-0.016 738 891 601 562 496 530 550 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 550 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
-0.016 738 891 601 562 496 530 550 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:

|-0.016 738 891 601 562 496 530 550 4| = 0.016 738 891 601 562 496 530 550 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: 0.016 738 891 601 562 496 530 550 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) 0.016 738 891 601 562 496 530 550 4 × 2 = 0 + 0.033 477 783 203 124 993 061 100 8;
  • 2) 0.033 477 783 203 124 993 061 100 8 × 2 = 0 + 0.066 955 566 406 249 986 122 201 6;
  • 3) 0.066 955 566 406 249 986 122 201 6 × 2 = 0 + 0.133 911 132 812 499 972 244 403 2;
  • 4) 0.133 911 132 812 499 972 244 403 2 × 2 = 0 + 0.267 822 265 624 999 944 488 806 4;
  • 5) 0.267 822 265 624 999 944 488 806 4 × 2 = 0 + 0.535 644 531 249 999 888 977 612 8;
  • 6) 0.535 644 531 249 999 888 977 612 8 × 2 = 1 + 0.071 289 062 499 999 777 955 225 6;
  • 7) 0.071 289 062 499 999 777 955 225 6 × 2 = 0 + 0.142 578 124 999 999 555 910 451 2;
  • 8) 0.142 578 124 999 999 555 910 451 2 × 2 = 0 + 0.285 156 249 999 999 111 820 902 4;
  • 9) 0.285 156 249 999 999 111 820 902 4 × 2 = 0 + 0.570 312 499 999 998 223 641 804 8;
  • 10) 0.570 312 499 999 998 223 641 804 8 × 2 = 1 + 0.140 624 999 999 996 447 283 609 6;
  • 11) 0.140 624 999 999 996 447 283 609 6 × 2 = 0 + 0.281 249 999 999 992 894 567 219 2;
  • 12) 0.281 249 999 999 992 894 567 219 2 × 2 = 0 + 0.562 499 999 999 985 789 134 438 4;
  • 13) 0.562 499 999 999 985 789 134 438 4 × 2 = 1 + 0.124 999 999 999 971 578 268 876 8;
  • 14) 0.124 999 999 999 971 578 268 876 8 × 2 = 0 + 0.249 999 999 999 943 156 537 753 6;
  • 15) 0.249 999 999 999 943 156 537 753 6 × 2 = 0 + 0.499 999 999 999 886 313 075 507 2;
  • 16) 0.499 999 999 999 886 313 075 507 2 × 2 = 0 + 0.999 999 999 999 772 626 151 014 4;
  • 17) 0.999 999 999 999 772 626 151 014 4 × 2 = 1 + 0.999 999 999 999 545 252 302 028 8;
  • 18) 0.999 999 999 999 545 252 302 028 8 × 2 = 1 + 0.999 999 999 999 090 504 604 057 6;
  • 19) 0.999 999 999 999 090 504 604 057 6 × 2 = 1 + 0.999 999 999 998 181 009 208 115 2;
  • 20) 0.999 999 999 998 181 009 208 115 2 × 2 = 1 + 0.999 999 999 996 362 018 416 230 4;
  • 21) 0.999 999 999 996 362 018 416 230 4 × 2 = 1 + 0.999 999 999 992 724 036 832 460 8;
  • 22) 0.999 999 999 992 724 036 832 460 8 × 2 = 1 + 0.999 999 999 985 448 073 664 921 6;
  • 23) 0.999 999 999 985 448 073 664 921 6 × 2 = 1 + 0.999 999 999 970 896 147 329 843 2;
  • 24) 0.999 999 999 970 896 147 329 843 2 × 2 = 1 + 0.999 999 999 941 792 294 659 686 4;
  • 25) 0.999 999 999 941 792 294 659 686 4 × 2 = 1 + 0.999 999 999 883 584 589 319 372 8;
  • 26) 0.999 999 999 883 584 589 319 372 8 × 2 = 1 + 0.999 999 999 767 169 178 638 745 6;
  • 27) 0.999 999 999 767 169 178 638 745 6 × 2 = 1 + 0.999 999 999 534 338 357 277 491 2;
  • 28) 0.999 999 999 534 338 357 277 491 2 × 2 = 1 + 0.999 999 999 068 676 714 554 982 4;
  • 29) 0.999 999 999 068 676 714 554 982 4 × 2 = 1 + 0.999 999 998 137 353 429 109 964 8;
  • 30) 0.999 999 998 137 353 429 109 964 8 × 2 = 1 + 0.999 999 996 274 706 858 219 929 6;
  • 31) 0.999 999 996 274 706 858 219 929 6 × 2 = 1 + 0.999 999 992 549 413 716 439 859 2;
  • 32) 0.999 999 992 549 413 716 439 859 2 × 2 = 1 + 0.999 999 985 098 827 432 879 718 4;
  • 33) 0.999 999 985 098 827 432 879 718 4 × 2 = 1 + 0.999 999 970 197 654 865 759 436 8;
  • 34) 0.999 999 970 197 654 865 759 436 8 × 2 = 1 + 0.999 999 940 395 309 731 518 873 6;
  • 35) 0.999 999 940 395 309 731 518 873 6 × 2 = 1 + 0.999 999 880 790 619 463 037 747 2;
  • 36) 0.999 999 880 790 619 463 037 747 2 × 2 = 1 + 0.999 999 761 581 238 926 075 494 4;
  • 37) 0.999 999 761 581 238 926 075 494 4 × 2 = 1 + 0.999 999 523 162 477 852 150 988 8;
  • 38) 0.999 999 523 162 477 852 150 988 8 × 2 = 1 + 0.999 999 046 324 955 704 301 977 6;
  • 39) 0.999 999 046 324 955 704 301 977 6 × 2 = 1 + 0.999 998 092 649 911 408 603 955 2;
  • 40) 0.999 998 092 649 911 408 603 955 2 × 2 = 1 + 0.999 996 185 299 822 817 207 910 4;
  • 41) 0.999 996 185 299 822 817 207 910 4 × 2 = 1 + 0.999 992 370 599 645 634 415 820 8;
  • 42) 0.999 992 370 599 645 634 415 820 8 × 2 = 1 + 0.999 984 741 199 291 268 831 641 6;
  • 43) 0.999 984 741 199 291 268 831 641 6 × 2 = 1 + 0.999 969 482 398 582 537 663 283 2;
  • 44) 0.999 969 482 398 582 537 663 283 2 × 2 = 1 + 0.999 938 964 797 165 075 326 566 4;
  • 45) 0.999 938 964 797 165 075 326 566 4 × 2 = 1 + 0.999 877 929 594 330 150 653 132 8;
  • 46) 0.999 877 929 594 330 150 653 132 8 × 2 = 1 + 0.999 755 859 188 660 301 306 265 6;
  • 47) 0.999 755 859 188 660 301 306 265 6 × 2 = 1 + 0.999 511 718 377 320 602 612 531 2;
  • 48) 0.999 511 718 377 320 602 612 531 2 × 2 = 1 + 0.999 023 436 754 641 205 225 062 4;
  • 49) 0.999 023 436 754 641 205 225 062 4 × 2 = 1 + 0.998 046 873 509 282 410 450 124 8;
  • 50) 0.998 046 873 509 282 410 450 124 8 × 2 = 1 + 0.996 093 747 018 564 820 900 249 6;
  • 51) 0.996 093 747 018 564 820 900 249 6 × 2 = 1 + 0.992 187 494 037 129 641 800 499 2;
  • 52) 0.992 187 494 037 129 641 800 499 2 × 2 = 1 + 0.984 374 988 074 259 283 600 998 4;
  • 53) 0.984 374 988 074 259 283 600 998 4 × 2 = 1 + 0.968 749 976 148 518 567 201 996 8;
  • 54) 0.968 749 976 148 518 567 201 996 8 × 2 = 1 + 0.937 499 952 297 037 134 403 993 6;
  • 55) 0.937 499 952 297 037 134 403 993 6 × 2 = 1 + 0.874 999 904 594 074 268 807 987 2;
  • 56) 0.874 999 904 594 074 268 807 987 2 × 2 = 1 + 0.749 999 809 188 148 537 615 974 4;
  • 57) 0.749 999 809 188 148 537 615 974 4 × 2 = 1 + 0.499 999 618 376 297 075 231 948 8;
  • 58) 0.499 999 618 376 297 075 231 948 8 × 2 = 0 + 0.999 999 236 752 594 150 463 897 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:


0.016 738 891 601 562 496 530 550 4(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 550 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:


0.016 738 891 601 562 496 530 550 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 -0.016 738 891 601 562 496 530 550 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