-0.000 000 000 631 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 631(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
-0.000 000 000 631(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:

|-0.000 000 000 631| = 0.000 000 000 631


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: 0.000 000 000 631.

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) 0.000 000 000 631 × 2 = 0 + 0.000 000 001 262;
  • 2) 0.000 000 001 262 × 2 = 0 + 0.000 000 002 524;
  • 3) 0.000 000 002 524 × 2 = 0 + 0.000 000 005 048;
  • 4) 0.000 000 005 048 × 2 = 0 + 0.000 000 010 096;
  • 5) 0.000 000 010 096 × 2 = 0 + 0.000 000 020 192;
  • 6) 0.000 000 020 192 × 2 = 0 + 0.000 000 040 384;
  • 7) 0.000 000 040 384 × 2 = 0 + 0.000 000 080 768;
  • 8) 0.000 000 080 768 × 2 = 0 + 0.000 000 161 536;
  • 9) 0.000 000 161 536 × 2 = 0 + 0.000 000 323 072;
  • 10) 0.000 000 323 072 × 2 = 0 + 0.000 000 646 144;
  • 11) 0.000 000 646 144 × 2 = 0 + 0.000 001 292 288;
  • 12) 0.000 001 292 288 × 2 = 0 + 0.000 002 584 576;
  • 13) 0.000 002 584 576 × 2 = 0 + 0.000 005 169 152;
  • 14) 0.000 005 169 152 × 2 = 0 + 0.000 010 338 304;
  • 15) 0.000 010 338 304 × 2 = 0 + 0.000 020 676 608;
  • 16) 0.000 020 676 608 × 2 = 0 + 0.000 041 353 216;
  • 17) 0.000 041 353 216 × 2 = 0 + 0.000 082 706 432;
  • 18) 0.000 082 706 432 × 2 = 0 + 0.000 165 412 864;
  • 19) 0.000 165 412 864 × 2 = 0 + 0.000 330 825 728;
  • 20) 0.000 330 825 728 × 2 = 0 + 0.000 661 651 456;
  • 21) 0.000 661 651 456 × 2 = 0 + 0.001 323 302 912;
  • 22) 0.001 323 302 912 × 2 = 0 + 0.002 646 605 824;
  • 23) 0.002 646 605 824 × 2 = 0 + 0.005 293 211 648;
  • 24) 0.005 293 211 648 × 2 = 0 + 0.010 586 423 296;
  • 25) 0.010 586 423 296 × 2 = 0 + 0.021 172 846 592;
  • 26) 0.021 172 846 592 × 2 = 0 + 0.042 345 693 184;
  • 27) 0.042 345 693 184 × 2 = 0 + 0.084 691 386 368;
  • 28) 0.084 691 386 368 × 2 = 0 + 0.169 382 772 736;
  • 29) 0.169 382 772 736 × 2 = 0 + 0.338 765 545 472;
  • 30) 0.338 765 545 472 × 2 = 0 + 0.677 531 090 944;
  • 31) 0.677 531 090 944 × 2 = 1 + 0.355 062 181 888;
  • 32) 0.355 062 181 888 × 2 = 0 + 0.710 124 363 776;
  • 33) 0.710 124 363 776 × 2 = 1 + 0.420 248 727 552;
  • 34) 0.420 248 727 552 × 2 = 0 + 0.840 497 455 104;
  • 35) 0.840 497 455 104 × 2 = 1 + 0.680 994 910 208;
  • 36) 0.680 994 910 208 × 2 = 1 + 0.361 989 820 416;
  • 37) 0.361 989 820 416 × 2 = 0 + 0.723 979 640 832;
  • 38) 0.723 979 640 832 × 2 = 1 + 0.447 959 281 664;
  • 39) 0.447 959 281 664 × 2 = 0 + 0.895 918 563 328;
  • 40) 0.895 918 563 328 × 2 = 1 + 0.791 837 126 656;
  • 41) 0.791 837 126 656 × 2 = 1 + 0.583 674 253 312;
  • 42) 0.583 674 253 312 × 2 = 1 + 0.167 348 506 624;
  • 43) 0.167 348 506 624 × 2 = 0 + 0.334 697 013 248;
  • 44) 0.334 697 013 248 × 2 = 0 + 0.669 394 026 496;
  • 45) 0.669 394 026 496 × 2 = 1 + 0.338 788 052 992;
  • 46) 0.338 788 052 992 × 2 = 0 + 0.677 576 105 984;
  • 47) 0.677 576 105 984 × 2 = 1 + 0.355 152 211 968;
  • 48) 0.355 152 211 968 × 2 = 0 + 0.710 304 423 936;
  • 49) 0.710 304 423 936 × 2 = 1 + 0.420 608 847 872;
  • 50) 0.420 608 847 872 × 2 = 0 + 0.841 217 695 744;
  • 51) 0.841 217 695 744 × 2 = 1 + 0.682 435 391 488;
  • 52) 0.682 435 391 488 × 2 = 1 + 0.364 870 782 976;
  • 53) 0.364 870 782 976 × 2 = 0 + 0.729 741 565 952;
  • 54) 0.729 741 565 952 × 2 = 1 + 0.459 483 131 904;

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:


0.000 000 000 631(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 0101 1100 1010 1011 01(2)

6. Positive number before normalization:

0.000 000 000 631(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 0101 1100 1010 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:


0.000 000 000 631(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 0101 1100 1010 1011 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 0101 1100 1010 1011 01(2) × 20 =

1.0101 1010 1110 0101 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):
1.0101 1010 1110 0101 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. 010 1101 0111 0010 1010 1101 =

010 1101 0111 0010 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) =
010 1101 0111 0010 1010 1101


Decimal number -0.000 000 000 631 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 010 1101 0111 0010 1010 1101

Share

» Convert decimal -0.000 000 000 591 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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