-0.145 067 813 487 901 050 860 08 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 860 08(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
-0.145 067 813 487 901 050 860 08(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:

|-0.145 067 813 487 901 050 860 08| = 0.145 067 813 487 901 050 860 08


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: 0.145 067 813 487 901 050 860 08.

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) 0.145 067 813 487 901 050 860 08 × 2 = 0 + 0.290 135 626 975 802 101 720 16;
  • 2) 0.290 135 626 975 802 101 720 16 × 2 = 0 + 0.580 271 253 951 604 203 440 32;
  • 3) 0.580 271 253 951 604 203 440 32 × 2 = 1 + 0.160 542 507 903 208 406 880 64;
  • 4) 0.160 542 507 903 208 406 880 64 × 2 = 0 + 0.321 085 015 806 416 813 761 28;
  • 5) 0.321 085 015 806 416 813 761 28 × 2 = 0 + 0.642 170 031 612 833 627 522 56;
  • 6) 0.642 170 031 612 833 627 522 56 × 2 = 1 + 0.284 340 063 225 667 255 045 12;
  • 7) 0.284 340 063 225 667 255 045 12 × 2 = 0 + 0.568 680 126 451 334 510 090 24;
  • 8) 0.568 680 126 451 334 510 090 24 × 2 = 1 + 0.137 360 252 902 669 020 180 48;
  • 9) 0.137 360 252 902 669 020 180 48 × 2 = 0 + 0.274 720 505 805 338 040 360 96;
  • 10) 0.274 720 505 805 338 040 360 96 × 2 = 0 + 0.549 441 011 610 676 080 721 92;
  • 11) 0.549 441 011 610 676 080 721 92 × 2 = 1 + 0.098 882 023 221 352 161 443 84;
  • 12) 0.098 882 023 221 352 161 443 84 × 2 = 0 + 0.197 764 046 442 704 322 887 68;
  • 13) 0.197 764 046 442 704 322 887 68 × 2 = 0 + 0.395 528 092 885 408 645 775 36;
  • 14) 0.395 528 092 885 408 645 775 36 × 2 = 0 + 0.791 056 185 770 817 291 550 72;
  • 15) 0.791 056 185 770 817 291 550 72 × 2 = 1 + 0.582 112 371 541 634 583 101 44;
  • 16) 0.582 112 371 541 634 583 101 44 × 2 = 1 + 0.164 224 743 083 269 166 202 88;
  • 17) 0.164 224 743 083 269 166 202 88 × 2 = 0 + 0.328 449 486 166 538 332 405 76;
  • 18) 0.328 449 486 166 538 332 405 76 × 2 = 0 + 0.656 898 972 333 076 664 811 52;
  • 19) 0.656 898 972 333 076 664 811 52 × 2 = 1 + 0.313 797 944 666 153 329 623 04;
  • 20) 0.313 797 944 666 153 329 623 04 × 2 = 0 + 0.627 595 889 332 306 659 246 08;
  • 21) 0.627 595 889 332 306 659 246 08 × 2 = 1 + 0.255 191 778 664 613 318 492 16;
  • 22) 0.255 191 778 664 613 318 492 16 × 2 = 0 + 0.510 383 557 329 226 636 984 32;
  • 23) 0.510 383 557 329 226 636 984 32 × 2 = 1 + 0.020 767 114 658 453 273 968 64;
  • 24) 0.020 767 114 658 453 273 968 64 × 2 = 0 + 0.041 534 229 316 906 547 937 28;
  • 25) 0.041 534 229 316 906 547 937 28 × 2 = 0 + 0.083 068 458 633 813 095 874 56;
  • 26) 0.083 068 458 633 813 095 874 56 × 2 = 0 + 0.166 136 917 267 626 191 749 12;
  • 27) 0.166 136 917 267 626 191 749 12 × 2 = 0 + 0.332 273 834 535 252 383 498 24;
  • 28) 0.332 273 834 535 252 383 498 24 × 2 = 0 + 0.664 547 669 070 504 766 996 48;
  • 29) 0.664 547 669 070 504 766 996 48 × 2 = 1 + 0.329 095 338 141 009 533 992 96;
  • 30) 0.329 095 338 141 009 533 992 96 × 2 = 0 + 0.658 190 676 282 019 067 985 92;
  • 31) 0.658 190 676 282 019 067 985 92 × 2 = 1 + 0.316 381 352 564 038 135 971 84;
  • 32) 0.316 381 352 564 038 135 971 84 × 2 = 0 + 0.632 762 705 128 076 271 943 68;
  • 33) 0.632 762 705 128 076 271 943 68 × 2 = 1 + 0.265 525 410 256 152 543 887 36;
  • 34) 0.265 525 410 256 152 543 887 36 × 2 = 0 + 0.531 050 820 512 305 087 774 72;
  • 35) 0.531 050 820 512 305 087 774 72 × 2 = 1 + 0.062 101 641 024 610 175 549 44;
  • 36) 0.062 101 641 024 610 175 549 44 × 2 = 0 + 0.124 203 282 049 220 351 098 88;
  • 37) 0.124 203 282 049 220 351 098 88 × 2 = 0 + 0.248 406 564 098 440 702 197 76;
  • 38) 0.248 406 564 098 440 702 197 76 × 2 = 0 + 0.496 813 128 196 881 404 395 52;
  • 39) 0.496 813 128 196 881 404 395 52 × 2 = 0 + 0.993 626 256 393 762 808 791 04;
  • 40) 0.993 626 256 393 762 808 791 04 × 2 = 1 + 0.987 252 512 787 525 617 582 08;
  • 41) 0.987 252 512 787 525 617 582 08 × 2 = 1 + 0.974 505 025 575 051 235 164 16;
  • 42) 0.974 505 025 575 051 235 164 16 × 2 = 1 + 0.949 010 051 150 102 470 328 32;
  • 43) 0.949 010 051 150 102 470 328 32 × 2 = 1 + 0.898 020 102 300 204 940 656 64;
  • 44) 0.898 020 102 300 204 940 656 64 × 2 = 1 + 0.796 040 204 600 409 881 313 28;
  • 45) 0.796 040 204 600 409 881 313 28 × 2 = 1 + 0.592 080 409 200 819 762 626 56;
  • 46) 0.592 080 409 200 819 762 626 56 × 2 = 1 + 0.184 160 818 401 639 525 253 12;
  • 47) 0.184 160 818 401 639 525 253 12 × 2 = 0 + 0.368 321 636 803 279 050 506 24;
  • 48) 0.368 321 636 803 279 050 506 24 × 2 = 0 + 0.736 643 273 606 558 101 012 48;
  • 49) 0.736 643 273 606 558 101 012 48 × 2 = 1 + 0.473 286 547 213 116 202 024 96;
  • 50) 0.473 286 547 213 116 202 024 96 × 2 = 0 + 0.946 573 094 426 232 404 049 92;
  • 51) 0.946 573 094 426 232 404 049 92 × 2 = 1 + 0.893 146 188 852 464 808 099 84;
  • 52) 0.893 146 188 852 464 808 099 84 × 2 = 1 + 0.786 292 377 704 929 616 199 68;
  • 53) 0.786 292 377 704 929 616 199 68 × 2 = 1 + 0.572 584 755 409 859 232 399 36;
  • 54) 0.572 584 755 409 859 232 399 36 × 2 = 1 + 0.145 169 510 819 718 464 798 72;
  • 55) 0.145 169 510 819 718 464 798 72 × 2 = 0 + 0.290 339 021 639 436 929 597 44;

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:


0.145 067 813 487 901 050 860 08(10) =

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

6. Positive number before normalization:

0.145 067 813 487 901 050 860 08(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:


0.145 067 813 487 901 050 860 08(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 -0.145 067 813 487 901 050 860 08 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