-0.000 000 000 791 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 791(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
-0.000 000 000 791(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:

|-0.000 000 000 791| = 0.000 000 000 791


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: 0.000 000 000 791.

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) 0.000 000 000 791 × 2 = 0 + 0.000 000 001 582;
  • 2) 0.000 000 001 582 × 2 = 0 + 0.000 000 003 164;
  • 3) 0.000 000 003 164 × 2 = 0 + 0.000 000 006 328;
  • 4) 0.000 000 006 328 × 2 = 0 + 0.000 000 012 656;
  • 5) 0.000 000 012 656 × 2 = 0 + 0.000 000 025 312;
  • 6) 0.000 000 025 312 × 2 = 0 + 0.000 000 050 624;
  • 7) 0.000 000 050 624 × 2 = 0 + 0.000 000 101 248;
  • 8) 0.000 000 101 248 × 2 = 0 + 0.000 000 202 496;
  • 9) 0.000 000 202 496 × 2 = 0 + 0.000 000 404 992;
  • 10) 0.000 000 404 992 × 2 = 0 + 0.000 000 809 984;
  • 11) 0.000 000 809 984 × 2 = 0 + 0.000 001 619 968;
  • 12) 0.000 001 619 968 × 2 = 0 + 0.000 003 239 936;
  • 13) 0.000 003 239 936 × 2 = 0 + 0.000 006 479 872;
  • 14) 0.000 006 479 872 × 2 = 0 + 0.000 012 959 744;
  • 15) 0.000 012 959 744 × 2 = 0 + 0.000 025 919 488;
  • 16) 0.000 025 919 488 × 2 = 0 + 0.000 051 838 976;
  • 17) 0.000 051 838 976 × 2 = 0 + 0.000 103 677 952;
  • 18) 0.000 103 677 952 × 2 = 0 + 0.000 207 355 904;
  • 19) 0.000 207 355 904 × 2 = 0 + 0.000 414 711 808;
  • 20) 0.000 414 711 808 × 2 = 0 + 0.000 829 423 616;
  • 21) 0.000 829 423 616 × 2 = 0 + 0.001 658 847 232;
  • 22) 0.001 658 847 232 × 2 = 0 + 0.003 317 694 464;
  • 23) 0.003 317 694 464 × 2 = 0 + 0.006 635 388 928;
  • 24) 0.006 635 388 928 × 2 = 0 + 0.013 270 777 856;
  • 25) 0.013 270 777 856 × 2 = 0 + 0.026 541 555 712;
  • 26) 0.026 541 555 712 × 2 = 0 + 0.053 083 111 424;
  • 27) 0.053 083 111 424 × 2 = 0 + 0.106 166 222 848;
  • 28) 0.106 166 222 848 × 2 = 0 + 0.212 332 445 696;
  • 29) 0.212 332 445 696 × 2 = 0 + 0.424 664 891 392;
  • 30) 0.424 664 891 392 × 2 = 0 + 0.849 329 782 784;
  • 31) 0.849 329 782 784 × 2 = 1 + 0.698 659 565 568;
  • 32) 0.698 659 565 568 × 2 = 1 + 0.397 319 131 136;
  • 33) 0.397 319 131 136 × 2 = 0 + 0.794 638 262 272;
  • 34) 0.794 638 262 272 × 2 = 1 + 0.589 276 524 544;
  • 35) 0.589 276 524 544 × 2 = 1 + 0.178 553 049 088;
  • 36) 0.178 553 049 088 × 2 = 0 + 0.357 106 098 176;
  • 37) 0.357 106 098 176 × 2 = 0 + 0.714 212 196 352;
  • 38) 0.714 212 196 352 × 2 = 1 + 0.428 424 392 704;
  • 39) 0.428 424 392 704 × 2 = 0 + 0.856 848 785 408;
  • 40) 0.856 848 785 408 × 2 = 1 + 0.713 697 570 816;
  • 41) 0.713 697 570 816 × 2 = 1 + 0.427 395 141 632;
  • 42) 0.427 395 141 632 × 2 = 0 + 0.854 790 283 264;
  • 43) 0.854 790 283 264 × 2 = 1 + 0.709 580 566 528;
  • 44) 0.709 580 566 528 × 2 = 1 + 0.419 161 133 056;
  • 45) 0.419 161 133 056 × 2 = 0 + 0.838 322 266 112;
  • 46) 0.838 322 266 112 × 2 = 1 + 0.676 644 532 224;
  • 47) 0.676 644 532 224 × 2 = 1 + 0.353 289 064 448;
  • 48) 0.353 289 064 448 × 2 = 0 + 0.706 578 128 896;
  • 49) 0.706 578 128 896 × 2 = 1 + 0.413 156 257 792;
  • 50) 0.413 156 257 792 × 2 = 0 + 0.826 312 515 584;
  • 51) 0.826 312 515 584 × 2 = 1 + 0.652 625 031 168;
  • 52) 0.652 625 031 168 × 2 = 1 + 0.305 250 062 336;
  • 53) 0.305 250 062 336 × 2 = 0 + 0.610 500 124 672;
  • 54) 0.610 500 124 672 × 2 = 1 + 0.221 000 249 344;

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:


0.000 000 000 791(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0101 1011 0110 1011 01(2)

6. Positive number before normalization:

0.000 000 000 791(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0101 1011 0110 1011 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:


0.000 000 000 791(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0101 1011 0110 1011 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0110 0101 1011 0110 1011 01(2) × 20 =

1.1011 0010 1101 1011 0101 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):
1.1011 0010 1101 1011 0101 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. 101 1001 0110 1101 1010 1101 =

101 1001 0110 1101 1010 1101


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) =
101 1001 0110 1101 1010 1101


Decimal number -0.000 000 000 791 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 1001 0110 1101 1010 1101

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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