-0.016 738 891 601 562 496 520 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 520 8(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
-0.016 738 891 601 562 496 520 8(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:

|-0.016 738 891 601 562 496 520 8| = 0.016 738 891 601 562 496 520 8


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: 0.016 738 891 601 562 496 520 8.

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) 0.016 738 891 601 562 496 520 8 × 2 = 0 + 0.033 477 783 203 124 993 041 6;
  • 2) 0.033 477 783 203 124 993 041 6 × 2 = 0 + 0.066 955 566 406 249 986 083 2;
  • 3) 0.066 955 566 406 249 986 083 2 × 2 = 0 + 0.133 911 132 812 499 972 166 4;
  • 4) 0.133 911 132 812 499 972 166 4 × 2 = 0 + 0.267 822 265 624 999 944 332 8;
  • 5) 0.267 822 265 624 999 944 332 8 × 2 = 0 + 0.535 644 531 249 999 888 665 6;
  • 6) 0.535 644 531 249 999 888 665 6 × 2 = 1 + 0.071 289 062 499 999 777 331 2;
  • 7) 0.071 289 062 499 999 777 331 2 × 2 = 0 + 0.142 578 124 999 999 554 662 4;
  • 8) 0.142 578 124 999 999 554 662 4 × 2 = 0 + 0.285 156 249 999 999 109 324 8;
  • 9) 0.285 156 249 999 999 109 324 8 × 2 = 0 + 0.570 312 499 999 998 218 649 6;
  • 10) 0.570 312 499 999 998 218 649 6 × 2 = 1 + 0.140 624 999 999 996 437 299 2;
  • 11) 0.140 624 999 999 996 437 299 2 × 2 = 0 + 0.281 249 999 999 992 874 598 4;
  • 12) 0.281 249 999 999 992 874 598 4 × 2 = 0 + 0.562 499 999 999 985 749 196 8;
  • 13) 0.562 499 999 999 985 749 196 8 × 2 = 1 + 0.124 999 999 999 971 498 393 6;
  • 14) 0.124 999 999 999 971 498 393 6 × 2 = 0 + 0.249 999 999 999 942 996 787 2;
  • 15) 0.249 999 999 999 942 996 787 2 × 2 = 0 + 0.499 999 999 999 885 993 574 4;
  • 16) 0.499 999 999 999 885 993 574 4 × 2 = 0 + 0.999 999 999 999 771 987 148 8;
  • 17) 0.999 999 999 999 771 987 148 8 × 2 = 1 + 0.999 999 999 999 543 974 297 6;
  • 18) 0.999 999 999 999 543 974 297 6 × 2 = 1 + 0.999 999 999 999 087 948 595 2;
  • 19) 0.999 999 999 999 087 948 595 2 × 2 = 1 + 0.999 999 999 998 175 897 190 4;
  • 20) 0.999 999 999 998 175 897 190 4 × 2 = 1 + 0.999 999 999 996 351 794 380 8;
  • 21) 0.999 999 999 996 351 794 380 8 × 2 = 1 + 0.999 999 999 992 703 588 761 6;
  • 22) 0.999 999 999 992 703 588 761 6 × 2 = 1 + 0.999 999 999 985 407 177 523 2;
  • 23) 0.999 999 999 985 407 177 523 2 × 2 = 1 + 0.999 999 999 970 814 355 046 4;
  • 24) 0.999 999 999 970 814 355 046 4 × 2 = 1 + 0.999 999 999 941 628 710 092 8;
  • 25) 0.999 999 999 941 628 710 092 8 × 2 = 1 + 0.999 999 999 883 257 420 185 6;
  • 26) 0.999 999 999 883 257 420 185 6 × 2 = 1 + 0.999 999 999 766 514 840 371 2;
  • 27) 0.999 999 999 766 514 840 371 2 × 2 = 1 + 0.999 999 999 533 029 680 742 4;
  • 28) 0.999 999 999 533 029 680 742 4 × 2 = 1 + 0.999 999 999 066 059 361 484 8;
  • 29) 0.999 999 999 066 059 361 484 8 × 2 = 1 + 0.999 999 998 132 118 722 969 6;
  • 30) 0.999 999 998 132 118 722 969 6 × 2 = 1 + 0.999 999 996 264 237 445 939 2;
  • 31) 0.999 999 996 264 237 445 939 2 × 2 = 1 + 0.999 999 992 528 474 891 878 4;
  • 32) 0.999 999 992 528 474 891 878 4 × 2 = 1 + 0.999 999 985 056 949 783 756 8;
  • 33) 0.999 999 985 056 949 783 756 8 × 2 = 1 + 0.999 999 970 113 899 567 513 6;
  • 34) 0.999 999 970 113 899 567 513 6 × 2 = 1 + 0.999 999 940 227 799 135 027 2;
  • 35) 0.999 999 940 227 799 135 027 2 × 2 = 1 + 0.999 999 880 455 598 270 054 4;
  • 36) 0.999 999 880 455 598 270 054 4 × 2 = 1 + 0.999 999 760 911 196 540 108 8;
  • 37) 0.999 999 760 911 196 540 108 8 × 2 = 1 + 0.999 999 521 822 393 080 217 6;
  • 38) 0.999 999 521 822 393 080 217 6 × 2 = 1 + 0.999 999 043 644 786 160 435 2;
  • 39) 0.999 999 043 644 786 160 435 2 × 2 = 1 + 0.999 998 087 289 572 320 870 4;
  • 40) 0.999 998 087 289 572 320 870 4 × 2 = 1 + 0.999 996 174 579 144 641 740 8;
  • 41) 0.999 996 174 579 144 641 740 8 × 2 = 1 + 0.999 992 349 158 289 283 481 6;
  • 42) 0.999 992 349 158 289 283 481 6 × 2 = 1 + 0.999 984 698 316 578 566 963 2;
  • 43) 0.999 984 698 316 578 566 963 2 × 2 = 1 + 0.999 969 396 633 157 133 926 4;
  • 44) 0.999 969 396 633 157 133 926 4 × 2 = 1 + 0.999 938 793 266 314 267 852 8;
  • 45) 0.999 938 793 266 314 267 852 8 × 2 = 1 + 0.999 877 586 532 628 535 705 6;
  • 46) 0.999 877 586 532 628 535 705 6 × 2 = 1 + 0.999 755 173 065 257 071 411 2;
  • 47) 0.999 755 173 065 257 071 411 2 × 2 = 1 + 0.999 510 346 130 514 142 822 4;
  • 48) 0.999 510 346 130 514 142 822 4 × 2 = 1 + 0.999 020 692 261 028 285 644 8;
  • 49) 0.999 020 692 261 028 285 644 8 × 2 = 1 + 0.998 041 384 522 056 571 289 6;
  • 50) 0.998 041 384 522 056 571 289 6 × 2 = 1 + 0.996 082 769 044 113 142 579 2;
  • 51) 0.996 082 769 044 113 142 579 2 × 2 = 1 + 0.992 165 538 088 226 285 158 4;
  • 52) 0.992 165 538 088 226 285 158 4 × 2 = 1 + 0.984 331 076 176 452 570 316 8;
  • 53) 0.984 331 076 176 452 570 316 8 × 2 = 1 + 0.968 662 152 352 905 140 633 6;
  • 54) 0.968 662 152 352 905 140 633 6 × 2 = 1 + 0.937 324 304 705 810 281 267 2;
  • 55) 0.937 324 304 705 810 281 267 2 × 2 = 1 + 0.874 648 609 411 620 562 534 4;
  • 56) 0.874 648 609 411 620 562 534 4 × 2 = 1 + 0.749 297 218 823 241 125 068 8;
  • 57) 0.749 297 218 823 241 125 068 8 × 2 = 1 + 0.498 594 437 646 482 250 137 6;
  • 58) 0.498 594 437 646 482 250 137 6 × 2 = 0 + 0.997 188 875 292 964 500 275 2;

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:


0.016 738 891 601 562 496 520 8(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 520 8(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:


0.016 738 891 601 562 496 520 8(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 -0.016 738 891 601 562 496 520 8 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