-0.145 067 813 487 901 050 894 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 894(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
-0.145 067 813 487 901 050 894(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:

|-0.145 067 813 487 901 050 894| = 0.145 067 813 487 901 050 894


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: 0.145 067 813 487 901 050 894.

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) 0.145 067 813 487 901 050 894 × 2 = 0 + 0.290 135 626 975 802 101 788;
  • 2) 0.290 135 626 975 802 101 788 × 2 = 0 + 0.580 271 253 951 604 203 576;
  • 3) 0.580 271 253 951 604 203 576 × 2 = 1 + 0.160 542 507 903 208 407 152;
  • 4) 0.160 542 507 903 208 407 152 × 2 = 0 + 0.321 085 015 806 416 814 304;
  • 5) 0.321 085 015 806 416 814 304 × 2 = 0 + 0.642 170 031 612 833 628 608;
  • 6) 0.642 170 031 612 833 628 608 × 2 = 1 + 0.284 340 063 225 667 257 216;
  • 7) 0.284 340 063 225 667 257 216 × 2 = 0 + 0.568 680 126 451 334 514 432;
  • 8) 0.568 680 126 451 334 514 432 × 2 = 1 + 0.137 360 252 902 669 028 864;
  • 9) 0.137 360 252 902 669 028 864 × 2 = 0 + 0.274 720 505 805 338 057 728;
  • 10) 0.274 720 505 805 338 057 728 × 2 = 0 + 0.549 441 011 610 676 115 456;
  • 11) 0.549 441 011 610 676 115 456 × 2 = 1 + 0.098 882 023 221 352 230 912;
  • 12) 0.098 882 023 221 352 230 912 × 2 = 0 + 0.197 764 046 442 704 461 824;
  • 13) 0.197 764 046 442 704 461 824 × 2 = 0 + 0.395 528 092 885 408 923 648;
  • 14) 0.395 528 092 885 408 923 648 × 2 = 0 + 0.791 056 185 770 817 847 296;
  • 15) 0.791 056 185 770 817 847 296 × 2 = 1 + 0.582 112 371 541 635 694 592;
  • 16) 0.582 112 371 541 635 694 592 × 2 = 1 + 0.164 224 743 083 271 389 184;
  • 17) 0.164 224 743 083 271 389 184 × 2 = 0 + 0.328 449 486 166 542 778 368;
  • 18) 0.328 449 486 166 542 778 368 × 2 = 0 + 0.656 898 972 333 085 556 736;
  • 19) 0.656 898 972 333 085 556 736 × 2 = 1 + 0.313 797 944 666 171 113 472;
  • 20) 0.313 797 944 666 171 113 472 × 2 = 0 + 0.627 595 889 332 342 226 944;
  • 21) 0.627 595 889 332 342 226 944 × 2 = 1 + 0.255 191 778 664 684 453 888;
  • 22) 0.255 191 778 664 684 453 888 × 2 = 0 + 0.510 383 557 329 368 907 776;
  • 23) 0.510 383 557 329 368 907 776 × 2 = 1 + 0.020 767 114 658 737 815 552;
  • 24) 0.020 767 114 658 737 815 552 × 2 = 0 + 0.041 534 229 317 475 631 104;
  • 25) 0.041 534 229 317 475 631 104 × 2 = 0 + 0.083 068 458 634 951 262 208;
  • 26) 0.083 068 458 634 951 262 208 × 2 = 0 + 0.166 136 917 269 902 524 416;
  • 27) 0.166 136 917 269 902 524 416 × 2 = 0 + 0.332 273 834 539 805 048 832;
  • 28) 0.332 273 834 539 805 048 832 × 2 = 0 + 0.664 547 669 079 610 097 664;
  • 29) 0.664 547 669 079 610 097 664 × 2 = 1 + 0.329 095 338 159 220 195 328;
  • 30) 0.329 095 338 159 220 195 328 × 2 = 0 + 0.658 190 676 318 440 390 656;
  • 31) 0.658 190 676 318 440 390 656 × 2 = 1 + 0.316 381 352 636 880 781 312;
  • 32) 0.316 381 352 636 880 781 312 × 2 = 0 + 0.632 762 705 273 761 562 624;
  • 33) 0.632 762 705 273 761 562 624 × 2 = 1 + 0.265 525 410 547 523 125 248;
  • 34) 0.265 525 410 547 523 125 248 × 2 = 0 + 0.531 050 821 095 046 250 496;
  • 35) 0.531 050 821 095 046 250 496 × 2 = 1 + 0.062 101 642 190 092 500 992;
  • 36) 0.062 101 642 190 092 500 992 × 2 = 0 + 0.124 203 284 380 185 001 984;
  • 37) 0.124 203 284 380 185 001 984 × 2 = 0 + 0.248 406 568 760 370 003 968;
  • 38) 0.248 406 568 760 370 003 968 × 2 = 0 + 0.496 813 137 520 740 007 936;
  • 39) 0.496 813 137 520 740 007 936 × 2 = 0 + 0.993 626 275 041 480 015 872;
  • 40) 0.993 626 275 041 480 015 872 × 2 = 1 + 0.987 252 550 082 960 031 744;
  • 41) 0.987 252 550 082 960 031 744 × 2 = 1 + 0.974 505 100 165 920 063 488;
  • 42) 0.974 505 100 165 920 063 488 × 2 = 1 + 0.949 010 200 331 840 126 976;
  • 43) 0.949 010 200 331 840 126 976 × 2 = 1 + 0.898 020 400 663 680 253 952;
  • 44) 0.898 020 400 663 680 253 952 × 2 = 1 + 0.796 040 801 327 360 507 904;
  • 45) 0.796 040 801 327 360 507 904 × 2 = 1 + 0.592 081 602 654 721 015 808;
  • 46) 0.592 081 602 654 721 015 808 × 2 = 1 + 0.184 163 205 309 442 031 616;
  • 47) 0.184 163 205 309 442 031 616 × 2 = 0 + 0.368 326 410 618 884 063 232;
  • 48) 0.368 326 410 618 884 063 232 × 2 = 0 + 0.736 652 821 237 768 126 464;
  • 49) 0.736 652 821 237 768 126 464 × 2 = 1 + 0.473 305 642 475 536 252 928;
  • 50) 0.473 305 642 475 536 252 928 × 2 = 0 + 0.946 611 284 951 072 505 856;
  • 51) 0.946 611 284 951 072 505 856 × 2 = 1 + 0.893 222 569 902 145 011 712;
  • 52) 0.893 222 569 902 145 011 712 × 2 = 1 + 0.786 445 139 804 290 023 424;
  • 53) 0.786 445 139 804 290 023 424 × 2 = 1 + 0.572 890 279 608 580 046 848;
  • 54) 0.572 890 279 608 580 046 848 × 2 = 1 + 0.145 780 559 217 160 093 696;
  • 55) 0.145 780 559 217 160 093 696 × 2 = 0 + 0.291 561 118 434 320 187 392;

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:


0.145 067 813 487 901 050 894(10) =

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

6. Positive number before normalization:

0.145 067 813 487 901 050 894(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:


0.145 067 813 487 901 050 894(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 -0.145 067 813 487 901 050 894 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

Share

» Convert decimal -0.145 067 813 487 901 050 82 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 minutes

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