-0.000 000 000 832 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 832(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
-0.000 000 000 832(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:

|-0.000 000 000 832| = 0.000 000 000 832


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: 0.000 000 000 832.

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) 0.000 000 000 832 × 2 = 0 + 0.000 000 001 664;
  • 2) 0.000 000 001 664 × 2 = 0 + 0.000 000 003 328;
  • 3) 0.000 000 003 328 × 2 = 0 + 0.000 000 006 656;
  • 4) 0.000 000 006 656 × 2 = 0 + 0.000 000 013 312;
  • 5) 0.000 000 013 312 × 2 = 0 + 0.000 000 026 624;
  • 6) 0.000 000 026 624 × 2 = 0 + 0.000 000 053 248;
  • 7) 0.000 000 053 248 × 2 = 0 + 0.000 000 106 496;
  • 8) 0.000 000 106 496 × 2 = 0 + 0.000 000 212 992;
  • 9) 0.000 000 212 992 × 2 = 0 + 0.000 000 425 984;
  • 10) 0.000 000 425 984 × 2 = 0 + 0.000 000 851 968;
  • 11) 0.000 000 851 968 × 2 = 0 + 0.000 001 703 936;
  • 12) 0.000 001 703 936 × 2 = 0 + 0.000 003 407 872;
  • 13) 0.000 003 407 872 × 2 = 0 + 0.000 006 815 744;
  • 14) 0.000 006 815 744 × 2 = 0 + 0.000 013 631 488;
  • 15) 0.000 013 631 488 × 2 = 0 + 0.000 027 262 976;
  • 16) 0.000 027 262 976 × 2 = 0 + 0.000 054 525 952;
  • 17) 0.000 054 525 952 × 2 = 0 + 0.000 109 051 904;
  • 18) 0.000 109 051 904 × 2 = 0 + 0.000 218 103 808;
  • 19) 0.000 218 103 808 × 2 = 0 + 0.000 436 207 616;
  • 20) 0.000 436 207 616 × 2 = 0 + 0.000 872 415 232;
  • 21) 0.000 872 415 232 × 2 = 0 + 0.001 744 830 464;
  • 22) 0.001 744 830 464 × 2 = 0 + 0.003 489 660 928;
  • 23) 0.003 489 660 928 × 2 = 0 + 0.006 979 321 856;
  • 24) 0.006 979 321 856 × 2 = 0 + 0.013 958 643 712;
  • 25) 0.013 958 643 712 × 2 = 0 + 0.027 917 287 424;
  • 26) 0.027 917 287 424 × 2 = 0 + 0.055 834 574 848;
  • 27) 0.055 834 574 848 × 2 = 0 + 0.111 669 149 696;
  • 28) 0.111 669 149 696 × 2 = 0 + 0.223 338 299 392;
  • 29) 0.223 338 299 392 × 2 = 0 + 0.446 676 598 784;
  • 30) 0.446 676 598 784 × 2 = 0 + 0.893 353 197 568;
  • 31) 0.893 353 197 568 × 2 = 1 + 0.786 706 395 136;
  • 32) 0.786 706 395 136 × 2 = 1 + 0.573 412 790 272;
  • 33) 0.573 412 790 272 × 2 = 1 + 0.146 825 580 544;
  • 34) 0.146 825 580 544 × 2 = 0 + 0.293 651 161 088;
  • 35) 0.293 651 161 088 × 2 = 0 + 0.587 302 322 176;
  • 36) 0.587 302 322 176 × 2 = 1 + 0.174 604 644 352;
  • 37) 0.174 604 644 352 × 2 = 0 + 0.349 209 288 704;
  • 38) 0.349 209 288 704 × 2 = 0 + 0.698 418 577 408;
  • 39) 0.698 418 577 408 × 2 = 1 + 0.396 837 154 816;
  • 40) 0.396 837 154 816 × 2 = 0 + 0.793 674 309 632;
  • 41) 0.793 674 309 632 × 2 = 1 + 0.587 348 619 264;
  • 42) 0.587 348 619 264 × 2 = 1 + 0.174 697 238 528;
  • 43) 0.174 697 238 528 × 2 = 0 + 0.349 394 477 056;
  • 44) 0.349 394 477 056 × 2 = 0 + 0.698 788 954 112;
  • 45) 0.698 788 954 112 × 2 = 1 + 0.397 577 908 224;
  • 46) 0.397 577 908 224 × 2 = 0 + 0.795 155 816 448;
  • 47) 0.795 155 816 448 × 2 = 1 + 0.590 311 632 896;
  • 48) 0.590 311 632 896 × 2 = 1 + 0.180 623 265 792;
  • 49) 0.180 623 265 792 × 2 = 0 + 0.361 246 531 584;
  • 50) 0.361 246 531 584 × 2 = 0 + 0.722 493 063 168;
  • 51) 0.722 493 063 168 × 2 = 1 + 0.444 986 126 336;
  • 52) 0.444 986 126 336 × 2 = 0 + 0.889 972 252 672;
  • 53) 0.889 972 252 672 × 2 = 1 + 0.779 944 505 344;
  • 54) 0.779 944 505 344 × 2 = 1 + 0.559 889 010 688;

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:


0.000 000 000 832(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 0010 1100 1011 0010 11(2)

6. Positive number before normalization:

0.000 000 000 832(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 0010 1100 1011 0010 11(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:


0.000 000 000 832(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 0010 1100 1011 0010 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 0010 1100 1011 0010 11(2) × 20 =

1.1100 1001 0110 0101 1001 011(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):
1.1100 1001 0110 0101 1001 011


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 0100 1011 0010 1100 1011 =

110 0100 1011 0010 1100 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) =
0110 0000

Mantissa (23 bits) =
110 0100 1011 0010 1100 1011


Decimal number -0.000 000 000 832 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 0100 1011 0010 1100 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