-0.016 738 891 601 562 496 530 549 32 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 549 32(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
-0.016 738 891 601 562 496 530 549 32(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:

|-0.016 738 891 601 562 496 530 549 32| = 0.016 738 891 601 562 496 530 549 32


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: 0.016 738 891 601 562 496 530 549 32.

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) 0.016 738 891 601 562 496 530 549 32 × 2 = 0 + 0.033 477 783 203 124 993 061 098 64;
  • 2) 0.033 477 783 203 124 993 061 098 64 × 2 = 0 + 0.066 955 566 406 249 986 122 197 28;
  • 3) 0.066 955 566 406 249 986 122 197 28 × 2 = 0 + 0.133 911 132 812 499 972 244 394 56;
  • 4) 0.133 911 132 812 499 972 244 394 56 × 2 = 0 + 0.267 822 265 624 999 944 488 789 12;
  • 5) 0.267 822 265 624 999 944 488 789 12 × 2 = 0 + 0.535 644 531 249 999 888 977 578 24;
  • 6) 0.535 644 531 249 999 888 977 578 24 × 2 = 1 + 0.071 289 062 499 999 777 955 156 48;
  • 7) 0.071 289 062 499 999 777 955 156 48 × 2 = 0 + 0.142 578 124 999 999 555 910 312 96;
  • 8) 0.142 578 124 999 999 555 910 312 96 × 2 = 0 + 0.285 156 249 999 999 111 820 625 92;
  • 9) 0.285 156 249 999 999 111 820 625 92 × 2 = 0 + 0.570 312 499 999 998 223 641 251 84;
  • 10) 0.570 312 499 999 998 223 641 251 84 × 2 = 1 + 0.140 624 999 999 996 447 282 503 68;
  • 11) 0.140 624 999 999 996 447 282 503 68 × 2 = 0 + 0.281 249 999 999 992 894 565 007 36;
  • 12) 0.281 249 999 999 992 894 565 007 36 × 2 = 0 + 0.562 499 999 999 985 789 130 014 72;
  • 13) 0.562 499 999 999 985 789 130 014 72 × 2 = 1 + 0.124 999 999 999 971 578 260 029 44;
  • 14) 0.124 999 999 999 971 578 260 029 44 × 2 = 0 + 0.249 999 999 999 943 156 520 058 88;
  • 15) 0.249 999 999 999 943 156 520 058 88 × 2 = 0 + 0.499 999 999 999 886 313 040 117 76;
  • 16) 0.499 999 999 999 886 313 040 117 76 × 2 = 0 + 0.999 999 999 999 772 626 080 235 52;
  • 17) 0.999 999 999 999 772 626 080 235 52 × 2 = 1 + 0.999 999 999 999 545 252 160 471 04;
  • 18) 0.999 999 999 999 545 252 160 471 04 × 2 = 1 + 0.999 999 999 999 090 504 320 942 08;
  • 19) 0.999 999 999 999 090 504 320 942 08 × 2 = 1 + 0.999 999 999 998 181 008 641 884 16;
  • 20) 0.999 999 999 998 181 008 641 884 16 × 2 = 1 + 0.999 999 999 996 362 017 283 768 32;
  • 21) 0.999 999 999 996 362 017 283 768 32 × 2 = 1 + 0.999 999 999 992 724 034 567 536 64;
  • 22) 0.999 999 999 992 724 034 567 536 64 × 2 = 1 + 0.999 999 999 985 448 069 135 073 28;
  • 23) 0.999 999 999 985 448 069 135 073 28 × 2 = 1 + 0.999 999 999 970 896 138 270 146 56;
  • 24) 0.999 999 999 970 896 138 270 146 56 × 2 = 1 + 0.999 999 999 941 792 276 540 293 12;
  • 25) 0.999 999 999 941 792 276 540 293 12 × 2 = 1 + 0.999 999 999 883 584 553 080 586 24;
  • 26) 0.999 999 999 883 584 553 080 586 24 × 2 = 1 + 0.999 999 999 767 169 106 161 172 48;
  • 27) 0.999 999 999 767 169 106 161 172 48 × 2 = 1 + 0.999 999 999 534 338 212 322 344 96;
  • 28) 0.999 999 999 534 338 212 322 344 96 × 2 = 1 + 0.999 999 999 068 676 424 644 689 92;
  • 29) 0.999 999 999 068 676 424 644 689 92 × 2 = 1 + 0.999 999 998 137 352 849 289 379 84;
  • 30) 0.999 999 998 137 352 849 289 379 84 × 2 = 1 + 0.999 999 996 274 705 698 578 759 68;
  • 31) 0.999 999 996 274 705 698 578 759 68 × 2 = 1 + 0.999 999 992 549 411 397 157 519 36;
  • 32) 0.999 999 992 549 411 397 157 519 36 × 2 = 1 + 0.999 999 985 098 822 794 315 038 72;
  • 33) 0.999 999 985 098 822 794 315 038 72 × 2 = 1 + 0.999 999 970 197 645 588 630 077 44;
  • 34) 0.999 999 970 197 645 588 630 077 44 × 2 = 1 + 0.999 999 940 395 291 177 260 154 88;
  • 35) 0.999 999 940 395 291 177 260 154 88 × 2 = 1 + 0.999 999 880 790 582 354 520 309 76;
  • 36) 0.999 999 880 790 582 354 520 309 76 × 2 = 1 + 0.999 999 761 581 164 709 040 619 52;
  • 37) 0.999 999 761 581 164 709 040 619 52 × 2 = 1 + 0.999 999 523 162 329 418 081 239 04;
  • 38) 0.999 999 523 162 329 418 081 239 04 × 2 = 1 + 0.999 999 046 324 658 836 162 478 08;
  • 39) 0.999 999 046 324 658 836 162 478 08 × 2 = 1 + 0.999 998 092 649 317 672 324 956 16;
  • 40) 0.999 998 092 649 317 672 324 956 16 × 2 = 1 + 0.999 996 185 298 635 344 649 912 32;
  • 41) 0.999 996 185 298 635 344 649 912 32 × 2 = 1 + 0.999 992 370 597 270 689 299 824 64;
  • 42) 0.999 992 370 597 270 689 299 824 64 × 2 = 1 + 0.999 984 741 194 541 378 599 649 28;
  • 43) 0.999 984 741 194 541 378 599 649 28 × 2 = 1 + 0.999 969 482 389 082 757 199 298 56;
  • 44) 0.999 969 482 389 082 757 199 298 56 × 2 = 1 + 0.999 938 964 778 165 514 398 597 12;
  • 45) 0.999 938 964 778 165 514 398 597 12 × 2 = 1 + 0.999 877 929 556 331 028 797 194 24;
  • 46) 0.999 877 929 556 331 028 797 194 24 × 2 = 1 + 0.999 755 859 112 662 057 594 388 48;
  • 47) 0.999 755 859 112 662 057 594 388 48 × 2 = 1 + 0.999 511 718 225 324 115 188 776 96;
  • 48) 0.999 511 718 225 324 115 188 776 96 × 2 = 1 + 0.999 023 436 450 648 230 377 553 92;
  • 49) 0.999 023 436 450 648 230 377 553 92 × 2 = 1 + 0.998 046 872 901 296 460 755 107 84;
  • 50) 0.998 046 872 901 296 460 755 107 84 × 2 = 1 + 0.996 093 745 802 592 921 510 215 68;
  • 51) 0.996 093 745 802 592 921 510 215 68 × 2 = 1 + 0.992 187 491 605 185 843 020 431 36;
  • 52) 0.992 187 491 605 185 843 020 431 36 × 2 = 1 + 0.984 374 983 210 371 686 040 862 72;
  • 53) 0.984 374 983 210 371 686 040 862 72 × 2 = 1 + 0.968 749 966 420 743 372 081 725 44;
  • 54) 0.968 749 966 420 743 372 081 725 44 × 2 = 1 + 0.937 499 932 841 486 744 163 450 88;
  • 55) 0.937 499 932 841 486 744 163 450 88 × 2 = 1 + 0.874 999 865 682 973 488 326 901 76;
  • 56) 0.874 999 865 682 973 488 326 901 76 × 2 = 1 + 0.749 999 731 365 946 976 653 803 52;
  • 57) 0.749 999 731 365 946 976 653 803 52 × 2 = 1 + 0.499 999 462 731 893 953 307 607 04;
  • 58) 0.499 999 462 731 893 953 307 607 04 × 2 = 0 + 0.999 998 925 463 787 906 615 214 08;

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:


0.016 738 891 601 562 496 530 549 32(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 549 32(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:


0.016 738 891 601 562 496 530 549 32(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 -0.016 738 891 601 562 496 530 549 32 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