-0.000 000 000 814 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 814(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
-0.000 000 000 814(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:

|-0.000 000 000 814| = 0.000 000 000 814


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: 0.000 000 000 814.

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) 0.000 000 000 814 × 2 = 0 + 0.000 000 001 628;
  • 2) 0.000 000 001 628 × 2 = 0 + 0.000 000 003 256;
  • 3) 0.000 000 003 256 × 2 = 0 + 0.000 000 006 512;
  • 4) 0.000 000 006 512 × 2 = 0 + 0.000 000 013 024;
  • 5) 0.000 000 013 024 × 2 = 0 + 0.000 000 026 048;
  • 6) 0.000 000 026 048 × 2 = 0 + 0.000 000 052 096;
  • 7) 0.000 000 052 096 × 2 = 0 + 0.000 000 104 192;
  • 8) 0.000 000 104 192 × 2 = 0 + 0.000 000 208 384;
  • 9) 0.000 000 208 384 × 2 = 0 + 0.000 000 416 768;
  • 10) 0.000 000 416 768 × 2 = 0 + 0.000 000 833 536;
  • 11) 0.000 000 833 536 × 2 = 0 + 0.000 001 667 072;
  • 12) 0.000 001 667 072 × 2 = 0 + 0.000 003 334 144;
  • 13) 0.000 003 334 144 × 2 = 0 + 0.000 006 668 288;
  • 14) 0.000 006 668 288 × 2 = 0 + 0.000 013 336 576;
  • 15) 0.000 013 336 576 × 2 = 0 + 0.000 026 673 152;
  • 16) 0.000 026 673 152 × 2 = 0 + 0.000 053 346 304;
  • 17) 0.000 053 346 304 × 2 = 0 + 0.000 106 692 608;
  • 18) 0.000 106 692 608 × 2 = 0 + 0.000 213 385 216;
  • 19) 0.000 213 385 216 × 2 = 0 + 0.000 426 770 432;
  • 20) 0.000 426 770 432 × 2 = 0 + 0.000 853 540 864;
  • 21) 0.000 853 540 864 × 2 = 0 + 0.001 707 081 728;
  • 22) 0.001 707 081 728 × 2 = 0 + 0.003 414 163 456;
  • 23) 0.003 414 163 456 × 2 = 0 + 0.006 828 326 912;
  • 24) 0.006 828 326 912 × 2 = 0 + 0.013 656 653 824;
  • 25) 0.013 656 653 824 × 2 = 0 + 0.027 313 307 648;
  • 26) 0.027 313 307 648 × 2 = 0 + 0.054 626 615 296;
  • 27) 0.054 626 615 296 × 2 = 0 + 0.109 253 230 592;
  • 28) 0.109 253 230 592 × 2 = 0 + 0.218 506 461 184;
  • 29) 0.218 506 461 184 × 2 = 0 + 0.437 012 922 368;
  • 30) 0.437 012 922 368 × 2 = 0 + 0.874 025 844 736;
  • 31) 0.874 025 844 736 × 2 = 1 + 0.748 051 689 472;
  • 32) 0.748 051 689 472 × 2 = 1 + 0.496 103 378 944;
  • 33) 0.496 103 378 944 × 2 = 0 + 0.992 206 757 888;
  • 34) 0.992 206 757 888 × 2 = 1 + 0.984 413 515 776;
  • 35) 0.984 413 515 776 × 2 = 1 + 0.968 827 031 552;
  • 36) 0.968 827 031 552 × 2 = 1 + 0.937 654 063 104;
  • 37) 0.937 654 063 104 × 2 = 1 + 0.875 308 126 208;
  • 38) 0.875 308 126 208 × 2 = 1 + 0.750 616 252 416;
  • 39) 0.750 616 252 416 × 2 = 1 + 0.501 232 504 832;
  • 40) 0.501 232 504 832 × 2 = 1 + 0.002 465 009 664;
  • 41) 0.002 465 009 664 × 2 = 0 + 0.004 930 019 328;
  • 42) 0.004 930 019 328 × 2 = 0 + 0.009 860 038 656;
  • 43) 0.009 860 038 656 × 2 = 0 + 0.019 720 077 312;
  • 44) 0.019 720 077 312 × 2 = 0 + 0.039 440 154 624;
  • 45) 0.039 440 154 624 × 2 = 0 + 0.078 880 309 248;
  • 46) 0.078 880 309 248 × 2 = 0 + 0.157 760 618 496;
  • 47) 0.157 760 618 496 × 2 = 0 + 0.315 521 236 992;
  • 48) 0.315 521 236 992 × 2 = 0 + 0.631 042 473 984;
  • 49) 0.631 042 473 984 × 2 = 1 + 0.262 084 947 968;
  • 50) 0.262 084 947 968 × 2 = 0 + 0.524 169 895 936;
  • 51) 0.524 169 895 936 × 2 = 1 + 0.048 339 791 872;
  • 52) 0.048 339 791 872 × 2 = 0 + 0.096 679 583 744;
  • 53) 0.096 679 583 744 × 2 = 0 + 0.193 359 167 488;
  • 54) 0.193 359 167 488 × 2 = 0 + 0.386 718 334 976;

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:


0.000 000 000 814(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1111 0000 0000 1010 00(2)

6. Positive number before normalization:

0.000 000 000 814(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1111 0000 0000 1010 00(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:


0.000 000 000 814(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1111 0000 0000 1010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1111 0000 0000 1010 00(2) × 20 =

1.1011 1111 1000 0000 0101 000(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 1111 1000 0000 0101 000


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 1111 1100 0000 0010 1000 =

101 1111 1100 0000 0010 1000


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 1111 1100 0000 0010 1000


Decimal number -0.000 000 000 814 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 1111 1100 0000 0010 1000

Share

» Convert decimal -0.000 000 000 841 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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