-0.000 000 000 836 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 836(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
-0.000 000 000 836(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:

|-0.000 000 000 836| = 0.000 000 000 836


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: 0.000 000 000 836.

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) 0.000 000 000 836 × 2 = 0 + 0.000 000 001 672;
  • 2) 0.000 000 001 672 × 2 = 0 + 0.000 000 003 344;
  • 3) 0.000 000 003 344 × 2 = 0 + 0.000 000 006 688;
  • 4) 0.000 000 006 688 × 2 = 0 + 0.000 000 013 376;
  • 5) 0.000 000 013 376 × 2 = 0 + 0.000 000 026 752;
  • 6) 0.000 000 026 752 × 2 = 0 + 0.000 000 053 504;
  • 7) 0.000 000 053 504 × 2 = 0 + 0.000 000 107 008;
  • 8) 0.000 000 107 008 × 2 = 0 + 0.000 000 214 016;
  • 9) 0.000 000 214 016 × 2 = 0 + 0.000 000 428 032;
  • 10) 0.000 000 428 032 × 2 = 0 + 0.000 000 856 064;
  • 11) 0.000 000 856 064 × 2 = 0 + 0.000 001 712 128;
  • 12) 0.000 001 712 128 × 2 = 0 + 0.000 003 424 256;
  • 13) 0.000 003 424 256 × 2 = 0 + 0.000 006 848 512;
  • 14) 0.000 006 848 512 × 2 = 0 + 0.000 013 697 024;
  • 15) 0.000 013 697 024 × 2 = 0 + 0.000 027 394 048;
  • 16) 0.000 027 394 048 × 2 = 0 + 0.000 054 788 096;
  • 17) 0.000 054 788 096 × 2 = 0 + 0.000 109 576 192;
  • 18) 0.000 109 576 192 × 2 = 0 + 0.000 219 152 384;
  • 19) 0.000 219 152 384 × 2 = 0 + 0.000 438 304 768;
  • 20) 0.000 438 304 768 × 2 = 0 + 0.000 876 609 536;
  • 21) 0.000 876 609 536 × 2 = 0 + 0.001 753 219 072;
  • 22) 0.001 753 219 072 × 2 = 0 + 0.003 506 438 144;
  • 23) 0.003 506 438 144 × 2 = 0 + 0.007 012 876 288;
  • 24) 0.007 012 876 288 × 2 = 0 + 0.014 025 752 576;
  • 25) 0.014 025 752 576 × 2 = 0 + 0.028 051 505 152;
  • 26) 0.028 051 505 152 × 2 = 0 + 0.056 103 010 304;
  • 27) 0.056 103 010 304 × 2 = 0 + 0.112 206 020 608;
  • 28) 0.112 206 020 608 × 2 = 0 + 0.224 412 041 216;
  • 29) 0.224 412 041 216 × 2 = 0 + 0.448 824 082 432;
  • 30) 0.448 824 082 432 × 2 = 0 + 0.897 648 164 864;
  • 31) 0.897 648 164 864 × 2 = 1 + 0.795 296 329 728;
  • 32) 0.795 296 329 728 × 2 = 1 + 0.590 592 659 456;
  • 33) 0.590 592 659 456 × 2 = 1 + 0.181 185 318 912;
  • 34) 0.181 185 318 912 × 2 = 0 + 0.362 370 637 824;
  • 35) 0.362 370 637 824 × 2 = 0 + 0.724 741 275 648;
  • 36) 0.724 741 275 648 × 2 = 1 + 0.449 482 551 296;
  • 37) 0.449 482 551 296 × 2 = 0 + 0.898 965 102 592;
  • 38) 0.898 965 102 592 × 2 = 1 + 0.797 930 205 184;
  • 39) 0.797 930 205 184 × 2 = 1 + 0.595 860 410 368;
  • 40) 0.595 860 410 368 × 2 = 1 + 0.191 720 820 736;
  • 41) 0.191 720 820 736 × 2 = 0 + 0.383 441 641 472;
  • 42) 0.383 441 641 472 × 2 = 0 + 0.766 883 282 944;
  • 43) 0.766 883 282 944 × 2 = 1 + 0.533 766 565 888;
  • 44) 0.533 766 565 888 × 2 = 1 + 0.067 533 131 776;
  • 45) 0.067 533 131 776 × 2 = 0 + 0.135 066 263 552;
  • 46) 0.135 066 263 552 × 2 = 0 + 0.270 132 527 104;
  • 47) 0.270 132 527 104 × 2 = 0 + 0.540 265 054 208;
  • 48) 0.540 265 054 208 × 2 = 1 + 0.080 530 108 416;
  • 49) 0.080 530 108 416 × 2 = 0 + 0.161 060 216 832;
  • 50) 0.161 060 216 832 × 2 = 0 + 0.322 120 433 664;
  • 51) 0.322 120 433 664 × 2 = 0 + 0.644 240 867 328;
  • 52) 0.644 240 867 328 × 2 = 1 + 0.288 481 734 656;
  • 53) 0.288 481 734 656 × 2 = 0 + 0.576 963 469 312;
  • 54) 0.576 963 469 312 × 2 = 1 + 0.153 926 938 624;

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:


0.000 000 000 836(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 0111 0011 0001 0001 01(2)

6. Positive number before normalization:

0.000 000 000 836(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 0111 0011 0001 0001 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:


0.000 000 000 836(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 0111 0011 0001 0001 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1001 0111 0011 0001 0001 01(2) × 20 =

1.1100 1011 1001 1000 1000 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):
1.1100 1011 1001 1000 1000 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 0101 1100 1100 0100 0101 =

110 0101 1100 1100 0100 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 0101 1100 1100 0100 0101


Decimal number -0.000 000 000 836 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 0101 1100 1100 0100 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