-0.000 000 000 741 61 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 61(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
-0.000 000 000 741 61(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:

|-0.000 000 000 741 61| = 0.000 000 000 741 61


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: 0.000 000 000 741 61.

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) 0.000 000 000 741 61 × 2 = 0 + 0.000 000 001 483 22;
  • 2) 0.000 000 001 483 22 × 2 = 0 + 0.000 000 002 966 44;
  • 3) 0.000 000 002 966 44 × 2 = 0 + 0.000 000 005 932 88;
  • 4) 0.000 000 005 932 88 × 2 = 0 + 0.000 000 011 865 76;
  • 5) 0.000 000 011 865 76 × 2 = 0 + 0.000 000 023 731 52;
  • 6) 0.000 000 023 731 52 × 2 = 0 + 0.000 000 047 463 04;
  • 7) 0.000 000 047 463 04 × 2 = 0 + 0.000 000 094 926 08;
  • 8) 0.000 000 094 926 08 × 2 = 0 + 0.000 000 189 852 16;
  • 9) 0.000 000 189 852 16 × 2 = 0 + 0.000 000 379 704 32;
  • 10) 0.000 000 379 704 32 × 2 = 0 + 0.000 000 759 408 64;
  • 11) 0.000 000 759 408 64 × 2 = 0 + 0.000 001 518 817 28;
  • 12) 0.000 001 518 817 28 × 2 = 0 + 0.000 003 037 634 56;
  • 13) 0.000 003 037 634 56 × 2 = 0 + 0.000 006 075 269 12;
  • 14) 0.000 006 075 269 12 × 2 = 0 + 0.000 012 150 538 24;
  • 15) 0.000 012 150 538 24 × 2 = 0 + 0.000 024 301 076 48;
  • 16) 0.000 024 301 076 48 × 2 = 0 + 0.000 048 602 152 96;
  • 17) 0.000 048 602 152 96 × 2 = 0 + 0.000 097 204 305 92;
  • 18) 0.000 097 204 305 92 × 2 = 0 + 0.000 194 408 611 84;
  • 19) 0.000 194 408 611 84 × 2 = 0 + 0.000 388 817 223 68;
  • 20) 0.000 388 817 223 68 × 2 = 0 + 0.000 777 634 447 36;
  • 21) 0.000 777 634 447 36 × 2 = 0 + 0.001 555 268 894 72;
  • 22) 0.001 555 268 894 72 × 2 = 0 + 0.003 110 537 789 44;
  • 23) 0.003 110 537 789 44 × 2 = 0 + 0.006 221 075 578 88;
  • 24) 0.006 221 075 578 88 × 2 = 0 + 0.012 442 151 157 76;
  • 25) 0.012 442 151 157 76 × 2 = 0 + 0.024 884 302 315 52;
  • 26) 0.024 884 302 315 52 × 2 = 0 + 0.049 768 604 631 04;
  • 27) 0.049 768 604 631 04 × 2 = 0 + 0.099 537 209 262 08;
  • 28) 0.099 537 209 262 08 × 2 = 0 + 0.199 074 418 524 16;
  • 29) 0.199 074 418 524 16 × 2 = 0 + 0.398 148 837 048 32;
  • 30) 0.398 148 837 048 32 × 2 = 0 + 0.796 297 674 096 64;
  • 31) 0.796 297 674 096 64 × 2 = 1 + 0.592 595 348 193 28;
  • 32) 0.592 595 348 193 28 × 2 = 1 + 0.185 190 696 386 56;
  • 33) 0.185 190 696 386 56 × 2 = 0 + 0.370 381 392 773 12;
  • 34) 0.370 381 392 773 12 × 2 = 0 + 0.740 762 785 546 24;
  • 35) 0.740 762 785 546 24 × 2 = 1 + 0.481 525 571 092 48;
  • 36) 0.481 525 571 092 48 × 2 = 0 + 0.963 051 142 184 96;
  • 37) 0.963 051 142 184 96 × 2 = 1 + 0.926 102 284 369 92;
  • 38) 0.926 102 284 369 92 × 2 = 1 + 0.852 204 568 739 84;
  • 39) 0.852 204 568 739 84 × 2 = 1 + 0.704 409 137 479 68;
  • 40) 0.704 409 137 479 68 × 2 = 1 + 0.408 818 274 959 36;
  • 41) 0.408 818 274 959 36 × 2 = 0 + 0.817 636 549 918 72;
  • 42) 0.817 636 549 918 72 × 2 = 1 + 0.635 273 099 837 44;
  • 43) 0.635 273 099 837 44 × 2 = 1 + 0.270 546 199 674 88;
  • 44) 0.270 546 199 674 88 × 2 = 0 + 0.541 092 399 349 76;
  • 45) 0.541 092 399 349 76 × 2 = 1 + 0.082 184 798 699 52;
  • 46) 0.082 184 798 699 52 × 2 = 0 + 0.164 369 597 399 04;
  • 47) 0.164 369 597 399 04 × 2 = 0 + 0.328 739 194 798 08;
  • 48) 0.328 739 194 798 08 × 2 = 0 + 0.657 478 389 596 16;
  • 49) 0.657 478 389 596 16 × 2 = 1 + 0.314 956 779 192 32;
  • 50) 0.314 956 779 192 32 × 2 = 0 + 0.629 913 558 384 64;
  • 51) 0.629 913 558 384 64 × 2 = 1 + 0.259 827 116 769 28;
  • 52) 0.259 827 116 769 28 × 2 = 0 + 0.519 654 233 538 56;
  • 53) 0.519 654 233 538 56 × 2 = 1 + 0.039 308 467 077 12;
  • 54) 0.039 308 467 077 12 × 2 = 0 + 0.078 616 934 154 24;

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:


0.000 000 000 741 61(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0110 1000 1010 10(2)

6. Positive number before normalization:

0.000 000 000 741 61(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0110 1000 1010 10(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:


0.000 000 000 741 61(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0110 1000 1010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 0110 1000 1010 10(2) × 20 =

1.1001 0111 1011 0100 0101 010(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):
1.1001 0111 1011 0100 0101 010


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. 100 1011 1101 1010 0010 1010 =

100 1011 1101 1010 0010 1010


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) =
100 1011 1101 1010 0010 1010


Decimal number -0.000 000 000 741 61 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1101 1010 0010 1010

Share

» Convert decimal -0.000 000 000 742 61 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 Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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