-0.042 419 142 328 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 328 5(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
-0.042 419 142 328 5(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:

|-0.042 419 142 328 5| = 0.042 419 142 328 5


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: 0.042 419 142 328 5.

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) 0.042 419 142 328 5 × 2 = 0 + 0.084 838 284 657;
  • 2) 0.084 838 284 657 × 2 = 0 + 0.169 676 569 314;
  • 3) 0.169 676 569 314 × 2 = 0 + 0.339 353 138 628;
  • 4) 0.339 353 138 628 × 2 = 0 + 0.678 706 277 256;
  • 5) 0.678 706 277 256 × 2 = 1 + 0.357 412 554 512;
  • 6) 0.357 412 554 512 × 2 = 0 + 0.714 825 109 024;
  • 7) 0.714 825 109 024 × 2 = 1 + 0.429 650 218 048;
  • 8) 0.429 650 218 048 × 2 = 0 + 0.859 300 436 096;
  • 9) 0.859 300 436 096 × 2 = 1 + 0.718 600 872 192;
  • 10) 0.718 600 872 192 × 2 = 1 + 0.437 201 744 384;
  • 11) 0.437 201 744 384 × 2 = 0 + 0.874 403 488 768;
  • 12) 0.874 403 488 768 × 2 = 1 + 0.748 806 977 536;
  • 13) 0.748 806 977 536 × 2 = 1 + 0.497 613 955 072;
  • 14) 0.497 613 955 072 × 2 = 0 + 0.995 227 910 144;
  • 15) 0.995 227 910 144 × 2 = 1 + 0.990 455 820 288;
  • 16) 0.990 455 820 288 × 2 = 1 + 0.980 911 640 576;
  • 17) 0.980 911 640 576 × 2 = 1 + 0.961 823 281 152;
  • 18) 0.961 823 281 152 × 2 = 1 + 0.923 646 562 304;
  • 19) 0.923 646 562 304 × 2 = 1 + 0.847 293 124 608;
  • 20) 0.847 293 124 608 × 2 = 1 + 0.694 586 249 216;
  • 21) 0.694 586 249 216 × 2 = 1 + 0.389 172 498 432;
  • 22) 0.389 172 498 432 × 2 = 0 + 0.778 344 996 864;
  • 23) 0.778 344 996 864 × 2 = 1 + 0.556 689 993 728;
  • 24) 0.556 689 993 728 × 2 = 1 + 0.113 379 987 456;
  • 25) 0.113 379 987 456 × 2 = 0 + 0.226 759 974 912;
  • 26) 0.226 759 974 912 × 2 = 0 + 0.453 519 949 824;
  • 27) 0.453 519 949 824 × 2 = 0 + 0.907 039 899 648;
  • 28) 0.907 039 899 648 × 2 = 1 + 0.814 079 799 296;
  • 29) 0.814 079 799 296 × 2 = 1 + 0.628 159 598 592;
  • 30) 0.628 159 598 592 × 2 = 1 + 0.256 319 197 184;
  • 31) 0.256 319 197 184 × 2 = 0 + 0.512 638 394 368;
  • 32) 0.512 638 394 368 × 2 = 1 + 0.025 276 788 736;
  • 33) 0.025 276 788 736 × 2 = 0 + 0.050 553 577 472;
  • 34) 0.050 553 577 472 × 2 = 0 + 0.101 107 154 944;
  • 35) 0.101 107 154 944 × 2 = 0 + 0.202 214 309 888;
  • 36) 0.202 214 309 888 × 2 = 0 + 0.404 428 619 776;
  • 37) 0.404 428 619 776 × 2 = 0 + 0.808 857 239 552;
  • 38) 0.808 857 239 552 × 2 = 1 + 0.617 714 479 104;
  • 39) 0.617 714 479 104 × 2 = 1 + 0.235 428 958 208;
  • 40) 0.235 428 958 208 × 2 = 0 + 0.470 857 916 416;
  • 41) 0.470 857 916 416 × 2 = 0 + 0.941 715 832 832;
  • 42) 0.941 715 832 832 × 2 = 1 + 0.883 431 665 664;
  • 43) 0.883 431 665 664 × 2 = 1 + 0.766 863 331 328;
  • 44) 0.766 863 331 328 × 2 = 1 + 0.533 726 662 656;
  • 45) 0.533 726 662 656 × 2 = 1 + 0.067 453 325 312;
  • 46) 0.067 453 325 312 × 2 = 0 + 0.134 906 650 624;
  • 47) 0.134 906 650 624 × 2 = 0 + 0.269 813 301 248;
  • 48) 0.269 813 301 248 × 2 = 0 + 0.539 626 602 496;
  • 49) 0.539 626 602 496 × 2 = 1 + 0.079 253 204 992;
  • 50) 0.079 253 204 992 × 2 = 0 + 0.158 506 409 984;
  • 51) 0.158 506 409 984 × 2 = 0 + 0.317 012 819 968;
  • 52) 0.317 012 819 968 × 2 = 0 + 0.634 025 639 936;
  • 53) 0.634 025 639 936 × 2 = 1 + 0.268 051 279 872;
  • 54) 0.268 051 279 872 × 2 = 0 + 0.536 102 559 744;
  • 55) 0.536 102 559 744 × 2 = 1 + 0.072 205 119 488;
  • 56) 0.072 205 119 488 × 2 = 0 + 0.144 410 238 976;
  • 57) 0.144 410 238 976 × 2 = 0 + 0.288 820 477 952;

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:


0.042 419 142 328 5(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 0110 0111 1000 1000 1010 0(2)

6. Positive number before normalization:

0.042 419 142 328 5(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 0110 0111 1000 1000 1010 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.042 419 142 328 5(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 0110 0111 1000 1000 1010 0(2) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 0110 0111 1000 1000 1010 0(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1010 0000 1100 1111 0001 0001 0100(2) × 2-5


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): -5

Mantissa (not normalized):
1.0101 1011 0111 1111 0110 0011 1010 0000 1100 1111 0001 0001 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1018(10) =

011 1111 1010(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. 0101 1011 0111 1111 0110 0011 1010 0000 1100 1111 0001 0001 0100 =

0101 1011 0111 1111 0110 0011 1010 0000 1100 1111 0001 0001 0100


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 1010

Mantissa (52 bits) =
0101 1011 0111 1111 0110 0011 1010 0000 1100 1111 0001 0001 0100


Decimal number -0.042 419 142 328 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0101 1011 0111 1111 0110 0011 1010 0000 1100 1111 0001 0001 0100

» Convert decimal -0.042 419 142 320 9 to 64 bit double precision IEEE 754 binary floating point representation standard

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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