-0.240 000 000 000 048 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.240 000 000 000 048(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
-0.240 000 000 000 048(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:

|-0.240 000 000 000 048| = 0.240 000 000 000 048


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: 0.240 000 000 000 048.

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) 0.240 000 000 000 048 × 2 = 0 + 0.480 000 000 000 096;
  • 2) 0.480 000 000 000 096 × 2 = 0 + 0.960 000 000 000 192;
  • 3) 0.960 000 000 000 192 × 2 = 1 + 0.920 000 000 000 384;
  • 4) 0.920 000 000 000 384 × 2 = 1 + 0.840 000 000 000 768;
  • 5) 0.840 000 000 000 768 × 2 = 1 + 0.680 000 000 001 536;
  • 6) 0.680 000 000 001 536 × 2 = 1 + 0.360 000 000 003 072;
  • 7) 0.360 000 000 003 072 × 2 = 0 + 0.720 000 000 006 144;
  • 8) 0.720 000 000 006 144 × 2 = 1 + 0.440 000 000 012 288;
  • 9) 0.440 000 000 012 288 × 2 = 0 + 0.880 000 000 024 576;
  • 10) 0.880 000 000 024 576 × 2 = 1 + 0.760 000 000 049 152;
  • 11) 0.760 000 000 049 152 × 2 = 1 + 0.520 000 000 098 304;
  • 12) 0.520 000 000 098 304 × 2 = 1 + 0.040 000 000 196 608;
  • 13) 0.040 000 000 196 608 × 2 = 0 + 0.080 000 000 393 216;
  • 14) 0.080 000 000 393 216 × 2 = 0 + 0.160 000 000 786 432;
  • 15) 0.160 000 000 786 432 × 2 = 0 + 0.320 000 001 572 864;
  • 16) 0.320 000 001 572 864 × 2 = 0 + 0.640 000 003 145 728;
  • 17) 0.640 000 003 145 728 × 2 = 1 + 0.280 000 006 291 456;
  • 18) 0.280 000 006 291 456 × 2 = 0 + 0.560 000 012 582 912;
  • 19) 0.560 000 012 582 912 × 2 = 1 + 0.120 000 025 165 824;
  • 20) 0.120 000 025 165 824 × 2 = 0 + 0.240 000 050 331 648;
  • 21) 0.240 000 050 331 648 × 2 = 0 + 0.480 000 100 663 296;
  • 22) 0.480 000 100 663 296 × 2 = 0 + 0.960 000 201 326 592;
  • 23) 0.960 000 201 326 592 × 2 = 1 + 0.920 000 402 653 184;
  • 24) 0.920 000 402 653 184 × 2 = 1 + 0.840 000 805 306 368;
  • 25) 0.840 000 805 306 368 × 2 = 1 + 0.680 001 610 612 736;
  • 26) 0.680 001 610 612 736 × 2 = 1 + 0.360 003 221 225 472;
  • 27) 0.360 003 221 225 472 × 2 = 0 + 0.720 006 442 450 944;
  • 28) 0.720 006 442 450 944 × 2 = 1 + 0.440 012 884 901 888;
  • 29) 0.440 012 884 901 888 × 2 = 0 + 0.880 025 769 803 776;
  • 30) 0.880 025 769 803 776 × 2 = 1 + 0.760 051 539 607 552;
  • 31) 0.760 051 539 607 552 × 2 = 1 + 0.520 103 079 215 104;
  • 32) 0.520 103 079 215 104 × 2 = 1 + 0.040 206 158 430 208;
  • 33) 0.040 206 158 430 208 × 2 = 0 + 0.080 412 316 860 416;
  • 34) 0.080 412 316 860 416 × 2 = 0 + 0.160 824 633 720 832;
  • 35) 0.160 824 633 720 832 × 2 = 0 + 0.321 649 267 441 664;
  • 36) 0.321 649 267 441 664 × 2 = 0 + 0.643 298 534 883 328;
  • 37) 0.643 298 534 883 328 × 2 = 1 + 0.286 597 069 766 656;
  • 38) 0.286 597 069 766 656 × 2 = 0 + 0.573 194 139 533 312;
  • 39) 0.573 194 139 533 312 × 2 = 1 + 0.146 388 279 066 624;
  • 40) 0.146 388 279 066 624 × 2 = 0 + 0.292 776 558 133 248;
  • 41) 0.292 776 558 133 248 × 2 = 0 + 0.585 553 116 266 496;
  • 42) 0.585 553 116 266 496 × 2 = 1 + 0.171 106 232 532 992;
  • 43) 0.171 106 232 532 992 × 2 = 0 + 0.342 212 465 065 984;
  • 44) 0.342 212 465 065 984 × 2 = 0 + 0.684 424 930 131 968;
  • 45) 0.684 424 930 131 968 × 2 = 1 + 0.368 849 860 263 936;
  • 46) 0.368 849 860 263 936 × 2 = 0 + 0.737 699 720 527 872;
  • 47) 0.737 699 720 527 872 × 2 = 1 + 0.475 399 441 055 744;
  • 48) 0.475 399 441 055 744 × 2 = 0 + 0.950 798 882 111 488;
  • 49) 0.950 798 882 111 488 × 2 = 1 + 0.901 597 764 222 976;
  • 50) 0.901 597 764 222 976 × 2 = 1 + 0.803 195 528 445 952;
  • 51) 0.803 195 528 445 952 × 2 = 1 + 0.606 391 056 891 904;
  • 52) 0.606 391 056 891 904 × 2 = 1 + 0.212 782 113 783 808;
  • 53) 0.212 782 113 783 808 × 2 = 0 + 0.425 564 227 567 616;
  • 54) 0.425 564 227 567 616 × 2 = 0 + 0.851 128 455 135 232;
  • 55) 0.851 128 455 135 232 × 2 = 1 + 0.702 256 910 270 464;

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:


0.240 000 000 000 048(10) =

0.0011 1101 0111 0000 1010 0011 1101 0111 0000 1010 0100 1010 1111 001(2)

6. Positive number before normalization:

0.240 000 000 000 048(10) =

0.0011 1101 0111 0000 1010 0011 1101 0111 0000 1010 0100 1010 1111 001(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:


0.240 000 000 000 048(10) =

0.0011 1101 0111 0000 1010 0011 1101 0111 0000 1010 0100 1010 1111 001(2) =

0.0011 1101 0111 0000 1010 0011 1101 0111 0000 1010 0100 1010 1111 001(2) × 20 =

1.1110 1011 1000 0101 0001 1110 1011 1000 0101 0010 0101 0111 1001(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):
1.1110 1011 1000 0101 0001 1110 1011 1000 0101 0010 0101 0111 1001


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. 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010 0101 0111 1001 =

1110 1011 1000 0101 0001 1110 1011 1000 0101 0010 0101 0111 1001


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) =
1110 1011 1000 0101 0001 1110 1011 1000 0101 0010 0101 0111 1001


Decimal number -0.240 000 000 000 048 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 1110 1011 1000 0101 0001 1110 1011 1000 0101 0010 0101 0111 1001

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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