-0.016 738 891 601 562 496 530 539 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 539 1(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
-0.016 738 891 601 562 496 530 539 1(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:

|-0.016 738 891 601 562 496 530 539 1| = 0.016 738 891 601 562 496 530 539 1


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: 0.016 738 891 601 562 496 530 539 1.

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) 0.016 738 891 601 562 496 530 539 1 × 2 = 0 + 0.033 477 783 203 124 993 061 078 2;
  • 2) 0.033 477 783 203 124 993 061 078 2 × 2 = 0 + 0.066 955 566 406 249 986 122 156 4;
  • 3) 0.066 955 566 406 249 986 122 156 4 × 2 = 0 + 0.133 911 132 812 499 972 244 312 8;
  • 4) 0.133 911 132 812 499 972 244 312 8 × 2 = 0 + 0.267 822 265 624 999 944 488 625 6;
  • 5) 0.267 822 265 624 999 944 488 625 6 × 2 = 0 + 0.535 644 531 249 999 888 977 251 2;
  • 6) 0.535 644 531 249 999 888 977 251 2 × 2 = 1 + 0.071 289 062 499 999 777 954 502 4;
  • 7) 0.071 289 062 499 999 777 954 502 4 × 2 = 0 + 0.142 578 124 999 999 555 909 004 8;
  • 8) 0.142 578 124 999 999 555 909 004 8 × 2 = 0 + 0.285 156 249 999 999 111 818 009 6;
  • 9) 0.285 156 249 999 999 111 818 009 6 × 2 = 0 + 0.570 312 499 999 998 223 636 019 2;
  • 10) 0.570 312 499 999 998 223 636 019 2 × 2 = 1 + 0.140 624 999 999 996 447 272 038 4;
  • 11) 0.140 624 999 999 996 447 272 038 4 × 2 = 0 + 0.281 249 999 999 992 894 544 076 8;
  • 12) 0.281 249 999 999 992 894 544 076 8 × 2 = 0 + 0.562 499 999 999 985 789 088 153 6;
  • 13) 0.562 499 999 999 985 789 088 153 6 × 2 = 1 + 0.124 999 999 999 971 578 176 307 2;
  • 14) 0.124 999 999 999 971 578 176 307 2 × 2 = 0 + 0.249 999 999 999 943 156 352 614 4;
  • 15) 0.249 999 999 999 943 156 352 614 4 × 2 = 0 + 0.499 999 999 999 886 312 705 228 8;
  • 16) 0.499 999 999 999 886 312 705 228 8 × 2 = 0 + 0.999 999 999 999 772 625 410 457 6;
  • 17) 0.999 999 999 999 772 625 410 457 6 × 2 = 1 + 0.999 999 999 999 545 250 820 915 2;
  • 18) 0.999 999 999 999 545 250 820 915 2 × 2 = 1 + 0.999 999 999 999 090 501 641 830 4;
  • 19) 0.999 999 999 999 090 501 641 830 4 × 2 = 1 + 0.999 999 999 998 181 003 283 660 8;
  • 20) 0.999 999 999 998 181 003 283 660 8 × 2 = 1 + 0.999 999 999 996 362 006 567 321 6;
  • 21) 0.999 999 999 996 362 006 567 321 6 × 2 = 1 + 0.999 999 999 992 724 013 134 643 2;
  • 22) 0.999 999 999 992 724 013 134 643 2 × 2 = 1 + 0.999 999 999 985 448 026 269 286 4;
  • 23) 0.999 999 999 985 448 026 269 286 4 × 2 = 1 + 0.999 999 999 970 896 052 538 572 8;
  • 24) 0.999 999 999 970 896 052 538 572 8 × 2 = 1 + 0.999 999 999 941 792 105 077 145 6;
  • 25) 0.999 999 999 941 792 105 077 145 6 × 2 = 1 + 0.999 999 999 883 584 210 154 291 2;
  • 26) 0.999 999 999 883 584 210 154 291 2 × 2 = 1 + 0.999 999 999 767 168 420 308 582 4;
  • 27) 0.999 999 999 767 168 420 308 582 4 × 2 = 1 + 0.999 999 999 534 336 840 617 164 8;
  • 28) 0.999 999 999 534 336 840 617 164 8 × 2 = 1 + 0.999 999 999 068 673 681 234 329 6;
  • 29) 0.999 999 999 068 673 681 234 329 6 × 2 = 1 + 0.999 999 998 137 347 362 468 659 2;
  • 30) 0.999 999 998 137 347 362 468 659 2 × 2 = 1 + 0.999 999 996 274 694 724 937 318 4;
  • 31) 0.999 999 996 274 694 724 937 318 4 × 2 = 1 + 0.999 999 992 549 389 449 874 636 8;
  • 32) 0.999 999 992 549 389 449 874 636 8 × 2 = 1 + 0.999 999 985 098 778 899 749 273 6;
  • 33) 0.999 999 985 098 778 899 749 273 6 × 2 = 1 + 0.999 999 970 197 557 799 498 547 2;
  • 34) 0.999 999 970 197 557 799 498 547 2 × 2 = 1 + 0.999 999 940 395 115 598 997 094 4;
  • 35) 0.999 999 940 395 115 598 997 094 4 × 2 = 1 + 0.999 999 880 790 231 197 994 188 8;
  • 36) 0.999 999 880 790 231 197 994 188 8 × 2 = 1 + 0.999 999 761 580 462 395 988 377 6;
  • 37) 0.999 999 761 580 462 395 988 377 6 × 2 = 1 + 0.999 999 523 160 924 791 976 755 2;
  • 38) 0.999 999 523 160 924 791 976 755 2 × 2 = 1 + 0.999 999 046 321 849 583 953 510 4;
  • 39) 0.999 999 046 321 849 583 953 510 4 × 2 = 1 + 0.999 998 092 643 699 167 907 020 8;
  • 40) 0.999 998 092 643 699 167 907 020 8 × 2 = 1 + 0.999 996 185 287 398 335 814 041 6;
  • 41) 0.999 996 185 287 398 335 814 041 6 × 2 = 1 + 0.999 992 370 574 796 671 628 083 2;
  • 42) 0.999 992 370 574 796 671 628 083 2 × 2 = 1 + 0.999 984 741 149 593 343 256 166 4;
  • 43) 0.999 984 741 149 593 343 256 166 4 × 2 = 1 + 0.999 969 482 299 186 686 512 332 8;
  • 44) 0.999 969 482 299 186 686 512 332 8 × 2 = 1 + 0.999 938 964 598 373 373 024 665 6;
  • 45) 0.999 938 964 598 373 373 024 665 6 × 2 = 1 + 0.999 877 929 196 746 746 049 331 2;
  • 46) 0.999 877 929 196 746 746 049 331 2 × 2 = 1 + 0.999 755 858 393 493 492 098 662 4;
  • 47) 0.999 755 858 393 493 492 098 662 4 × 2 = 1 + 0.999 511 716 786 986 984 197 324 8;
  • 48) 0.999 511 716 786 986 984 197 324 8 × 2 = 1 + 0.999 023 433 573 973 968 394 649 6;
  • 49) 0.999 023 433 573 973 968 394 649 6 × 2 = 1 + 0.998 046 867 147 947 936 789 299 2;
  • 50) 0.998 046 867 147 947 936 789 299 2 × 2 = 1 + 0.996 093 734 295 895 873 578 598 4;
  • 51) 0.996 093 734 295 895 873 578 598 4 × 2 = 1 + 0.992 187 468 591 791 747 157 196 8;
  • 52) 0.992 187 468 591 791 747 157 196 8 × 2 = 1 + 0.984 374 937 183 583 494 314 393 6;
  • 53) 0.984 374 937 183 583 494 314 393 6 × 2 = 1 + 0.968 749 874 367 166 988 628 787 2;
  • 54) 0.968 749 874 367 166 988 628 787 2 × 2 = 1 + 0.937 499 748 734 333 977 257 574 4;
  • 55) 0.937 499 748 734 333 977 257 574 4 × 2 = 1 + 0.874 999 497 468 667 954 515 148 8;
  • 56) 0.874 999 497 468 667 954 515 148 8 × 2 = 1 + 0.749 998 994 937 335 909 030 297 6;
  • 57) 0.749 998 994 937 335 909 030 297 6 × 2 = 1 + 0.499 997 989 874 671 818 060 595 2;
  • 58) 0.499 997 989 874 671 818 060 595 2 × 2 = 0 + 0.999 995 979 749 343 636 121 190 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:


0.016 738 891 601 562 496 530 539 1(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 539 1(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:


0.016 738 891 601 562 496 530 539 1(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 -0.016 738 891 601 562 496 530 539 1 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