-0.016 738 891 601 562 496 530 552 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 552 15(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
-0.016 738 891 601 562 496 530 552 15(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:

|-0.016 738 891 601 562 496 530 552 15| = 0.016 738 891 601 562 496 530 552 15


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: 0.016 738 891 601 562 496 530 552 15.

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) 0.016 738 891 601 562 496 530 552 15 × 2 = 0 + 0.033 477 783 203 124 993 061 104 3;
  • 2) 0.033 477 783 203 124 993 061 104 3 × 2 = 0 + 0.066 955 566 406 249 986 122 208 6;
  • 3) 0.066 955 566 406 249 986 122 208 6 × 2 = 0 + 0.133 911 132 812 499 972 244 417 2;
  • 4) 0.133 911 132 812 499 972 244 417 2 × 2 = 0 + 0.267 822 265 624 999 944 488 834 4;
  • 5) 0.267 822 265 624 999 944 488 834 4 × 2 = 0 + 0.535 644 531 249 999 888 977 668 8;
  • 6) 0.535 644 531 249 999 888 977 668 8 × 2 = 1 + 0.071 289 062 499 999 777 955 337 6;
  • 7) 0.071 289 062 499 999 777 955 337 6 × 2 = 0 + 0.142 578 124 999 999 555 910 675 2;
  • 8) 0.142 578 124 999 999 555 910 675 2 × 2 = 0 + 0.285 156 249 999 999 111 821 350 4;
  • 9) 0.285 156 249 999 999 111 821 350 4 × 2 = 0 + 0.570 312 499 999 998 223 642 700 8;
  • 10) 0.570 312 499 999 998 223 642 700 8 × 2 = 1 + 0.140 624 999 999 996 447 285 401 6;
  • 11) 0.140 624 999 999 996 447 285 401 6 × 2 = 0 + 0.281 249 999 999 992 894 570 803 2;
  • 12) 0.281 249 999 999 992 894 570 803 2 × 2 = 0 + 0.562 499 999 999 985 789 141 606 4;
  • 13) 0.562 499 999 999 985 789 141 606 4 × 2 = 1 + 0.124 999 999 999 971 578 283 212 8;
  • 14) 0.124 999 999 999 971 578 283 212 8 × 2 = 0 + 0.249 999 999 999 943 156 566 425 6;
  • 15) 0.249 999 999 999 943 156 566 425 6 × 2 = 0 + 0.499 999 999 999 886 313 132 851 2;
  • 16) 0.499 999 999 999 886 313 132 851 2 × 2 = 0 + 0.999 999 999 999 772 626 265 702 4;
  • 17) 0.999 999 999 999 772 626 265 702 4 × 2 = 1 + 0.999 999 999 999 545 252 531 404 8;
  • 18) 0.999 999 999 999 545 252 531 404 8 × 2 = 1 + 0.999 999 999 999 090 505 062 809 6;
  • 19) 0.999 999 999 999 090 505 062 809 6 × 2 = 1 + 0.999 999 999 998 181 010 125 619 2;
  • 20) 0.999 999 999 998 181 010 125 619 2 × 2 = 1 + 0.999 999 999 996 362 020 251 238 4;
  • 21) 0.999 999 999 996 362 020 251 238 4 × 2 = 1 + 0.999 999 999 992 724 040 502 476 8;
  • 22) 0.999 999 999 992 724 040 502 476 8 × 2 = 1 + 0.999 999 999 985 448 081 004 953 6;
  • 23) 0.999 999 999 985 448 081 004 953 6 × 2 = 1 + 0.999 999 999 970 896 162 009 907 2;
  • 24) 0.999 999 999 970 896 162 009 907 2 × 2 = 1 + 0.999 999 999 941 792 324 019 814 4;
  • 25) 0.999 999 999 941 792 324 019 814 4 × 2 = 1 + 0.999 999 999 883 584 648 039 628 8;
  • 26) 0.999 999 999 883 584 648 039 628 8 × 2 = 1 + 0.999 999 999 767 169 296 079 257 6;
  • 27) 0.999 999 999 767 169 296 079 257 6 × 2 = 1 + 0.999 999 999 534 338 592 158 515 2;
  • 28) 0.999 999 999 534 338 592 158 515 2 × 2 = 1 + 0.999 999 999 068 677 184 317 030 4;
  • 29) 0.999 999 999 068 677 184 317 030 4 × 2 = 1 + 0.999 999 998 137 354 368 634 060 8;
  • 30) 0.999 999 998 137 354 368 634 060 8 × 2 = 1 + 0.999 999 996 274 708 737 268 121 6;
  • 31) 0.999 999 996 274 708 737 268 121 6 × 2 = 1 + 0.999 999 992 549 417 474 536 243 2;
  • 32) 0.999 999 992 549 417 474 536 243 2 × 2 = 1 + 0.999 999 985 098 834 949 072 486 4;
  • 33) 0.999 999 985 098 834 949 072 486 4 × 2 = 1 + 0.999 999 970 197 669 898 144 972 8;
  • 34) 0.999 999 970 197 669 898 144 972 8 × 2 = 1 + 0.999 999 940 395 339 796 289 945 6;
  • 35) 0.999 999 940 395 339 796 289 945 6 × 2 = 1 + 0.999 999 880 790 679 592 579 891 2;
  • 36) 0.999 999 880 790 679 592 579 891 2 × 2 = 1 + 0.999 999 761 581 359 185 159 782 4;
  • 37) 0.999 999 761 581 359 185 159 782 4 × 2 = 1 + 0.999 999 523 162 718 370 319 564 8;
  • 38) 0.999 999 523 162 718 370 319 564 8 × 2 = 1 + 0.999 999 046 325 436 740 639 129 6;
  • 39) 0.999 999 046 325 436 740 639 129 6 × 2 = 1 + 0.999 998 092 650 873 481 278 259 2;
  • 40) 0.999 998 092 650 873 481 278 259 2 × 2 = 1 + 0.999 996 185 301 746 962 556 518 4;
  • 41) 0.999 996 185 301 746 962 556 518 4 × 2 = 1 + 0.999 992 370 603 493 925 113 036 8;
  • 42) 0.999 992 370 603 493 925 113 036 8 × 2 = 1 + 0.999 984 741 206 987 850 226 073 6;
  • 43) 0.999 984 741 206 987 850 226 073 6 × 2 = 1 + 0.999 969 482 413 975 700 452 147 2;
  • 44) 0.999 969 482 413 975 700 452 147 2 × 2 = 1 + 0.999 938 964 827 951 400 904 294 4;
  • 45) 0.999 938 964 827 951 400 904 294 4 × 2 = 1 + 0.999 877 929 655 902 801 808 588 8;
  • 46) 0.999 877 929 655 902 801 808 588 8 × 2 = 1 + 0.999 755 859 311 805 603 617 177 6;
  • 47) 0.999 755 859 311 805 603 617 177 6 × 2 = 1 + 0.999 511 718 623 611 207 234 355 2;
  • 48) 0.999 511 718 623 611 207 234 355 2 × 2 = 1 + 0.999 023 437 247 222 414 468 710 4;
  • 49) 0.999 023 437 247 222 414 468 710 4 × 2 = 1 + 0.998 046 874 494 444 828 937 420 8;
  • 50) 0.998 046 874 494 444 828 937 420 8 × 2 = 1 + 0.996 093 748 988 889 657 874 841 6;
  • 51) 0.996 093 748 988 889 657 874 841 6 × 2 = 1 + 0.992 187 497 977 779 315 749 683 2;
  • 52) 0.992 187 497 977 779 315 749 683 2 × 2 = 1 + 0.984 374 995 955 558 631 499 366 4;
  • 53) 0.984 374 995 955 558 631 499 366 4 × 2 = 1 + 0.968 749 991 911 117 262 998 732 8;
  • 54) 0.968 749 991 911 117 262 998 732 8 × 2 = 1 + 0.937 499 983 822 234 525 997 465 6;
  • 55) 0.937 499 983 822 234 525 997 465 6 × 2 = 1 + 0.874 999 967 644 469 051 994 931 2;
  • 56) 0.874 999 967 644 469 051 994 931 2 × 2 = 1 + 0.749 999 935 288 938 103 989 862 4;
  • 57) 0.749 999 935 288 938 103 989 862 4 × 2 = 1 + 0.499 999 870 577 876 207 979 724 8;
  • 58) 0.499 999 870 577 876 207 979 724 8 × 2 = 0 + 0.999 999 741 155 752 415 959 449 6;

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:


0.016 738 891 601 562 496 530 552 15(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 552 15(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:


0.016 738 891 601 562 496 530 552 15(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 -0.016 738 891 601 562 496 530 552 15 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