-0.016 738 891 601 562 496 530 549 33 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 549 33(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
-0.016 738 891 601 562 496 530 549 33(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:

|-0.016 738 891 601 562 496 530 549 33| = 0.016 738 891 601 562 496 530 549 33


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: 0.016 738 891 601 562 496 530 549 33.

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) 0.016 738 891 601 562 496 530 549 33 × 2 = 0 + 0.033 477 783 203 124 993 061 098 66;
  • 2) 0.033 477 783 203 124 993 061 098 66 × 2 = 0 + 0.066 955 566 406 249 986 122 197 32;
  • 3) 0.066 955 566 406 249 986 122 197 32 × 2 = 0 + 0.133 911 132 812 499 972 244 394 64;
  • 4) 0.133 911 132 812 499 972 244 394 64 × 2 = 0 + 0.267 822 265 624 999 944 488 789 28;
  • 5) 0.267 822 265 624 999 944 488 789 28 × 2 = 0 + 0.535 644 531 249 999 888 977 578 56;
  • 6) 0.535 644 531 249 999 888 977 578 56 × 2 = 1 + 0.071 289 062 499 999 777 955 157 12;
  • 7) 0.071 289 062 499 999 777 955 157 12 × 2 = 0 + 0.142 578 124 999 999 555 910 314 24;
  • 8) 0.142 578 124 999 999 555 910 314 24 × 2 = 0 + 0.285 156 249 999 999 111 820 628 48;
  • 9) 0.285 156 249 999 999 111 820 628 48 × 2 = 0 + 0.570 312 499 999 998 223 641 256 96;
  • 10) 0.570 312 499 999 998 223 641 256 96 × 2 = 1 + 0.140 624 999 999 996 447 282 513 92;
  • 11) 0.140 624 999 999 996 447 282 513 92 × 2 = 0 + 0.281 249 999 999 992 894 565 027 84;
  • 12) 0.281 249 999 999 992 894 565 027 84 × 2 = 0 + 0.562 499 999 999 985 789 130 055 68;
  • 13) 0.562 499 999 999 985 789 130 055 68 × 2 = 1 + 0.124 999 999 999 971 578 260 111 36;
  • 14) 0.124 999 999 999 971 578 260 111 36 × 2 = 0 + 0.249 999 999 999 943 156 520 222 72;
  • 15) 0.249 999 999 999 943 156 520 222 72 × 2 = 0 + 0.499 999 999 999 886 313 040 445 44;
  • 16) 0.499 999 999 999 886 313 040 445 44 × 2 = 0 + 0.999 999 999 999 772 626 080 890 88;
  • 17) 0.999 999 999 999 772 626 080 890 88 × 2 = 1 + 0.999 999 999 999 545 252 161 781 76;
  • 18) 0.999 999 999 999 545 252 161 781 76 × 2 = 1 + 0.999 999 999 999 090 504 323 563 52;
  • 19) 0.999 999 999 999 090 504 323 563 52 × 2 = 1 + 0.999 999 999 998 181 008 647 127 04;
  • 20) 0.999 999 999 998 181 008 647 127 04 × 2 = 1 + 0.999 999 999 996 362 017 294 254 08;
  • 21) 0.999 999 999 996 362 017 294 254 08 × 2 = 1 + 0.999 999 999 992 724 034 588 508 16;
  • 22) 0.999 999 999 992 724 034 588 508 16 × 2 = 1 + 0.999 999 999 985 448 069 177 016 32;
  • 23) 0.999 999 999 985 448 069 177 016 32 × 2 = 1 + 0.999 999 999 970 896 138 354 032 64;
  • 24) 0.999 999 999 970 896 138 354 032 64 × 2 = 1 + 0.999 999 999 941 792 276 708 065 28;
  • 25) 0.999 999 999 941 792 276 708 065 28 × 2 = 1 + 0.999 999 999 883 584 553 416 130 56;
  • 26) 0.999 999 999 883 584 553 416 130 56 × 2 = 1 + 0.999 999 999 767 169 106 832 261 12;
  • 27) 0.999 999 999 767 169 106 832 261 12 × 2 = 1 + 0.999 999 999 534 338 213 664 522 24;
  • 28) 0.999 999 999 534 338 213 664 522 24 × 2 = 1 + 0.999 999 999 068 676 427 329 044 48;
  • 29) 0.999 999 999 068 676 427 329 044 48 × 2 = 1 + 0.999 999 998 137 352 854 658 088 96;
  • 30) 0.999 999 998 137 352 854 658 088 96 × 2 = 1 + 0.999 999 996 274 705 709 316 177 92;
  • 31) 0.999 999 996 274 705 709 316 177 92 × 2 = 1 + 0.999 999 992 549 411 418 632 355 84;
  • 32) 0.999 999 992 549 411 418 632 355 84 × 2 = 1 + 0.999 999 985 098 822 837 264 711 68;
  • 33) 0.999 999 985 098 822 837 264 711 68 × 2 = 1 + 0.999 999 970 197 645 674 529 423 36;
  • 34) 0.999 999 970 197 645 674 529 423 36 × 2 = 1 + 0.999 999 940 395 291 349 058 846 72;
  • 35) 0.999 999 940 395 291 349 058 846 72 × 2 = 1 + 0.999 999 880 790 582 698 117 693 44;
  • 36) 0.999 999 880 790 582 698 117 693 44 × 2 = 1 + 0.999 999 761 581 165 396 235 386 88;
  • 37) 0.999 999 761 581 165 396 235 386 88 × 2 = 1 + 0.999 999 523 162 330 792 470 773 76;
  • 38) 0.999 999 523 162 330 792 470 773 76 × 2 = 1 + 0.999 999 046 324 661 584 941 547 52;
  • 39) 0.999 999 046 324 661 584 941 547 52 × 2 = 1 + 0.999 998 092 649 323 169 883 095 04;
  • 40) 0.999 998 092 649 323 169 883 095 04 × 2 = 1 + 0.999 996 185 298 646 339 766 190 08;
  • 41) 0.999 996 185 298 646 339 766 190 08 × 2 = 1 + 0.999 992 370 597 292 679 532 380 16;
  • 42) 0.999 992 370 597 292 679 532 380 16 × 2 = 1 + 0.999 984 741 194 585 359 064 760 32;
  • 43) 0.999 984 741 194 585 359 064 760 32 × 2 = 1 + 0.999 969 482 389 170 718 129 520 64;
  • 44) 0.999 969 482 389 170 718 129 520 64 × 2 = 1 + 0.999 938 964 778 341 436 259 041 28;
  • 45) 0.999 938 964 778 341 436 259 041 28 × 2 = 1 + 0.999 877 929 556 682 872 518 082 56;
  • 46) 0.999 877 929 556 682 872 518 082 56 × 2 = 1 + 0.999 755 859 113 365 745 036 165 12;
  • 47) 0.999 755 859 113 365 745 036 165 12 × 2 = 1 + 0.999 511 718 226 731 490 072 330 24;
  • 48) 0.999 511 718 226 731 490 072 330 24 × 2 = 1 + 0.999 023 436 453 462 980 144 660 48;
  • 49) 0.999 023 436 453 462 980 144 660 48 × 2 = 1 + 0.998 046 872 906 925 960 289 320 96;
  • 50) 0.998 046 872 906 925 960 289 320 96 × 2 = 1 + 0.996 093 745 813 851 920 578 641 92;
  • 51) 0.996 093 745 813 851 920 578 641 92 × 2 = 1 + 0.992 187 491 627 703 841 157 283 84;
  • 52) 0.992 187 491 627 703 841 157 283 84 × 2 = 1 + 0.984 374 983 255 407 682 314 567 68;
  • 53) 0.984 374 983 255 407 682 314 567 68 × 2 = 1 + 0.968 749 966 510 815 364 629 135 36;
  • 54) 0.968 749 966 510 815 364 629 135 36 × 2 = 1 + 0.937 499 933 021 630 729 258 270 72;
  • 55) 0.937 499 933 021 630 729 258 270 72 × 2 = 1 + 0.874 999 866 043 261 458 516 541 44;
  • 56) 0.874 999 866 043 261 458 516 541 44 × 2 = 1 + 0.749 999 732 086 522 917 033 082 88;
  • 57) 0.749 999 732 086 522 917 033 082 88 × 2 = 1 + 0.499 999 464 173 045 834 066 165 76;
  • 58) 0.499 999 464 173 045 834 066 165 76 × 2 = 0 + 0.999 998 928 346 091 668 132 331 52;

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:


0.016 738 891 601 562 496 530 549 33(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 549 33(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:


0.016 738 891 601 562 496 530 549 33(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 -0.016 738 891 601 562 496 530 549 33 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