-0.016 738 891 601 562 496 530 554 53 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 554 53(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
-0.016 738 891 601 562 496 530 554 53(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:

|-0.016 738 891 601 562 496 530 554 53| = 0.016 738 891 601 562 496 530 554 53


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: 0.016 738 891 601 562 496 530 554 53.

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) 0.016 738 891 601 562 496 530 554 53 × 2 = 0 + 0.033 477 783 203 124 993 061 109 06;
  • 2) 0.033 477 783 203 124 993 061 109 06 × 2 = 0 + 0.066 955 566 406 249 986 122 218 12;
  • 3) 0.066 955 566 406 249 986 122 218 12 × 2 = 0 + 0.133 911 132 812 499 972 244 436 24;
  • 4) 0.133 911 132 812 499 972 244 436 24 × 2 = 0 + 0.267 822 265 624 999 944 488 872 48;
  • 5) 0.267 822 265 624 999 944 488 872 48 × 2 = 0 + 0.535 644 531 249 999 888 977 744 96;
  • 6) 0.535 644 531 249 999 888 977 744 96 × 2 = 1 + 0.071 289 062 499 999 777 955 489 92;
  • 7) 0.071 289 062 499 999 777 955 489 92 × 2 = 0 + 0.142 578 124 999 999 555 910 979 84;
  • 8) 0.142 578 124 999 999 555 910 979 84 × 2 = 0 + 0.285 156 249 999 999 111 821 959 68;
  • 9) 0.285 156 249 999 999 111 821 959 68 × 2 = 0 + 0.570 312 499 999 998 223 643 919 36;
  • 10) 0.570 312 499 999 998 223 643 919 36 × 2 = 1 + 0.140 624 999 999 996 447 287 838 72;
  • 11) 0.140 624 999 999 996 447 287 838 72 × 2 = 0 + 0.281 249 999 999 992 894 575 677 44;
  • 12) 0.281 249 999 999 992 894 575 677 44 × 2 = 0 + 0.562 499 999 999 985 789 151 354 88;
  • 13) 0.562 499 999 999 985 789 151 354 88 × 2 = 1 + 0.124 999 999 999 971 578 302 709 76;
  • 14) 0.124 999 999 999 971 578 302 709 76 × 2 = 0 + 0.249 999 999 999 943 156 605 419 52;
  • 15) 0.249 999 999 999 943 156 605 419 52 × 2 = 0 + 0.499 999 999 999 886 313 210 839 04;
  • 16) 0.499 999 999 999 886 313 210 839 04 × 2 = 0 + 0.999 999 999 999 772 626 421 678 08;
  • 17) 0.999 999 999 999 772 626 421 678 08 × 2 = 1 + 0.999 999 999 999 545 252 843 356 16;
  • 18) 0.999 999 999 999 545 252 843 356 16 × 2 = 1 + 0.999 999 999 999 090 505 686 712 32;
  • 19) 0.999 999 999 999 090 505 686 712 32 × 2 = 1 + 0.999 999 999 998 181 011 373 424 64;
  • 20) 0.999 999 999 998 181 011 373 424 64 × 2 = 1 + 0.999 999 999 996 362 022 746 849 28;
  • 21) 0.999 999 999 996 362 022 746 849 28 × 2 = 1 + 0.999 999 999 992 724 045 493 698 56;
  • 22) 0.999 999 999 992 724 045 493 698 56 × 2 = 1 + 0.999 999 999 985 448 090 987 397 12;
  • 23) 0.999 999 999 985 448 090 987 397 12 × 2 = 1 + 0.999 999 999 970 896 181 974 794 24;
  • 24) 0.999 999 999 970 896 181 974 794 24 × 2 = 1 + 0.999 999 999 941 792 363 949 588 48;
  • 25) 0.999 999 999 941 792 363 949 588 48 × 2 = 1 + 0.999 999 999 883 584 727 899 176 96;
  • 26) 0.999 999 999 883 584 727 899 176 96 × 2 = 1 + 0.999 999 999 767 169 455 798 353 92;
  • 27) 0.999 999 999 767 169 455 798 353 92 × 2 = 1 + 0.999 999 999 534 338 911 596 707 84;
  • 28) 0.999 999 999 534 338 911 596 707 84 × 2 = 1 + 0.999 999 999 068 677 823 193 415 68;
  • 29) 0.999 999 999 068 677 823 193 415 68 × 2 = 1 + 0.999 999 998 137 355 646 386 831 36;
  • 30) 0.999 999 998 137 355 646 386 831 36 × 2 = 1 + 0.999 999 996 274 711 292 773 662 72;
  • 31) 0.999 999 996 274 711 292 773 662 72 × 2 = 1 + 0.999 999 992 549 422 585 547 325 44;
  • 32) 0.999 999 992 549 422 585 547 325 44 × 2 = 1 + 0.999 999 985 098 845 171 094 650 88;
  • 33) 0.999 999 985 098 845 171 094 650 88 × 2 = 1 + 0.999 999 970 197 690 342 189 301 76;
  • 34) 0.999 999 970 197 690 342 189 301 76 × 2 = 1 + 0.999 999 940 395 380 684 378 603 52;
  • 35) 0.999 999 940 395 380 684 378 603 52 × 2 = 1 + 0.999 999 880 790 761 368 757 207 04;
  • 36) 0.999 999 880 790 761 368 757 207 04 × 2 = 1 + 0.999 999 761 581 522 737 514 414 08;
  • 37) 0.999 999 761 581 522 737 514 414 08 × 2 = 1 + 0.999 999 523 163 045 475 028 828 16;
  • 38) 0.999 999 523 163 045 475 028 828 16 × 2 = 1 + 0.999 999 046 326 090 950 057 656 32;
  • 39) 0.999 999 046 326 090 950 057 656 32 × 2 = 1 + 0.999 998 092 652 181 900 115 312 64;
  • 40) 0.999 998 092 652 181 900 115 312 64 × 2 = 1 + 0.999 996 185 304 363 800 230 625 28;
  • 41) 0.999 996 185 304 363 800 230 625 28 × 2 = 1 + 0.999 992 370 608 727 600 461 250 56;
  • 42) 0.999 992 370 608 727 600 461 250 56 × 2 = 1 + 0.999 984 741 217 455 200 922 501 12;
  • 43) 0.999 984 741 217 455 200 922 501 12 × 2 = 1 + 0.999 969 482 434 910 401 845 002 24;
  • 44) 0.999 969 482 434 910 401 845 002 24 × 2 = 1 + 0.999 938 964 869 820 803 690 004 48;
  • 45) 0.999 938 964 869 820 803 690 004 48 × 2 = 1 + 0.999 877 929 739 641 607 380 008 96;
  • 46) 0.999 877 929 739 641 607 380 008 96 × 2 = 1 + 0.999 755 859 479 283 214 760 017 92;
  • 47) 0.999 755 859 479 283 214 760 017 92 × 2 = 1 + 0.999 511 718 958 566 429 520 035 84;
  • 48) 0.999 511 718 958 566 429 520 035 84 × 2 = 1 + 0.999 023 437 917 132 859 040 071 68;
  • 49) 0.999 023 437 917 132 859 040 071 68 × 2 = 1 + 0.998 046 875 834 265 718 080 143 36;
  • 50) 0.998 046 875 834 265 718 080 143 36 × 2 = 1 + 0.996 093 751 668 531 436 160 286 72;
  • 51) 0.996 093 751 668 531 436 160 286 72 × 2 = 1 + 0.992 187 503 337 062 872 320 573 44;
  • 52) 0.992 187 503 337 062 872 320 573 44 × 2 = 1 + 0.984 375 006 674 125 744 641 146 88;
  • 53) 0.984 375 006 674 125 744 641 146 88 × 2 = 1 + 0.968 750 013 348 251 489 282 293 76;
  • 54) 0.968 750 013 348 251 489 282 293 76 × 2 = 1 + 0.937 500 026 696 502 978 564 587 52;
  • 55) 0.937 500 026 696 502 978 564 587 52 × 2 = 1 + 0.875 000 053 393 005 957 129 175 04;
  • 56) 0.875 000 053 393 005 957 129 175 04 × 2 = 1 + 0.750 000 106 786 011 914 258 350 08;
  • 57) 0.750 000 106 786 011 914 258 350 08 × 2 = 1 + 0.500 000 213 572 023 828 516 700 16;
  • 58) 0.500 000 213 572 023 828 516 700 16 × 2 = 1 + 0.000 000 427 144 047 657 033 400 32;

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:


0.016 738 891 601 562 496 530 554 53(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 554 53(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:


0.016 738 891 601 562 496 530 554 53(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 -0.016 738 891 601 562 496 530 554 53 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