-0.016 738 891 601 562 496 530 538 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 538 4(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
-0.016 738 891 601 562 496 530 538 4(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:

|-0.016 738 891 601 562 496 530 538 4| = 0.016 738 891 601 562 496 530 538 4


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: 0.016 738 891 601 562 496 530 538 4.

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) 0.016 738 891 601 562 496 530 538 4 × 2 = 0 + 0.033 477 783 203 124 993 061 076 8;
  • 2) 0.033 477 783 203 124 993 061 076 8 × 2 = 0 + 0.066 955 566 406 249 986 122 153 6;
  • 3) 0.066 955 566 406 249 986 122 153 6 × 2 = 0 + 0.133 911 132 812 499 972 244 307 2;
  • 4) 0.133 911 132 812 499 972 244 307 2 × 2 = 0 + 0.267 822 265 624 999 944 488 614 4;
  • 5) 0.267 822 265 624 999 944 488 614 4 × 2 = 0 + 0.535 644 531 249 999 888 977 228 8;
  • 6) 0.535 644 531 249 999 888 977 228 8 × 2 = 1 + 0.071 289 062 499 999 777 954 457 6;
  • 7) 0.071 289 062 499 999 777 954 457 6 × 2 = 0 + 0.142 578 124 999 999 555 908 915 2;
  • 8) 0.142 578 124 999 999 555 908 915 2 × 2 = 0 + 0.285 156 249 999 999 111 817 830 4;
  • 9) 0.285 156 249 999 999 111 817 830 4 × 2 = 0 + 0.570 312 499 999 998 223 635 660 8;
  • 10) 0.570 312 499 999 998 223 635 660 8 × 2 = 1 + 0.140 624 999 999 996 447 271 321 6;
  • 11) 0.140 624 999 999 996 447 271 321 6 × 2 = 0 + 0.281 249 999 999 992 894 542 643 2;
  • 12) 0.281 249 999 999 992 894 542 643 2 × 2 = 0 + 0.562 499 999 999 985 789 085 286 4;
  • 13) 0.562 499 999 999 985 789 085 286 4 × 2 = 1 + 0.124 999 999 999 971 578 170 572 8;
  • 14) 0.124 999 999 999 971 578 170 572 8 × 2 = 0 + 0.249 999 999 999 943 156 341 145 6;
  • 15) 0.249 999 999 999 943 156 341 145 6 × 2 = 0 + 0.499 999 999 999 886 312 682 291 2;
  • 16) 0.499 999 999 999 886 312 682 291 2 × 2 = 0 + 0.999 999 999 999 772 625 364 582 4;
  • 17) 0.999 999 999 999 772 625 364 582 4 × 2 = 1 + 0.999 999 999 999 545 250 729 164 8;
  • 18) 0.999 999 999 999 545 250 729 164 8 × 2 = 1 + 0.999 999 999 999 090 501 458 329 6;
  • 19) 0.999 999 999 999 090 501 458 329 6 × 2 = 1 + 0.999 999 999 998 181 002 916 659 2;
  • 20) 0.999 999 999 998 181 002 916 659 2 × 2 = 1 + 0.999 999 999 996 362 005 833 318 4;
  • 21) 0.999 999 999 996 362 005 833 318 4 × 2 = 1 + 0.999 999 999 992 724 011 666 636 8;
  • 22) 0.999 999 999 992 724 011 666 636 8 × 2 = 1 + 0.999 999 999 985 448 023 333 273 6;
  • 23) 0.999 999 999 985 448 023 333 273 6 × 2 = 1 + 0.999 999 999 970 896 046 666 547 2;
  • 24) 0.999 999 999 970 896 046 666 547 2 × 2 = 1 + 0.999 999 999 941 792 093 333 094 4;
  • 25) 0.999 999 999 941 792 093 333 094 4 × 2 = 1 + 0.999 999 999 883 584 186 666 188 8;
  • 26) 0.999 999 999 883 584 186 666 188 8 × 2 = 1 + 0.999 999 999 767 168 373 332 377 6;
  • 27) 0.999 999 999 767 168 373 332 377 6 × 2 = 1 + 0.999 999 999 534 336 746 664 755 2;
  • 28) 0.999 999 999 534 336 746 664 755 2 × 2 = 1 + 0.999 999 999 068 673 493 329 510 4;
  • 29) 0.999 999 999 068 673 493 329 510 4 × 2 = 1 + 0.999 999 998 137 346 986 659 020 8;
  • 30) 0.999 999 998 137 346 986 659 020 8 × 2 = 1 + 0.999 999 996 274 693 973 318 041 6;
  • 31) 0.999 999 996 274 693 973 318 041 6 × 2 = 1 + 0.999 999 992 549 387 946 636 083 2;
  • 32) 0.999 999 992 549 387 946 636 083 2 × 2 = 1 + 0.999 999 985 098 775 893 272 166 4;
  • 33) 0.999 999 985 098 775 893 272 166 4 × 2 = 1 + 0.999 999 970 197 551 786 544 332 8;
  • 34) 0.999 999 970 197 551 786 544 332 8 × 2 = 1 + 0.999 999 940 395 103 573 088 665 6;
  • 35) 0.999 999 940 395 103 573 088 665 6 × 2 = 1 + 0.999 999 880 790 207 146 177 331 2;
  • 36) 0.999 999 880 790 207 146 177 331 2 × 2 = 1 + 0.999 999 761 580 414 292 354 662 4;
  • 37) 0.999 999 761 580 414 292 354 662 4 × 2 = 1 + 0.999 999 523 160 828 584 709 324 8;
  • 38) 0.999 999 523 160 828 584 709 324 8 × 2 = 1 + 0.999 999 046 321 657 169 418 649 6;
  • 39) 0.999 999 046 321 657 169 418 649 6 × 2 = 1 + 0.999 998 092 643 314 338 837 299 2;
  • 40) 0.999 998 092 643 314 338 837 299 2 × 2 = 1 + 0.999 996 185 286 628 677 674 598 4;
  • 41) 0.999 996 185 286 628 677 674 598 4 × 2 = 1 + 0.999 992 370 573 257 355 349 196 8;
  • 42) 0.999 992 370 573 257 355 349 196 8 × 2 = 1 + 0.999 984 741 146 514 710 698 393 6;
  • 43) 0.999 984 741 146 514 710 698 393 6 × 2 = 1 + 0.999 969 482 293 029 421 396 787 2;
  • 44) 0.999 969 482 293 029 421 396 787 2 × 2 = 1 + 0.999 938 964 586 058 842 793 574 4;
  • 45) 0.999 938 964 586 058 842 793 574 4 × 2 = 1 + 0.999 877 929 172 117 685 587 148 8;
  • 46) 0.999 877 929 172 117 685 587 148 8 × 2 = 1 + 0.999 755 858 344 235 371 174 297 6;
  • 47) 0.999 755 858 344 235 371 174 297 6 × 2 = 1 + 0.999 511 716 688 470 742 348 595 2;
  • 48) 0.999 511 716 688 470 742 348 595 2 × 2 = 1 + 0.999 023 433 376 941 484 697 190 4;
  • 49) 0.999 023 433 376 941 484 697 190 4 × 2 = 1 + 0.998 046 866 753 882 969 394 380 8;
  • 50) 0.998 046 866 753 882 969 394 380 8 × 2 = 1 + 0.996 093 733 507 765 938 788 761 6;
  • 51) 0.996 093 733 507 765 938 788 761 6 × 2 = 1 + 0.992 187 467 015 531 877 577 523 2;
  • 52) 0.992 187 467 015 531 877 577 523 2 × 2 = 1 + 0.984 374 934 031 063 755 155 046 4;
  • 53) 0.984 374 934 031 063 755 155 046 4 × 2 = 1 + 0.968 749 868 062 127 510 310 092 8;
  • 54) 0.968 749 868 062 127 510 310 092 8 × 2 = 1 + 0.937 499 736 124 255 020 620 185 6;
  • 55) 0.937 499 736 124 255 020 620 185 6 × 2 = 1 + 0.874 999 472 248 510 041 240 371 2;
  • 56) 0.874 999 472 248 510 041 240 371 2 × 2 = 1 + 0.749 998 944 497 020 082 480 742 4;
  • 57) 0.749 998 944 497 020 082 480 742 4 × 2 = 1 + 0.499 997 888 994 040 164 961 484 8;
  • 58) 0.499 997 888 994 040 164 961 484 8 × 2 = 0 + 0.999 995 777 988 080 329 922 969 6;

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:


0.016 738 891 601 562 496 530 538 4(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 538 4(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:


0.016 738 891 601 562 496 530 538 4(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 -0.016 738 891 601 562 496 530 538 4 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