-0.016 738 891 601 562 496 530 531 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 531 3(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
-0.016 738 891 601 562 496 530 531 3(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:

|-0.016 738 891 601 562 496 530 531 3| = 0.016 738 891 601 562 496 530 531 3


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: 0.016 738 891 601 562 496 530 531 3.

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) 0.016 738 891 601 562 496 530 531 3 × 2 = 0 + 0.033 477 783 203 124 993 061 062 6;
  • 2) 0.033 477 783 203 124 993 061 062 6 × 2 = 0 + 0.066 955 566 406 249 986 122 125 2;
  • 3) 0.066 955 566 406 249 986 122 125 2 × 2 = 0 + 0.133 911 132 812 499 972 244 250 4;
  • 4) 0.133 911 132 812 499 972 244 250 4 × 2 = 0 + 0.267 822 265 624 999 944 488 500 8;
  • 5) 0.267 822 265 624 999 944 488 500 8 × 2 = 0 + 0.535 644 531 249 999 888 977 001 6;
  • 6) 0.535 644 531 249 999 888 977 001 6 × 2 = 1 + 0.071 289 062 499 999 777 954 003 2;
  • 7) 0.071 289 062 499 999 777 954 003 2 × 2 = 0 + 0.142 578 124 999 999 555 908 006 4;
  • 8) 0.142 578 124 999 999 555 908 006 4 × 2 = 0 + 0.285 156 249 999 999 111 816 012 8;
  • 9) 0.285 156 249 999 999 111 816 012 8 × 2 = 0 + 0.570 312 499 999 998 223 632 025 6;
  • 10) 0.570 312 499 999 998 223 632 025 6 × 2 = 1 + 0.140 624 999 999 996 447 264 051 2;
  • 11) 0.140 624 999 999 996 447 264 051 2 × 2 = 0 + 0.281 249 999 999 992 894 528 102 4;
  • 12) 0.281 249 999 999 992 894 528 102 4 × 2 = 0 + 0.562 499 999 999 985 789 056 204 8;
  • 13) 0.562 499 999 999 985 789 056 204 8 × 2 = 1 + 0.124 999 999 999 971 578 112 409 6;
  • 14) 0.124 999 999 999 971 578 112 409 6 × 2 = 0 + 0.249 999 999 999 943 156 224 819 2;
  • 15) 0.249 999 999 999 943 156 224 819 2 × 2 = 0 + 0.499 999 999 999 886 312 449 638 4;
  • 16) 0.499 999 999 999 886 312 449 638 4 × 2 = 0 + 0.999 999 999 999 772 624 899 276 8;
  • 17) 0.999 999 999 999 772 624 899 276 8 × 2 = 1 + 0.999 999 999 999 545 249 798 553 6;
  • 18) 0.999 999 999 999 545 249 798 553 6 × 2 = 1 + 0.999 999 999 999 090 499 597 107 2;
  • 19) 0.999 999 999 999 090 499 597 107 2 × 2 = 1 + 0.999 999 999 998 180 999 194 214 4;
  • 20) 0.999 999 999 998 180 999 194 214 4 × 2 = 1 + 0.999 999 999 996 361 998 388 428 8;
  • 21) 0.999 999 999 996 361 998 388 428 8 × 2 = 1 + 0.999 999 999 992 723 996 776 857 6;
  • 22) 0.999 999 999 992 723 996 776 857 6 × 2 = 1 + 0.999 999 999 985 447 993 553 715 2;
  • 23) 0.999 999 999 985 447 993 553 715 2 × 2 = 1 + 0.999 999 999 970 895 987 107 430 4;
  • 24) 0.999 999 999 970 895 987 107 430 4 × 2 = 1 + 0.999 999 999 941 791 974 214 860 8;
  • 25) 0.999 999 999 941 791 974 214 860 8 × 2 = 1 + 0.999 999 999 883 583 948 429 721 6;
  • 26) 0.999 999 999 883 583 948 429 721 6 × 2 = 1 + 0.999 999 999 767 167 896 859 443 2;
  • 27) 0.999 999 999 767 167 896 859 443 2 × 2 = 1 + 0.999 999 999 534 335 793 718 886 4;
  • 28) 0.999 999 999 534 335 793 718 886 4 × 2 = 1 + 0.999 999 999 068 671 587 437 772 8;
  • 29) 0.999 999 999 068 671 587 437 772 8 × 2 = 1 + 0.999 999 998 137 343 174 875 545 6;
  • 30) 0.999 999 998 137 343 174 875 545 6 × 2 = 1 + 0.999 999 996 274 686 349 751 091 2;
  • 31) 0.999 999 996 274 686 349 751 091 2 × 2 = 1 + 0.999 999 992 549 372 699 502 182 4;
  • 32) 0.999 999 992 549 372 699 502 182 4 × 2 = 1 + 0.999 999 985 098 745 399 004 364 8;
  • 33) 0.999 999 985 098 745 399 004 364 8 × 2 = 1 + 0.999 999 970 197 490 798 008 729 6;
  • 34) 0.999 999 970 197 490 798 008 729 6 × 2 = 1 + 0.999 999 940 394 981 596 017 459 2;
  • 35) 0.999 999 940 394 981 596 017 459 2 × 2 = 1 + 0.999 999 880 789 963 192 034 918 4;
  • 36) 0.999 999 880 789 963 192 034 918 4 × 2 = 1 + 0.999 999 761 579 926 384 069 836 8;
  • 37) 0.999 999 761 579 926 384 069 836 8 × 2 = 1 + 0.999 999 523 159 852 768 139 673 6;
  • 38) 0.999 999 523 159 852 768 139 673 6 × 2 = 1 + 0.999 999 046 319 705 536 279 347 2;
  • 39) 0.999 999 046 319 705 536 279 347 2 × 2 = 1 + 0.999 998 092 639 411 072 558 694 4;
  • 40) 0.999 998 092 639 411 072 558 694 4 × 2 = 1 + 0.999 996 185 278 822 145 117 388 8;
  • 41) 0.999 996 185 278 822 145 117 388 8 × 2 = 1 + 0.999 992 370 557 644 290 234 777 6;
  • 42) 0.999 992 370 557 644 290 234 777 6 × 2 = 1 + 0.999 984 741 115 288 580 469 555 2;
  • 43) 0.999 984 741 115 288 580 469 555 2 × 2 = 1 + 0.999 969 482 230 577 160 939 110 4;
  • 44) 0.999 969 482 230 577 160 939 110 4 × 2 = 1 + 0.999 938 964 461 154 321 878 220 8;
  • 45) 0.999 938 964 461 154 321 878 220 8 × 2 = 1 + 0.999 877 928 922 308 643 756 441 6;
  • 46) 0.999 877 928 922 308 643 756 441 6 × 2 = 1 + 0.999 755 857 844 617 287 512 883 2;
  • 47) 0.999 755 857 844 617 287 512 883 2 × 2 = 1 + 0.999 511 715 689 234 575 025 766 4;
  • 48) 0.999 511 715 689 234 575 025 766 4 × 2 = 1 + 0.999 023 431 378 469 150 051 532 8;
  • 49) 0.999 023 431 378 469 150 051 532 8 × 2 = 1 + 0.998 046 862 756 938 300 103 065 6;
  • 50) 0.998 046 862 756 938 300 103 065 6 × 2 = 1 + 0.996 093 725 513 876 600 206 131 2;
  • 51) 0.996 093 725 513 876 600 206 131 2 × 2 = 1 + 0.992 187 451 027 753 200 412 262 4;
  • 52) 0.992 187 451 027 753 200 412 262 4 × 2 = 1 + 0.984 374 902 055 506 400 824 524 8;
  • 53) 0.984 374 902 055 506 400 824 524 8 × 2 = 1 + 0.968 749 804 111 012 801 649 049 6;
  • 54) 0.968 749 804 111 012 801 649 049 6 × 2 = 1 + 0.937 499 608 222 025 603 298 099 2;
  • 55) 0.937 499 608 222 025 603 298 099 2 × 2 = 1 + 0.874 999 216 444 051 206 596 198 4;
  • 56) 0.874 999 216 444 051 206 596 198 4 × 2 = 1 + 0.749 998 432 888 102 413 192 396 8;
  • 57) 0.749 998 432 888 102 413 192 396 8 × 2 = 1 + 0.499 996 865 776 204 826 384 793 6;
  • 58) 0.499 996 865 776 204 826 384 793 6 × 2 = 0 + 0.999 993 731 552 409 652 769 587 2;

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:


0.016 738 891 601 562 496 530 531 3(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 531 3(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:


0.016 738 891 601 562 496 530 531 3(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 -0.016 738 891 601 562 496 530 531 3 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