856 417 28;
  • 22) 0.999 999 999 990 856 417 28 × 2 = 1 + 0.999 999 999 981 712 834 56;
  • 23) 0.999 999 999 981 712 834 56 × 2 = 1 + 0.999 999 999 963 425 669 12;
  • 24) 0.999 999 999 963 425 669 12 × 2 = 1 + 0.999 999 999 926 851 338 24;
  • 25) 0.999 999 999 926 851 338 24 × 2 = 1 + 0.999 999 999 853 702 676 48;
  • 26) 0.999 999 999 853 702 676 48 × 2 = 1 + 0.999 999 999 707 405 352 96;
  • 27) 0.999 999 999 707 405 352 96 × 2 = 1 + 0.999 999 999 414 810 705 92;
  • 28) 0.999 999 999 414 810 705 92 × 2 = 1 + 0.999 999 998 829 621 411 84;
  • 29) 0.999 999 998 829 621 411 84 × 2 = 1 + 0.999 999 997 659 242 823 68;
  • 30) 0.999 999 997 659 242 823 68 × 2 = 1 + 0.999 999 995 318 485 647 36;
  • 31) 0.999 999 995 318 485 647 36 × 2 = 1 + 0.999 999 990 636 971 294 72;
  • 32) 0.999 999 990 636 971 294 72 × 2 = 1 + 0.999 999 981 273 942 589 44;
  • 33) 0.999 999 981 273 942 589 44 × 2 = 1 + 0.999 999 962 547 885 178 88;
  • 34) 0.999 999 962 547 885 178 88 × 2 = 1 + 0.999 999 925 095 770 357 76;
  • 35) 0.999 999 925 095 770 357 76 × 2 = 1 + 0.999 999 850 191 540 715 52;
  • 36) 0.999 999 850 191 540 715 52 × 2 = 1 + 0.999 999 700 383 081 431 04;
  • 37) 0.999 999 700 383 081 431 04 × 2 = 1 + 0.999 999 400 766 162 862 08;
  • 38) 0.999 999 400 766 162 862 08 × 2 = 1 + 0.999 998 801 532 325 724 16;
  • 39) 0.999 998 801 532 325 724 16 × 2 = 1 + 0.999 997 603 064 651 448 32;
  • 40) 0.999 997 603 064 651 448 32 × 2 = 1 + 0.999 995 206 129 302 896 64;
  • 41) 0.999 995 206 129 302 896 64 × 2 = 1 + 0.999 990 412 258 605 793 28;
  • 42) 0.999 990 412 258 605 793 28 × 2 = 1 + 0.999 980 824 517 211 586 56;
  • 43) 0.999 980 824 517 211 586 56 × 2 = 1 + 0.999 961 649 034 423 173 12;
  • 44) 0.999 961 649 034 423 173 12 × 2 = 1 + 0.999 923 298 068 846 346 24;
  • 45) 0.999 923 298 068 846 346 24 × 2 = 1 + 0.999 846 596 137 692 692 48;
  • 46) 0.999 846 596 137 692 692 48 × 2 = 1 + 0.999 693 192 275 385 384 96;
  • 47) 0.999 693 192 275 385 384 96 × 2 = 1 + 0.999 386 384 550 770 769 92;
  • 48) 0.999 386 384 550 770 769 92 × 2 = 1 + 0.998 772 769 101 541 539 84;
  • 49) 0.998 772 769 101 541 539 84 × 2 = 1 + 0.997 545 538 203 083 079 68;
  • 50) 0.997 545 538 203 083 079 68 × 2 = 1 + 0.995 091 076 406 166 159 36;
  • 51) 0.995 091 076 406 166 159 36 × 2 = 1 + 0.990 182 152 812 332 318 72;
  • 52) 0.990 182 152 812 332 318 72 × 2 = 1 + 0.980 364 305 624 664 637 44;
  • 53) 0.980 364 305 624 664 637 44 × 2 = 1 + 0.960 728 611 249 329 274 88;
  • 54) 0.960 728 611 249 329 274 88 × 2 = 1 + 0.921 457 222 498 658 549 76;
  • 55) 0.921 457 222 498 658 549 76 × 2 = 1 + 0.842 914 444 997 317 099 52;
  • 56) 0.842 914 444 997 317 099 52 × 2 = 1 + 0.685 828 889 994 634 199 04;
  • 57) 0.685 828 889 994 634 199 04 × 2 = 1 + 0.371 657 779 989 268 398 08;
  • 58) 0.371 657 779 989 268 398 08 × 2 = 0 + 0.743 315 559 978 536 796 16;

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 64(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 64(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 64(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 64 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