-0.145 067 813 487 901 050 666 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 666(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
-0.145 067 813 487 901 050 666(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:

|-0.145 067 813 487 901 050 666| = 0.145 067 813 487 901 050 666


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: 0.145 067 813 487 901 050 666.

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) 0.145 067 813 487 901 050 666 × 2 = 0 + 0.290 135 626 975 802 101 332;
  • 2) 0.290 135 626 975 802 101 332 × 2 = 0 + 0.580 271 253 951 604 202 664;
  • 3) 0.580 271 253 951 604 202 664 × 2 = 1 + 0.160 542 507 903 208 405 328;
  • 4) 0.160 542 507 903 208 405 328 × 2 = 0 + 0.321 085 015 806 416 810 656;
  • 5) 0.321 085 015 806 416 810 656 × 2 = 0 + 0.642 170 031 612 833 621 312;
  • 6) 0.642 170 031 612 833 621 312 × 2 = 1 + 0.284 340 063 225 667 242 624;
  • 7) 0.284 340 063 225 667 242 624 × 2 = 0 + 0.568 680 126 451 334 485 248;
  • 8) 0.568 680 126 451 334 485 248 × 2 = 1 + 0.137 360 252 902 668 970 496;
  • 9) 0.137 360 252 902 668 970 496 × 2 = 0 + 0.274 720 505 805 337 940 992;
  • 10) 0.274 720 505 805 337 940 992 × 2 = 0 + 0.549 441 011 610 675 881 984;
  • 11) 0.549 441 011 610 675 881 984 × 2 = 1 + 0.098 882 023 221 351 763 968;
  • 12) 0.098 882 023 221 351 763 968 × 2 = 0 + 0.197 764 046 442 703 527 936;
  • 13) 0.197 764 046 442 703 527 936 × 2 = 0 + 0.395 528 092 885 407 055 872;
  • 14) 0.395 528 092 885 407 055 872 × 2 = 0 + 0.791 056 185 770 814 111 744;
  • 15) 0.791 056 185 770 814 111 744 × 2 = 1 + 0.582 112 371 541 628 223 488;
  • 16) 0.582 112 371 541 628 223 488 × 2 = 1 + 0.164 224 743 083 256 446 976;
  • 17) 0.164 224 743 083 256 446 976 × 2 = 0 + 0.328 449 486 166 512 893 952;
  • 18) 0.328 449 486 166 512 893 952 × 2 = 0 + 0.656 898 972 333 025 787 904;
  • 19) 0.656 898 972 333 025 787 904 × 2 = 1 + 0.313 797 944 666 051 575 808;
  • 20) 0.313 797 944 666 051 575 808 × 2 = 0 + 0.627 595 889 332 103 151 616;
  • 21) 0.627 595 889 332 103 151 616 × 2 = 1 + 0.255 191 778 664 206 303 232;
  • 22) 0.255 191 778 664 206 303 232 × 2 = 0 + 0.510 383 557 328 412 606 464;
  • 23) 0.510 383 557 328 412 606 464 × 2 = 1 + 0.020 767 114 656 825 212 928;
  • 24) 0.020 767 114 656 825 212 928 × 2 = 0 + 0.041 534 229 313 650 425 856;
  • 25) 0.041 534 229 313 650 425 856 × 2 = 0 + 0.083 068 458 627 300 851 712;
  • 26) 0.083 068 458 627 300 851 712 × 2 = 0 + 0.166 136 917 254 601 703 424;
  • 27) 0.166 136 917 254 601 703 424 × 2 = 0 + 0.332 273 834 509 203 406 848;
  • 28) 0.332 273 834 509 203 406 848 × 2 = 0 + 0.664 547 669 018 406 813 696;
  • 29) 0.664 547 669 018 406 813 696 × 2 = 1 + 0.329 095 338 036 813 627 392;
  • 30) 0.329 095 338 036 813 627 392 × 2 = 0 + 0.658 190 676 073 627 254 784;
  • 31) 0.658 190 676 073 627 254 784 × 2 = 1 + 0.316 381 352 147 254 509 568;
  • 32) 0.316 381 352 147 254 509 568 × 2 = 0 + 0.632 762 704 294 509 019 136;
  • 33) 0.632 762 704 294 509 019 136 × 2 = 1 + 0.265 525 408 589 018 038 272;
  • 34) 0.265 525 408 589 018 038 272 × 2 = 0 + 0.531 050 817 178 036 076 544;
  • 35) 0.531 050 817 178 036 076 544 × 2 = 1 + 0.062 101 634 356 072 153 088;
  • 36) 0.062 101 634 356 072 153 088 × 2 = 0 + 0.124 203 268 712 144 306 176;
  • 37) 0.124 203 268 712 144 306 176 × 2 = 0 + 0.248 406 537 424 288 612 352;
  • 38) 0.248 406 537 424 288 612 352 × 2 = 0 + 0.496 813 074 848 577 224 704;
  • 39) 0.496 813 074 848 577 224 704 × 2 = 0 + 0.993 626 149 697 154 449 408;
  • 40) 0.993 626 149 697 154 449 408 × 2 = 1 + 0.987 252 299 394 308 898 816;
  • 41) 0.987 252 299 394 308 898 816 × 2 = 1 + 0.974 504 598 788 617 797 632;
  • 42) 0.974 504 598 788 617 797 632 × 2 = 1 + 0.949 009 197 577 235 595 264;
  • 43) 0.949 009 197 577 235 595 264 × 2 = 1 + 0.898 018 395 154 471 190 528;
  • 44) 0.898 018 395 154 471 190 528 × 2 = 1 + 0.796 036 790 308 942 381 056;
  • 45) 0.796 036 790 308 942 381 056 × 2 = 1 + 0.592 073 580 617 884 762 112;
  • 46) 0.592 073 580 617 884 762 112 × 2 = 1 + 0.184 147 161 235 769 524 224;
  • 47) 0.184 147 161 235 769 524 224 × 2 = 0 + 0.368 294 322 471 539 048 448;
  • 48) 0.368 294 322 471 539 048 448 × 2 = 0 + 0.736 588 644 943 078 096 896;
  • 49) 0.736 588 644 943 078 096 896 × 2 = 1 + 0.473 177 289 886 156 193 792;
  • 50) 0.473 177 289 886 156 193 792 × 2 = 0 + 0.946 354 579 772 312 387 584;
  • 51) 0.946 354 579 772 312 387 584 × 2 = 1 + 0.892 709 159 544 624 775 168;
  • 52) 0.892 709 159 544 624 775 168 × 2 = 1 + 0.785 418 319 089 249 550 336;
  • 53) 0.785 418 319 089 249 550 336 × 2 = 1 + 0.570 836 638 178 499 100 672;
  • 54) 0.570 836 638 178 499 100 672 × 2 = 1 + 0.141 673 276 356 998 201 344;
  • 55) 0.141 673 276 356 998 201 344 × 2 = 0 + 0.283 346 552 713 996 402 688;

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:


0.145 067 813 487 901 050 666(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2)

6. Positive number before normalization:

0.145 067 813 487 901 050 666(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.145 067 813 487 901 050 666(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2) × 20 =

1.0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110(2) × 2-3


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): -3

Mantissa (not normalized):
1.0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1020(10) =

011 1111 1100(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. 0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110 =

0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 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 1100

Mantissa (52 bits) =
0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110


Decimal number -0.145 067 813 487 901 050 666 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 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