-0.000 000 000 856 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 856(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
-0.000 000 000 856(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:

|-0.000 000 000 856| = 0.000 000 000 856


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: 0.000 000 000 856.

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) 0.000 000 000 856 × 2 = 0 + 0.000 000 001 712;
  • 2) 0.000 000 001 712 × 2 = 0 + 0.000 000 003 424;
  • 3) 0.000 000 003 424 × 2 = 0 + 0.000 000 006 848;
  • 4) 0.000 000 006 848 × 2 = 0 + 0.000 000 013 696;
  • 5) 0.000 000 013 696 × 2 = 0 + 0.000 000 027 392;
  • 6) 0.000 000 027 392 × 2 = 0 + 0.000 000 054 784;
  • 7) 0.000 000 054 784 × 2 = 0 + 0.000 000 109 568;
  • 8) 0.000 000 109 568 × 2 = 0 + 0.000 000 219 136;
  • 9) 0.000 000 219 136 × 2 = 0 + 0.000 000 438 272;
  • 10) 0.000 000 438 272 × 2 = 0 + 0.000 000 876 544;
  • 11) 0.000 000 876 544 × 2 = 0 + 0.000 001 753 088;
  • 12) 0.000 001 753 088 × 2 = 0 + 0.000 003 506 176;
  • 13) 0.000 003 506 176 × 2 = 0 + 0.000 007 012 352;
  • 14) 0.000 007 012 352 × 2 = 0 + 0.000 014 024 704;
  • 15) 0.000 014 024 704 × 2 = 0 + 0.000 028 049 408;
  • 16) 0.000 028 049 408 × 2 = 0 + 0.000 056 098 816;
  • 17) 0.000 056 098 816 × 2 = 0 + 0.000 112 197 632;
  • 18) 0.000 112 197 632 × 2 = 0 + 0.000 224 395 264;
  • 19) 0.000 224 395 264 × 2 = 0 + 0.000 448 790 528;
  • 20) 0.000 448 790 528 × 2 = 0 + 0.000 897 581 056;
  • 21) 0.000 897 581 056 × 2 = 0 + 0.001 795 162 112;
  • 22) 0.001 795 162 112 × 2 = 0 + 0.003 590 324 224;
  • 23) 0.003 590 324 224 × 2 = 0 + 0.007 180 648 448;
  • 24) 0.007 180 648 448 × 2 = 0 + 0.014 361 296 896;
  • 25) 0.014 361 296 896 × 2 = 0 + 0.028 722 593 792;
  • 26) 0.028 722 593 792 × 2 = 0 + 0.057 445 187 584;
  • 27) 0.057 445 187 584 × 2 = 0 + 0.114 890 375 168;
  • 28) 0.114 890 375 168 × 2 = 0 + 0.229 780 750 336;
  • 29) 0.229 780 750 336 × 2 = 0 + 0.459 561 500 672;
  • 30) 0.459 561 500 672 × 2 = 0 + 0.919 123 001 344;
  • 31) 0.919 123 001 344 × 2 = 1 + 0.838 246 002 688;
  • 32) 0.838 246 002 688 × 2 = 1 + 0.676 492 005 376;
  • 33) 0.676 492 005 376 × 2 = 1 + 0.352 984 010 752;
  • 34) 0.352 984 010 752 × 2 = 0 + 0.705 968 021 504;
  • 35) 0.705 968 021 504 × 2 = 1 + 0.411 936 043 008;
  • 36) 0.411 936 043 008 × 2 = 0 + 0.823 872 086 016;
  • 37) 0.823 872 086 016 × 2 = 1 + 0.647 744 172 032;
  • 38) 0.647 744 172 032 × 2 = 1 + 0.295 488 344 064;
  • 39) 0.295 488 344 064 × 2 = 0 + 0.590 976 688 128;
  • 40) 0.590 976 688 128 × 2 = 1 + 0.181 953 376 256;
  • 41) 0.181 953 376 256 × 2 = 0 + 0.363 906 752 512;
  • 42) 0.363 906 752 512 × 2 = 0 + 0.727 813 505 024;
  • 43) 0.727 813 505 024 × 2 = 1 + 0.455 627 010 048;
  • 44) 0.455 627 010 048 × 2 = 0 + 0.911 254 020 096;
  • 45) 0.911 254 020 096 × 2 = 1 + 0.822 508 040 192;
  • 46) 0.822 508 040 192 × 2 = 1 + 0.645 016 080 384;
  • 47) 0.645 016 080 384 × 2 = 1 + 0.290 032 160 768;
  • 48) 0.290 032 160 768 × 2 = 0 + 0.580 064 321 536;
  • 49) 0.580 064 321 536 × 2 = 1 + 0.160 128 643 072;
  • 50) 0.160 128 643 072 × 2 = 0 + 0.320 257 286 144;
  • 51) 0.320 257 286 144 × 2 = 0 + 0.640 514 572 288;
  • 52) 0.640 514 572 288 × 2 = 1 + 0.281 029 144 576;
  • 53) 0.281 029 144 576 × 2 = 0 + 0.562 058 289 152;
  • 54) 0.562 058 289 152 × 2 = 1 + 0.124 116 578 304;

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:


0.000 000 000 856(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 1101 0010 1110 1001 01(2)

6. Positive number before normalization:

0.000 000 000 856(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 1101 0010 1110 1001 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 856(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 1101 0010 1110 1001 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1010 1101 0010 1110 1001 01(2) × 20 =

1.1101 0110 1001 0111 0100 101(2) × 2-31


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): -31

Mantissa (not normalized):
1.1101 0110 1001 0111 0100 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

96(10) =

0110 0000(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. 110 1011 0100 1011 1010 0101 =

110 1011 0100 1011 1010 0101


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) =
0110 0000

Mantissa (23 bits) =
110 1011 0100 1011 1010 0101


Decimal number -0.000 000 000 856 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 1011 0100 1011 1010 0101

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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