-0.016 738 891 601 562 496 530 555 33 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 555 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
-0.016 738 891 601 562 496 530 555 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:

|-0.016 738 891 601 562 496 530 555 33| = 0.016 738 891 601 562 496 530 555 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: 0.016 738 891 601 562 496 530 555 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) 0.016 738 891 601 562 496 530 555 33 × 2 = 0 + 0.033 477 783 203 124 993 061 110 66;
  • 2) 0.033 477 783 203 124 993 061 110 66 × 2 = 0 + 0.066 955 566 406 249 986 122 221 32;
  • 3) 0.066 955 566 406 249 986 122 221 32 × 2 = 0 + 0.133 911 132 812 499 972 244 442 64;
  • 4) 0.133 911 132 812 499 972 244 442 64 × 2 = 0 + 0.267 822 265 624 999 944 488 885 28;
  • 5) 0.267 822 265 624 999 944 488 885 28 × 2 = 0 + 0.535 644 531 249 999 888 977 770 56;
  • 6) 0.535 644 531 249 999 888 977 770 56 × 2 = 1 + 0.071 289 062 499 999 777 955 541 12;
  • 7) 0.071 289 062 499 999 777 955 541 12 × 2 = 0 + 0.142 578 124 999 999 555 911 082 24;
  • 8) 0.142 578 124 999 999 555 911 082 24 × 2 = 0 + 0.285 156 249 999 999 111 822 164 48;
  • 9) 0.285 156 249 999 999 111 822 164 48 × 2 = 0 + 0.570 312 499 999 998 223 644 328 96;
  • 10) 0.570 312 499 999 998 223 644 328 96 × 2 = 1 + 0.140 624 999 999 996 447 288 657 92;
  • 11) 0.140 624 999 999 996 447 288 657 92 × 2 = 0 + 0.281 249 999 999 992 894 577 315 84;
  • 12) 0.281 249 999 999 992 894 577 315 84 × 2 = 0 + 0.562 499 999 999 985 789 154 631 68;
  • 13) 0.562 499 999 999 985 789 154 631 68 × 2 = 1 + 0.124 999 999 999 971 578 309 263 36;
  • 14) 0.124 999 999 999 971 578 309 263 36 × 2 = 0 + 0.249 999 999 999 943 156 618 526 72;
  • 15) 0.249 999 999 999 943 156 618 526 72 × 2 = 0 + 0.499 999 999 999 886 313 237 053 44;
  • 16) 0.499 999 999 999 886 313 237 053 44 × 2 = 0 + 0.999 999 999 999 772 626 474 106 88;
  • 17) 0.999 999 999 999 772 626 474 106 88 × 2 = 1 + 0.999 999 999 999 545 252 948 213 76;
  • 18) 0.999 999 999 999 545 252 948 213 76 × 2 = 1 + 0.999 999 999 999 090 505 896 427 52;
  • 19) 0.999 999 999 999 090 505 896 427 52 × 2 = 1 + 0.999 999 999 998 181 011 792 855 04;
  • 20) 0.999 999 999 998 181 011 792 855 04 × 2 = 1 + 0.999 999 999 996 362 023 585 710 08;
  • 21) 0.999 999 999 996 362 023 585 710 08 × 2 = 1 + 0.999 999 999 992 724 047 171 420 16;
  • 22) 0.999 999 999 992 724 047 171 420 16 × 2 = 1 + 0.999 999 999 985 448 094 342 840 32;
  • 23) 0.999 999 999 985 448 094 342 840 32 × 2 = 1 + 0.999 999 999 970 896 188 685 680 64;
  • 24) 0.999 999 999 970 896 188 685 680 64 × 2 = 1 + 0.999 999 999 941 792 377 371 361 28;
  • 25) 0.999 999 999 941 792 377 371 361 28 × 2 = 1 + 0.999 999 999 883 584 754 742 722 56;
  • 26) 0.999 999 999 883 584 754 742 722 56 × 2 = 1 + 0.999 999 999 767 169 509 485 445 12;
  • 27) 0.999 999 999 767 169 509 485 445 12 × 2 = 1 + 0.999 999 999 534 339 018 970 890 24;
  • 28) 0.999 999 999 534 339 018 970 890 24 × 2 = 1 + 0.999 999 999 068 678 037 941 780 48;
  • 29) 0.999 999 999 068 678 037 941 780 48 × 2 = 1 + 0.999 999 998 137 356 075 883 560 96;
  • 30) 0.999 999 998 137 356 075 883 560 96 × 2 = 1 + 0.999 999 996 274 712 151 767 121 92;
  • 31) 0.999 999 996 274 712 151 767 121 92 × 2 = 1 + 0.999 999 992 549 424 303 534 243 84;
  • 32) 0.999 999 992 549 424 303 534 243 84 × 2 = 1 + 0.999 999 985 098 848 607 068 487 68;
  • 33) 0.999 999 985 098 848 607 068 487 68 × 2 = 1 + 0.999 999 970 197 697 214 136 975 36;
  • 34) 0.999 999 970 197 697 214 136 975 36 × 2 = 1 + 0.999 999 940 395 394 428 273 950 72;
  • 35) 0.999 999 940 395 394 428 273 950 72 × 2 = 1 + 0.999 999 880 790 788 856 547 901 44;
  • 36) 0.999 999 880 790 788 856 547 901 44 × 2 = 1 + 0.999 999 761 581 577 713 095 802 88;
  • 37) 0.999 999 761 581 577 713 095 802 88 × 2 = 1 + 0.999 999 523 163 155 426 191 605 76;
  • 38) 0.999 999 523 163 155 426 191 605 76 × 2 = 1 + 0.999 999 046 326 310 852 383 211 52;
  • 39) 0.999 999 046 326 310 852 383 211 52 × 2 = 1 + 0.999 998 092 652 621 704 766 423 04;
  • 40) 0.999 998 092 652 621 704 766 423 04 × 2 = 1 + 0.999 996 185 305 243 409 532 846 08;
  • 41) 0.999 996 185 305 243 409 532 846 08 × 2 = 1 + 0.999 992 370 610 486 819 065 692 16;
  • 42) 0.999 992 370 610 486 819 065 692 16 × 2 = 1 + 0.999 984 741 220 973 638 131 384 32;
  • 43) 0.999 984 741 220 973 638 131 384 32 × 2 = 1 + 0.999 969 482 441 947 276 262 768 64;
  • 44) 0.999 969 482 441 947 276 262 768 64 × 2 = 1 + 0.999 938 964 883 894 552 525 537 28;
  • 45) 0.999 938 964 883 894 552 525 537 28 × 2 = 1 + 0.999 877 929 767 789 105 051 074 56;
  • 46) 0.999 877 929 767 789 105 051 074 56 × 2 = 1 + 0.999 755 859 535 578 210 102 149 12;
  • 47) 0.999 755 859 535 578 210 102 149 12 × 2 = 1 + 0.999 511 719 071 156 420 204 298 24;
  • 48) 0.999 511 719 071 156 420 204 298 24 × 2 = 1 + 0.999 023 438 142 312 840 408 596 48;
  • 49) 0.999 023 438 142 312 840 408 596 48 × 2 = 1 + 0.998 046 876 284 625 680 817 192 96;
  • 50) 0.998 046 876 284 625 680 817 192 96 × 2 = 1 + 0.996 093 752 569 251 361 634 385 92;
  • 51) 0.996 093 752 569 251 361 634 385 92 × 2 = 1 + 0.992 187 505 138 502 723 268 771 84;
  • 52) 0.992 187 505 138 502 723 268 771 84 × 2 = 1 + 0.984 375 010 277 005 446 537 543 68;
  • 53) 0.984 375 010 277 005 446 537 543 68 × 2 = 1 + 0.968 750 020 554 010 893 075 087 36;
  • 54) 0.968 750 020 554 010 893 075 087 36 × 2 = 1 + 0.937 500 041 108 021 786 150 174 72;
  • 55) 0.937 500 041 108 021 786 150 174 72 × 2 = 1 + 0.875 000 082 216 043 572 300 349 44;
  • 56) 0.875 000 082 216 043 572 300 349 44 × 2 = 1 + 0.750 000 164 432 087 144 600 698 88;
  • 57) 0.750 000 164 432 087 144 600 698 88 × 2 = 1 + 0.500 000 328 864 174 289 201 397 76;
  • 58) 0.500 000 328 864 174 289 201 397 76 × 2 = 1 + 0.000 000 657 728 348 578 402 795 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:


0.016 738 891 601 562 496 530 555 33(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 555 33(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:


0.016 738 891 601 562 496 530 555 33(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 -0.016 738 891 601 562 496 530 555 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 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