-0.016 738 891 601 562 496 530 552 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 552 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
-0.016 738 891 601 562 496 530 552 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:

|-0.016 738 891 601 562 496 530 552 1| = 0.016 738 891 601 562 496 530 552 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: 0.016 738 891 601 562 496 530 552 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) 0.016 738 891 601 562 496 530 552 1 × 2 = 0 + 0.033 477 783 203 124 993 061 104 2;
  • 2) 0.033 477 783 203 124 993 061 104 2 × 2 = 0 + 0.066 955 566 406 249 986 122 208 4;
  • 3) 0.066 955 566 406 249 986 122 208 4 × 2 = 0 + 0.133 911 132 812 499 972 244 416 8;
  • 4) 0.133 911 132 812 499 972 244 416 8 × 2 = 0 + 0.267 822 265 624 999 944 488 833 6;
  • 5) 0.267 822 265 624 999 944 488 833 6 × 2 = 0 + 0.535 644 531 249 999 888 977 667 2;
  • 6) 0.535 644 531 249 999 888 977 667 2 × 2 = 1 + 0.071 289 062 499 999 777 955 334 4;
  • 7) 0.071 289 062 499 999 777 955 334 4 × 2 = 0 + 0.142 578 124 999 999 555 910 668 8;
  • 8) 0.142 578 124 999 999 555 910 668 8 × 2 = 0 + 0.285 156 249 999 999 111 821 337 6;
  • 9) 0.285 156 249 999 999 111 821 337 6 × 2 = 0 + 0.570 312 499 999 998 223 642 675 2;
  • 10) 0.570 312 499 999 998 223 642 675 2 × 2 = 1 + 0.140 624 999 999 996 447 285 350 4;
  • 11) 0.140 624 999 999 996 447 285 350 4 × 2 = 0 + 0.281 249 999 999 992 894 570 700 8;
  • 12) 0.281 249 999 999 992 894 570 700 8 × 2 = 0 + 0.562 499 999 999 985 789 141 401 6;
  • 13) 0.562 499 999 999 985 789 141 401 6 × 2 = 1 + 0.124 999 999 999 971 578 282 803 2;
  • 14) 0.124 999 999 999 971 578 282 803 2 × 2 = 0 + 0.249 999 999 999 943 156 565 606 4;
  • 15) 0.249 999 999 999 943 156 565 606 4 × 2 = 0 + 0.499 999 999 999 886 313 131 212 8;
  • 16) 0.499 999 999 999 886 313 131 212 8 × 2 = 0 + 0.999 999 999 999 772 626 262 425 6;
  • 17) 0.999 999 999 999 772 626 262 425 6 × 2 = 1 + 0.999 999 999 999 545 252 524 851 2;
  • 18) 0.999 999 999 999 545 252 524 851 2 × 2 = 1 + 0.999 999 999 999 090 505 049 702 4;
  • 19) 0.999 999 999 999 090 505 049 702 4 × 2 = 1 + 0.999 999 999 998 181 010 099 404 8;
  • 20) 0.999 999 999 998 181 010 099 404 8 × 2 = 1 + 0.999 999 999 996 362 020 198 809 6;
  • 21) 0.999 999 999 996 362 020 198 809 6 × 2 = 1 + 0.999 999 999 992 724 040 397 619 2;
  • 22) 0.999 999 999 992 724 040 397 619 2 × 2 = 1 + 0.999 999 999 985 448 080 795 238 4;
  • 23) 0.999 999 999 985 448 080 795 238 4 × 2 = 1 + 0.999 999 999 970 896 161 590 476 8;
  • 24) 0.999 999 999 970 896 161 590 476 8 × 2 = 1 + 0.999 999 999 941 792 323 180 953 6;
  • 25) 0.999 999 999 941 792 323 180 953 6 × 2 = 1 + 0.999 999 999 883 584 646 361 907 2;
  • 26) 0.999 999 999 883 584 646 361 907 2 × 2 = 1 + 0.999 999 999 767 169 292 723 814 4;
  • 27) 0.999 999 999 767 169 292 723 814 4 × 2 = 1 + 0.999 999 999 534 338 585 447 628 8;
  • 28) 0.999 999 999 534 338 585 447 628 8 × 2 = 1 + 0.999 999 999 068 677 170 895 257 6;
  • 29) 0.999 999 999 068 677 170 895 257 6 × 2 = 1 + 0.999 999 998 137 354 341 790 515 2;
  • 30) 0.999 999 998 137 354 341 790 515 2 × 2 = 1 + 0.999 999 996 274 708 683 581 030 4;
  • 31) 0.999 999 996 274 708 683 581 030 4 × 2 = 1 + 0.999 999 992 549 417 367 162 060 8;
  • 32) 0.999 999 992 549 417 367 162 060 8 × 2 = 1 + 0.999 999 985 098 834 734 324 121 6;
  • 33) 0.999 999 985 098 834 734 324 121 6 × 2 = 1 + 0.999 999 970 197 669 468 648 243 2;
  • 34) 0.999 999 970 197 669 468 648 243 2 × 2 = 1 + 0.999 999 940 395 338 937 296 486 4;
  • 35) 0.999 999 940 395 338 937 296 486 4 × 2 = 1 + 0.999 999 880 790 677 874 592 972 8;
  • 36) 0.999 999 880 790 677 874 592 972 8 × 2 = 1 + 0.999 999 761 581 355 749 185 945 6;
  • 37) 0.999 999 761 581 355 749 185 945 6 × 2 = 1 + 0.999 999 523 162 711 498 371 891 2;
  • 38) 0.999 999 523 162 711 498 371 891 2 × 2 = 1 + 0.999 999 046 325 422 996 743 782 4;
  • 39) 0.999 999 046 325 422 996 743 782 4 × 2 = 1 + 0.999 998 092 650 845 993 487 564 8;
  • 40) 0.999 998 092 650 845 993 487 564 8 × 2 = 1 + 0.999 996 185 301 691 986 975 129 6;
  • 41) 0.999 996 185 301 691 986 975 129 6 × 2 = 1 + 0.999 992 370 603 383 973 950 259 2;
  • 42) 0.999 992 370 603 383 973 950 259 2 × 2 = 1 + 0.999 984 741 206 767 947 900 518 4;
  • 43) 0.999 984 741 206 767 947 900 518 4 × 2 = 1 + 0.999 969 482 413 535 895 801 036 8;
  • 44) 0.999 969 482 413 535 895 801 036 8 × 2 = 1 + 0.999 938 964 827 071 791 602 073 6;
  • 45) 0.999 938 964 827 071 791 602 073 6 × 2 = 1 + 0.999 877 929 654 143 583 204 147 2;
  • 46) 0.999 877 929 654 143 583 204 147 2 × 2 = 1 + 0.999 755 859 308 287 166 408 294 4;
  • 47) 0.999 755 859 308 287 166 408 294 4 × 2 = 1 + 0.999 511 718 616 574 332 816 588 8;
  • 48) 0.999 511 718 616 574 332 816 588 8 × 2 = 1 + 0.999 023 437 233 148 665 633 177 6;
  • 49) 0.999 023 437 233 148 665 633 177 6 × 2 = 1 + 0.998 046 874 466 297 331 266 355 2;
  • 50) 0.998 046 874 466 297 331 266 355 2 × 2 = 1 + 0.996 093 748 932 594 662 532 710 4;
  • 51) 0.996 093 748 932 594 662 532 710 4 × 2 = 1 + 0.992 187 497 865 189 325 065 420 8;
  • 52) 0.992 187 497 865 189 325 065 420 8 × 2 = 1 + 0.984 374 995 730 378 650 130 841 6;
  • 53) 0.984 374 995 730 378 650 130 841 6 × 2 = 1 + 0.968 749 991 460 757 300 261 683 2;
  • 54) 0.968 749 991 460 757 300 261 683 2 × 2 = 1 + 0.937 499 982 921 514 600 523 366 4;
  • 55) 0.937 499 982 921 514 600 523 366 4 × 2 = 1 + 0.874 999 965 843 029 201 046 732 8;
  • 56) 0.874 999 965 843 029 201 046 732 8 × 2 = 1 + 0.749 999 931 686 058 402 093 465 6;
  • 57) 0.749 999 931 686 058 402 093 465 6 × 2 = 1 + 0.499 999 863 372 116 804 186 931 2;
  • 58) 0.499 999 863 372 116 804 186 931 2 × 2 = 0 + 0.999 999 726 744 233 608 373 862 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:


0.016 738 891 601 562 496 530 552 1(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 552 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:


0.016 738 891 601 562 496 530 552 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 -0.016 738 891 601 562 496 530 552 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