-0.000 000 000 366 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 366(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 366(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 366| = 0.000 000 000 366


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: 0.000 000 000 366.

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) 0.000 000 000 366 × 2 = 0 + 0.000 000 000 732;
  • 2) 0.000 000 000 732 × 2 = 0 + 0.000 000 001 464;
  • 3) 0.000 000 001 464 × 2 = 0 + 0.000 000 002 928;
  • 4) 0.000 000 002 928 × 2 = 0 + 0.000 000 005 856;
  • 5) 0.000 000 005 856 × 2 = 0 + 0.000 000 011 712;
  • 6) 0.000 000 011 712 × 2 = 0 + 0.000 000 023 424;
  • 7) 0.000 000 023 424 × 2 = 0 + 0.000 000 046 848;
  • 8) 0.000 000 046 848 × 2 = 0 + 0.000 000 093 696;
  • 9) 0.000 000 093 696 × 2 = 0 + 0.000 000 187 392;
  • 10) 0.000 000 187 392 × 2 = 0 + 0.000 000 374 784;
  • 11) 0.000 000 374 784 × 2 = 0 + 0.000 000 749 568;
  • 12) 0.000 000 749 568 × 2 = 0 + 0.000 001 499 136;
  • 13) 0.000 001 499 136 × 2 = 0 + 0.000 002 998 272;
  • 14) 0.000 002 998 272 × 2 = 0 + 0.000 005 996 544;
  • 15) 0.000 005 996 544 × 2 = 0 + 0.000 011 993 088;
  • 16) 0.000 011 993 088 × 2 = 0 + 0.000 023 986 176;
  • 17) 0.000 023 986 176 × 2 = 0 + 0.000 047 972 352;
  • 18) 0.000 047 972 352 × 2 = 0 + 0.000 095 944 704;
  • 19) 0.000 095 944 704 × 2 = 0 + 0.000 191 889 408;
  • 20) 0.000 191 889 408 × 2 = 0 + 0.000 383 778 816;
  • 21) 0.000 383 778 816 × 2 = 0 + 0.000 767 557 632;
  • 22) 0.000 767 557 632 × 2 = 0 + 0.001 535 115 264;
  • 23) 0.001 535 115 264 × 2 = 0 + 0.003 070 230 528;
  • 24) 0.003 070 230 528 × 2 = 0 + 0.006 140 461 056;
  • 25) 0.006 140 461 056 × 2 = 0 + 0.012 280 922 112;
  • 26) 0.012 280 922 112 × 2 = 0 + 0.024 561 844 224;
  • 27) 0.024 561 844 224 × 2 = 0 + 0.049 123 688 448;
  • 28) 0.049 123 688 448 × 2 = 0 + 0.098 247 376 896;
  • 29) 0.098 247 376 896 × 2 = 0 + 0.196 494 753 792;
  • 30) 0.196 494 753 792 × 2 = 0 + 0.392 989 507 584;
  • 31) 0.392 989 507 584 × 2 = 0 + 0.785 979 015 168;
  • 32) 0.785 979 015 168 × 2 = 1 + 0.571 958 030 336;
  • 33) 0.571 958 030 336 × 2 = 1 + 0.143 916 060 672;
  • 34) 0.143 916 060 672 × 2 = 0 + 0.287 832 121 344;
  • 35) 0.287 832 121 344 × 2 = 0 + 0.575 664 242 688;
  • 36) 0.575 664 242 688 × 2 = 1 + 0.151 328 485 376;
  • 37) 0.151 328 485 376 × 2 = 0 + 0.302 656 970 752;
  • 38) 0.302 656 970 752 × 2 = 0 + 0.605 313 941 504;
  • 39) 0.605 313 941 504 × 2 = 1 + 0.210 627 883 008;
  • 40) 0.210 627 883 008 × 2 = 0 + 0.421 255 766 016;
  • 41) 0.421 255 766 016 × 2 = 0 + 0.842 511 532 032;
  • 42) 0.842 511 532 032 × 2 = 1 + 0.685 023 064 064;
  • 43) 0.685 023 064 064 × 2 = 1 + 0.370 046 128 128;
  • 44) 0.370 046 128 128 × 2 = 0 + 0.740 092 256 256;
  • 45) 0.740 092 256 256 × 2 = 1 + 0.480 184 512 512;
  • 46) 0.480 184 512 512 × 2 = 0 + 0.960 369 025 024;
  • 47) 0.960 369 025 024 × 2 = 1 + 0.920 738 050 048;
  • 48) 0.920 738 050 048 × 2 = 1 + 0.841 476 100 096;
  • 49) 0.841 476 100 096 × 2 = 1 + 0.682 952 200 192;
  • 50) 0.682 952 200 192 × 2 = 1 + 0.365 904 400 384;
  • 51) 0.365 904 400 384 × 2 = 0 + 0.731 808 800 768;
  • 52) 0.731 808 800 768 × 2 = 1 + 0.463 617 601 536;
  • 53) 0.463 617 601 536 × 2 = 0 + 0.927 235 203 072;
  • 54) 0.927 235 203 072 × 2 = 1 + 0.854 470 406 144;
  • 55) 0.854 470 406 144 × 2 = 1 + 0.708 940 812 288;

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:


0.000 000 000 366(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1001 0010 0110 1011 1101 011(2)

6. Positive number before normalization:

0.000 000 000 366(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1001 0010 0110 1011 1101 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 366(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1001 0010 0110 1011 1101 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1001 0010 0110 1011 1101 011(2) × 20 =

1.1001 0010 0110 1011 1101 011(2) × 2-32


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -32

Mantissa (not normalized):
1.1001 0010 0110 1011 1101 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-32 + 2(8-1) - 1 =

(-32 + 127)(10) =

95(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

95(10) =

0101 1111(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1001 0011 0101 1110 1011 =

100 1001 0011 0101 1110 1011


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0101 1111

Mantissa (23 bits) =
100 1001 0011 0101 1110 1011


Decimal number -0.000 000 000 366 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1111 - 100 1001 0011 0101 1110 1011

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111