-0.145 067 813 487 901 050 882 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 882(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
-0.145 067 813 487 901 050 882(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:

|-0.145 067 813 487 901 050 882| = 0.145 067 813 487 901 050 882


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: 0.145 067 813 487 901 050 882.

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) 0.145 067 813 487 901 050 882 × 2 = 0 + 0.290 135 626 975 802 101 764;
  • 2) 0.290 135 626 975 802 101 764 × 2 = 0 + 0.580 271 253 951 604 203 528;
  • 3) 0.580 271 253 951 604 203 528 × 2 = 1 + 0.160 542 507 903 208 407 056;
  • 4) 0.160 542 507 903 208 407 056 × 2 = 0 + 0.321 085 015 806 416 814 112;
  • 5) 0.321 085 015 806 416 814 112 × 2 = 0 + 0.642 170 031 612 833 628 224;
  • 6) 0.642 170 031 612 833 628 224 × 2 = 1 + 0.284 340 063 225 667 256 448;
  • 7) 0.284 340 063 225 667 256 448 × 2 = 0 + 0.568 680 126 451 334 512 896;
  • 8) 0.568 680 126 451 334 512 896 × 2 = 1 + 0.137 360 252 902 669 025 792;
  • 9) 0.137 360 252 902 669 025 792 × 2 = 0 + 0.274 720 505 805 338 051 584;
  • 10) 0.274 720 505 805 338 051 584 × 2 = 0 + 0.549 441 011 610 676 103 168;
  • 11) 0.549 441 011 610 676 103 168 × 2 = 1 + 0.098 882 023 221 352 206 336;
  • 12) 0.098 882 023 221 352 206 336 × 2 = 0 + 0.197 764 046 442 704 412 672;
  • 13) 0.197 764 046 442 704 412 672 × 2 = 0 + 0.395 528 092 885 408 825 344;
  • 14) 0.395 528 092 885 408 825 344 × 2 = 0 + 0.791 056 185 770 817 650 688;
  • 15) 0.791 056 185 770 817 650 688 × 2 = 1 + 0.582 112 371 541 635 301 376;
  • 16) 0.582 112 371 541 635 301 376 × 2 = 1 + 0.164 224 743 083 270 602 752;
  • 17) 0.164 224 743 083 270 602 752 × 2 = 0 + 0.328 449 486 166 541 205 504;
  • 18) 0.328 449 486 166 541 205 504 × 2 = 0 + 0.656 898 972 333 082 411 008;
  • 19) 0.656 898 972 333 082 411 008 × 2 = 1 + 0.313 797 944 666 164 822 016;
  • 20) 0.313 797 944 666 164 822 016 × 2 = 0 + 0.627 595 889 332 329 644 032;
  • 21) 0.627 595 889 332 329 644 032 × 2 = 1 + 0.255 191 778 664 659 288 064;
  • 22) 0.255 191 778 664 659 288 064 × 2 = 0 + 0.510 383 557 329 318 576 128;
  • 23) 0.510 383 557 329 318 576 128 × 2 = 1 + 0.020 767 114 658 637 152 256;
  • 24) 0.020 767 114 658 637 152 256 × 2 = 0 + 0.041 534 229 317 274 304 512;
  • 25) 0.041 534 229 317 274 304 512 × 2 = 0 + 0.083 068 458 634 548 609 024;
  • 26) 0.083 068 458 634 548 609 024 × 2 = 0 + 0.166 136 917 269 097 218 048;
  • 27) 0.166 136 917 269 097 218 048 × 2 = 0 + 0.332 273 834 538 194 436 096;
  • 28) 0.332 273 834 538 194 436 096 × 2 = 0 + 0.664 547 669 076 388 872 192;
  • 29) 0.664 547 669 076 388 872 192 × 2 = 1 + 0.329 095 338 152 777 744 384;
  • 30) 0.329 095 338 152 777 744 384 × 2 = 0 + 0.658 190 676 305 555 488 768;
  • 31) 0.658 190 676 305 555 488 768 × 2 = 1 + 0.316 381 352 611 110 977 536;
  • 32) 0.316 381 352 611 110 977 536 × 2 = 0 + 0.632 762 705 222 221 955 072;
  • 33) 0.632 762 705 222 221 955 072 × 2 = 1 + 0.265 525 410 444 443 910 144;
  • 34) 0.265 525 410 444 443 910 144 × 2 = 0 + 0.531 050 820 888 887 820 288;
  • 35) 0.531 050 820 888 887 820 288 × 2 = 1 + 0.062 101 641 777 775 640 576;
  • 36) 0.062 101 641 777 775 640 576 × 2 = 0 + 0.124 203 283 555 551 281 152;
  • 37) 0.124 203 283 555 551 281 152 × 2 = 0 + 0.248 406 567 111 102 562 304;
  • 38) 0.248 406 567 111 102 562 304 × 2 = 0 + 0.496 813 134 222 205 124 608;
  • 39) 0.496 813 134 222 205 124 608 × 2 = 0 + 0.993 626 268 444 410 249 216;
  • 40) 0.993 626 268 444 410 249 216 × 2 = 1 + 0.987 252 536 888 820 498 432;
  • 41) 0.987 252 536 888 820 498 432 × 2 = 1 + 0.974 505 073 777 640 996 864;
  • 42) 0.974 505 073 777 640 996 864 × 2 = 1 + 0.949 010 147 555 281 993 728;
  • 43) 0.949 010 147 555 281 993 728 × 2 = 1 + 0.898 020 295 110 563 987 456;
  • 44) 0.898 020 295 110 563 987 456 × 2 = 1 + 0.796 040 590 221 127 974 912;
  • 45) 0.796 040 590 221 127 974 912 × 2 = 1 + 0.592 081 180 442 255 949 824;
  • 46) 0.592 081 180 442 255 949 824 × 2 = 1 + 0.184 162 360 884 511 899 648;
  • 47) 0.184 162 360 884 511 899 648 × 2 = 0 + 0.368 324 721 769 023 799 296;
  • 48) 0.368 324 721 769 023 799 296 × 2 = 0 + 0.736 649 443 538 047 598 592;
  • 49) 0.736 649 443 538 047 598 592 × 2 = 1 + 0.473 298 887 076 095 197 184;
  • 50) 0.473 298 887 076 095 197 184 × 2 = 0 + 0.946 597 774 152 190 394 368;
  • 51) 0.946 597 774 152 190 394 368 × 2 = 1 + 0.893 195 548 304 380 788 736;
  • 52) 0.893 195 548 304 380 788 736 × 2 = 1 + 0.786 391 096 608 761 577 472;
  • 53) 0.786 391 096 608 761 577 472 × 2 = 1 + 0.572 782 193 217 523 154 944;
  • 54) 0.572 782 193 217 523 154 944 × 2 = 1 + 0.145 564 386 435 046 309 888;
  • 55) 0.145 564 386 435 046 309 888 × 2 = 0 + 0.291 128 772 870 092 619 776;

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:


0.145 067 813 487 901 050 882(10) =

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

6. Positive number before normalization:

0.145 067 813 487 901 050 882(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:


0.145 067 813 487 901 050 882(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 -0.145 067 813 487 901 050 882 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