-0.016 738 891 601 562 496 529 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 529 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
-0.016 738 891 601 562 496 529 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:

|-0.016 738 891 601 562 496 529 8| = 0.016 738 891 601 562 496 529 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: 0.016 738 891 601 562 496 529 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) 0.016 738 891 601 562 496 529 8 × 2 = 0 + 0.033 477 783 203 124 993 059 6;
  • 2) 0.033 477 783 203 124 993 059 6 × 2 = 0 + 0.066 955 566 406 249 986 119 2;
  • 3) 0.066 955 566 406 249 986 119 2 × 2 = 0 + 0.133 911 132 812 499 972 238 4;
  • 4) 0.133 911 132 812 499 972 238 4 × 2 = 0 + 0.267 822 265 624 999 944 476 8;
  • 5) 0.267 822 265 624 999 944 476 8 × 2 = 0 + 0.535 644 531 249 999 888 953 6;
  • 6) 0.535 644 531 249 999 888 953 6 × 2 = 1 + 0.071 289 062 499 999 777 907 2;
  • 7) 0.071 289 062 499 999 777 907 2 × 2 = 0 + 0.142 578 124 999 999 555 814 4;
  • 8) 0.142 578 124 999 999 555 814 4 × 2 = 0 + 0.285 156 249 999 999 111 628 8;
  • 9) 0.285 156 249 999 999 111 628 8 × 2 = 0 + 0.570 312 499 999 998 223 257 6;
  • 10) 0.570 312 499 999 998 223 257 6 × 2 = 1 + 0.140 624 999 999 996 446 515 2;
  • 11) 0.140 624 999 999 996 446 515 2 × 2 = 0 + 0.281 249 999 999 992 893 030 4;
  • 12) 0.281 249 999 999 992 893 030 4 × 2 = 0 + 0.562 499 999 999 985 786 060 8;
  • 13) 0.562 499 999 999 985 786 060 8 × 2 = 1 + 0.124 999 999 999 971 572 121 6;
  • 14) 0.124 999 999 999 971 572 121 6 × 2 = 0 + 0.249 999 999 999 943 144 243 2;
  • 15) 0.249 999 999 999 943 144 243 2 × 2 = 0 + 0.499 999 999 999 886 288 486 4;
  • 16) 0.499 999 999 999 886 288 486 4 × 2 = 0 + 0.999 999 999 999 772 576 972 8;
  • 17) 0.999 999 999 999 772 576 972 8 × 2 = 1 + 0.999 999 999 999 545 153 945 6;
  • 18) 0.999 999 999 999 545 153 945 6 × 2 = 1 + 0.999 999 999 999 090 307 891 2;
  • 19) 0.999 999 999 999 090 307 891 2 × 2 = 1 + 0.999 999 999 998 180 615 782 4;
  • 20) 0.999 999 999 998 180 615 782 4 × 2 = 1 + 0.999 999 999 996 361 231 564 8;
  • 21) 0.999 999 999 996 361 231 564 8 × 2 = 1 + 0.999 999 999 992 722 463 129 6;
  • 22) 0.999 999 999 992 722 463 129 6 × 2 = 1 + 0.999 999 999 985 444 926 259 2;
  • 23) 0.999 999 999 985 444 926 259 2 × 2 = 1 + 0.999 999 999 970 889 852 518 4;
  • 24) 0.999 999 999 970 889 852 518 4 × 2 = 1 + 0.999 999 999 941 779 705 036 8;
  • 25) 0.999 999 999 941 779 705 036 8 × 2 = 1 + 0.999 999 999 883 559 410 073 6;
  • 26) 0.999 999 999 883 559 410 073 6 × 2 = 1 + 0.999 999 999 767 118 820 147 2;
  • 27) 0.999 999 999 767 118 820 147 2 × 2 = 1 + 0.999 999 999 534 237 640 294 4;
  • 28) 0.999 999 999 534 237 640 294 4 × 2 = 1 + 0.999 999 999 068 475 280 588 8;
  • 29) 0.999 999 999 068 475 280 588 8 × 2 = 1 + 0.999 999 998 136 950 561 177 6;
  • 30) 0.999 999 998 136 950 561 177 6 × 2 = 1 + 0.999 999 996 273 901 122 355 2;
  • 31) 0.999 999 996 273 901 122 355 2 × 2 = 1 + 0.999 999 992 547 802 244 710 4;
  • 32) 0.999 999 992 547 802 244 710 4 × 2 = 1 + 0.999 999 985 095 604 489 420 8;
  • 33) 0.999 999 985 095 604 489 420 8 × 2 = 1 + 0.999 999 970 191 208 978 841 6;
  • 34) 0.999 999 970 191 208 978 841 6 × 2 = 1 + 0.999 999 940 382 417 957 683 2;
  • 35) 0.999 999 940 382 417 957 683 2 × 2 = 1 + 0.999 999 880 764 835 915 366 4;
  • 36) 0.999 999 880 764 835 915 366 4 × 2 = 1 + 0.999 999 761 529 671 830 732 8;
  • 37) 0.999 999 761 529 671 830 732 8 × 2 = 1 + 0.999 999 523 059 343 661 465 6;
  • 38) 0.999 999 523 059 343 661 465 6 × 2 = 1 + 0.999 999 046 118 687 322 931 2;
  • 39) 0.999 999 046 118 687 322 931 2 × 2 = 1 + 0.999 998 092 237 374 645 862 4;
  • 40) 0.999 998 092 237 374 645 862 4 × 2 = 1 + 0.999 996 184 474 749 291 724 8;
  • 41) 0.999 996 184 474 749 291 724 8 × 2 = 1 + 0.999 992 368 949 498 583 449 6;
  • 42) 0.999 992 368 949 498 583 449 6 × 2 = 1 + 0.999 984 737 898 997 166 899 2;
  • 43) 0.999 984 737 898 997 166 899 2 × 2 = 1 + 0.999 969 475 797 994 333 798 4;
  • 44) 0.999 969 475 797 994 333 798 4 × 2 = 1 + 0.999 938 951 595 988 667 596 8;
  • 45) 0.999 938 951 595 988 667 596 8 × 2 = 1 + 0.999 877 903 191 977 335 193 6;
  • 46) 0.999 877 903 191 977 335 193 6 × 2 = 1 + 0.999 755 806 383 954 670 387 2;
  • 47) 0.999 755 806 383 954 670 387 2 × 2 = 1 + 0.999 511 612 767 909 340 774 4;
  • 48) 0.999 511 612 767 909 340 774 4 × 2 = 1 + 0.999 023 225 535 818 681 548 8;
  • 49) 0.999 023 225 535 818 681 548 8 × 2 = 1 + 0.998 046 451 071 637 363 097 6;
  • 50) 0.998 046 451 071 637 363 097 6 × 2 = 1 + 0.996 092 902 143 274 726 195 2;
  • 51) 0.996 092 902 143 274 726 195 2 × 2 = 1 + 0.992 185 804 286 549 452 390 4;
  • 52) 0.992 185 804 286 549 452 390 4 × 2 = 1 + 0.984 371 608 573 098 904 780 8;
  • 53) 0.984 371 608 573 098 904 780 8 × 2 = 1 + 0.968 743 217 146 197 809 561 6;
  • 54) 0.968 743 217 146 197 809 561 6 × 2 = 1 + 0.937 486 434 292 395 619 123 2;
  • 55) 0.937 486 434 292 395 619 123 2 × 2 = 1 + 0.874 972 868 584 791 238 246 4;
  • 56) 0.874 972 868 584 791 238 246 4 × 2 = 1 + 0.749 945 737 169 582 476 492 8;
  • 57) 0.749 945 737 169 582 476 492 8 × 2 = 1 + 0.499 891 474 339 164 952 985 6;
  • 58) 0.499 891 474 339 164 952 985 6 × 2 = 0 + 0.999 782 948 678 329 905 971 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:


0.016 738 891 601 562 496 529 8(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 529 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:


0.016 738 891 601 562 496 529 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 -0.016 738 891 601 562 496 529 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