-0.016 738 891 601 562 496 530 566 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 566 6(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
-0.016 738 891 601 562 496 530 566 6(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:

|-0.016 738 891 601 562 496 530 566 6| = 0.016 738 891 601 562 496 530 566 6


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: 0.016 738 891 601 562 496 530 566 6.

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) 0.016 738 891 601 562 496 530 566 6 × 2 = 0 + 0.033 477 783 203 124 993 061 133 2;
  • 2) 0.033 477 783 203 124 993 061 133 2 × 2 = 0 + 0.066 955 566 406 249 986 122 266 4;
  • 3) 0.066 955 566 406 249 986 122 266 4 × 2 = 0 + 0.133 911 132 812 499 972 244 532 8;
  • 4) 0.133 911 132 812 499 972 244 532 8 × 2 = 0 + 0.267 822 265 624 999 944 489 065 6;
  • 5) 0.267 822 265 624 999 944 489 065 6 × 2 = 0 + 0.535 644 531 249 999 888 978 131 2;
  • 6) 0.535 644 531 249 999 888 978 131 2 × 2 = 1 + 0.071 289 062 499 999 777 956 262 4;
  • 7) 0.071 289 062 499 999 777 956 262 4 × 2 = 0 + 0.142 578 124 999 999 555 912 524 8;
  • 8) 0.142 578 124 999 999 555 912 524 8 × 2 = 0 + 0.285 156 249 999 999 111 825 049 6;
  • 9) 0.285 156 249 999 999 111 825 049 6 × 2 = 0 + 0.570 312 499 999 998 223 650 099 2;
  • 10) 0.570 312 499 999 998 223 650 099 2 × 2 = 1 + 0.140 624 999 999 996 447 300 198 4;
  • 11) 0.140 624 999 999 996 447 300 198 4 × 2 = 0 + 0.281 249 999 999 992 894 600 396 8;
  • 12) 0.281 249 999 999 992 894 600 396 8 × 2 = 0 + 0.562 499 999 999 985 789 200 793 6;
  • 13) 0.562 499 999 999 985 789 200 793 6 × 2 = 1 + 0.124 999 999 999 971 578 401 587 2;
  • 14) 0.124 999 999 999 971 578 401 587 2 × 2 = 0 + 0.249 999 999 999 943 156 803 174 4;
  • 15) 0.249 999 999 999 943 156 803 174 4 × 2 = 0 + 0.499 999 999 999 886 313 606 348 8;
  • 16) 0.499 999 999 999 886 313 606 348 8 × 2 = 0 + 0.999 999 999 999 772 627 212 697 6;
  • 17) 0.999 999 999 999 772 627 212 697 6 × 2 = 1 + 0.999 999 999 999 545 254 425 395 2;
  • 18) 0.999 999 999 999 545 254 425 395 2 × 2 = 1 + 0.999 999 999 999 090 508 850 790 4;
  • 19) 0.999 999 999 999 090 508 850 790 4 × 2 = 1 + 0.999 999 999 998 181 017 701 580 8;
  • 20) 0.999 999 999 998 181 017 701 580 8 × 2 = 1 + 0.999 999 999 996 362 035 403 161 6;
  • 21) 0.999 999 999 996 362 035 403 161 6 × 2 = 1 + 0.999 999 999 992 724 070 806 323 2;
  • 22) 0.999 999 999 992 724 070 806 323 2 × 2 = 1 + 0.999 999 999 985 448 141 612 646 4;
  • 23) 0.999 999 999 985 448 141 612 646 4 × 2 = 1 + 0.999 999 999 970 896 283 225 292 8;
  • 24) 0.999 999 999 970 896 283 225 292 8 × 2 = 1 + 0.999 999 999 941 792 566 450 585 6;
  • 25) 0.999 999 999 941 792 566 450 585 6 × 2 = 1 + 0.999 999 999 883 585 132 901 171 2;
  • 26) 0.999 999 999 883 585 132 901 171 2 × 2 = 1 + 0.999 999 999 767 170 265 802 342 4;
  • 27) 0.999 999 999 767 170 265 802 342 4 × 2 = 1 + 0.999 999 999 534 340 531 604 684 8;
  • 28) 0.999 999 999 534 340 531 604 684 8 × 2 = 1 + 0.999 999 999 068 681 063 209 369 6;
  • 29) 0.999 999 999 068 681 063 209 369 6 × 2 = 1 + 0.999 999 998 137 362 126 418 739 2;
  • 30) 0.999 999 998 137 362 126 418 739 2 × 2 = 1 + 0.999 999 996 274 724 252 837 478 4;
  • 31) 0.999 999 996 274 724 252 837 478 4 × 2 = 1 + 0.999 999 992 549 448 505 674 956 8;
  • 32) 0.999 999 992 549 448 505 674 956 8 × 2 = 1 + 0.999 999 985 098 897 011 349 913 6;
  • 33) 0.999 999 985 098 897 011 349 913 6 × 2 = 1 + 0.999 999 970 197 794 022 699 827 2;
  • 34) 0.999 999 970 197 794 022 699 827 2 × 2 = 1 + 0.999 999 940 395 588 045 399 654 4;
  • 35) 0.999 999 940 395 588 045 399 654 4 × 2 = 1 + 0.999 999 880 791 176 090 799 308 8;
  • 36) 0.999 999 880 791 176 090 799 308 8 × 2 = 1 + 0.999 999 761 582 352 181 598 617 6;
  • 37) 0.999 999 761 582 352 181 598 617 6 × 2 = 1 + 0.999 999 523 164 704 363 197 235 2;
  • 38) 0.999 999 523 164 704 363 197 235 2 × 2 = 1 + 0.999 999 046 329 408 726 394 470 4;
  • 39) 0.999 999 046 329 408 726 394 470 4 × 2 = 1 + 0.999 998 092 658 817 452 788 940 8;
  • 40) 0.999 998 092 658 817 452 788 940 8 × 2 = 1 + 0.999 996 185 317 634 905 577 881 6;
  • 41) 0.999 996 185 317 634 905 577 881 6 × 2 = 1 + 0.999 992 370 635 269 811 155 763 2;
  • 42) 0.999 992 370 635 269 811 155 763 2 × 2 = 1 + 0.999 984 741 270 539 622 311 526 4;
  • 43) 0.999 984 741 270 539 622 311 526 4 × 2 = 1 + 0.999 969 482 541 079 244 623 052 8;
  • 44) 0.999 969 482 541 079 244 623 052 8 × 2 = 1 + 0.999 938 965 082 158 489 246 105 6;
  • 45) 0.999 938 965 082 158 489 246 105 6 × 2 = 1 + 0.999 877 930 164 316 978 492 211 2;
  • 46) 0.999 877 930 164 316 978 492 211 2 × 2 = 1 + 0.999 755 860 328 633 956 984 422 4;
  • 47) 0.999 755 860 328 633 956 984 422 4 × 2 = 1 + 0.999 511 720 657 267 913 968 844 8;
  • 48) 0.999 511 720 657 267 913 968 844 8 × 2 = 1 + 0.999 023 441 314 535 827 937 689 6;
  • 49) 0.999 023 441 314 535 827 937 689 6 × 2 = 1 + 0.998 046 882 629 071 655 875 379 2;
  • 50) 0.998 046 882 629 071 655 875 379 2 × 2 = 1 + 0.996 093 765 258 143 311 750 758 4;
  • 51) 0.996 093 765 258 143 311 750 758 4 × 2 = 1 + 0.992 187 530 516 286 623 501 516 8;
  • 52) 0.992 187 530 516 286 623 501 516 8 × 2 = 1 + 0.984 375 061 032 573 247 003 033 6;
  • 53) 0.984 375 061 032 573 247 003 033 6 × 2 = 1 + 0.968 750 122 065 146 494 006 067 2;
  • 54) 0.968 750 122 065 146 494 006 067 2 × 2 = 1 + 0.937 500 244 130 292 988 012 134 4;
  • 55) 0.937 500 244 130 292 988 012 134 4 × 2 = 1 + 0.875 000 488 260 585 976 024 268 8;
  • 56) 0.875 000 488 260 585 976 024 268 8 × 2 = 1 + 0.750 000 976 521 171 952 048 537 6;
  • 57) 0.750 000 976 521 171 952 048 537 6 × 2 = 1 + 0.500 001 953 042 343 904 097 075 2;
  • 58) 0.500 001 953 042 343 904 097 075 2 × 2 = 1 + 0.000 003 906 084 687 808 194 150 4;

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:


0.016 738 891 601 562 496 530 566 6(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 566 6(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(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:


0.016 738 891 601 562 496 530 566 6(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(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 1111


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 1111 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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 1111


Decimal number -0.016 738 891 601 562 496 530 566 6 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 1111

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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