-0.000 000 000 742 206 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 206(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
-0.000 000 000 742 206(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:

|-0.000 000 000 742 206| = 0.000 000 000 742 206


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: 0.000 000 000 742 206.

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) 0.000 000 000 742 206 × 2 = 0 + 0.000 000 001 484 412;
  • 2) 0.000 000 001 484 412 × 2 = 0 + 0.000 000 002 968 824;
  • 3) 0.000 000 002 968 824 × 2 = 0 + 0.000 000 005 937 648;
  • 4) 0.000 000 005 937 648 × 2 = 0 + 0.000 000 011 875 296;
  • 5) 0.000 000 011 875 296 × 2 = 0 + 0.000 000 023 750 592;
  • 6) 0.000 000 023 750 592 × 2 = 0 + 0.000 000 047 501 184;
  • 7) 0.000 000 047 501 184 × 2 = 0 + 0.000 000 095 002 368;
  • 8) 0.000 000 095 002 368 × 2 = 0 + 0.000 000 190 004 736;
  • 9) 0.000 000 190 004 736 × 2 = 0 + 0.000 000 380 009 472;
  • 10) 0.000 000 380 009 472 × 2 = 0 + 0.000 000 760 018 944;
  • 11) 0.000 000 760 018 944 × 2 = 0 + 0.000 001 520 037 888;
  • 12) 0.000 001 520 037 888 × 2 = 0 + 0.000 003 040 075 776;
  • 13) 0.000 003 040 075 776 × 2 = 0 + 0.000 006 080 151 552;
  • 14) 0.000 006 080 151 552 × 2 = 0 + 0.000 012 160 303 104;
  • 15) 0.000 012 160 303 104 × 2 = 0 + 0.000 024 320 606 208;
  • 16) 0.000 024 320 606 208 × 2 = 0 + 0.000 048 641 212 416;
  • 17) 0.000 048 641 212 416 × 2 = 0 + 0.000 097 282 424 832;
  • 18) 0.000 097 282 424 832 × 2 = 0 + 0.000 194 564 849 664;
  • 19) 0.000 194 564 849 664 × 2 = 0 + 0.000 389 129 699 328;
  • 20) 0.000 389 129 699 328 × 2 = 0 + 0.000 778 259 398 656;
  • 21) 0.000 778 259 398 656 × 2 = 0 + 0.001 556 518 797 312;
  • 22) 0.001 556 518 797 312 × 2 = 0 + 0.003 113 037 594 624;
  • 23) 0.003 113 037 594 624 × 2 = 0 + 0.006 226 075 189 248;
  • 24) 0.006 226 075 189 248 × 2 = 0 + 0.012 452 150 378 496;
  • 25) 0.012 452 150 378 496 × 2 = 0 + 0.024 904 300 756 992;
  • 26) 0.024 904 300 756 992 × 2 = 0 + 0.049 808 601 513 984;
  • 27) 0.049 808 601 513 984 × 2 = 0 + 0.099 617 203 027 968;
  • 28) 0.099 617 203 027 968 × 2 = 0 + 0.199 234 406 055 936;
  • 29) 0.199 234 406 055 936 × 2 = 0 + 0.398 468 812 111 872;
  • 30) 0.398 468 812 111 872 × 2 = 0 + 0.796 937 624 223 744;
  • 31) 0.796 937 624 223 744 × 2 = 1 + 0.593 875 248 447 488;
  • 32) 0.593 875 248 447 488 × 2 = 1 + 0.187 750 496 894 976;
  • 33) 0.187 750 496 894 976 × 2 = 0 + 0.375 500 993 789 952;
  • 34) 0.375 500 993 789 952 × 2 = 0 + 0.751 001 987 579 904;
  • 35) 0.751 001 987 579 904 × 2 = 1 + 0.502 003 975 159 808;
  • 36) 0.502 003 975 159 808 × 2 = 1 + 0.004 007 950 319 616;
  • 37) 0.004 007 950 319 616 × 2 = 0 + 0.008 015 900 639 232;
  • 38) 0.008 015 900 639 232 × 2 = 0 + 0.016 031 801 278 464;
  • 39) 0.016 031 801 278 464 × 2 = 0 + 0.032 063 602 556 928;
  • 40) 0.032 063 602 556 928 × 2 = 0 + 0.064 127 205 113 856;
  • 41) 0.064 127 205 113 856 × 2 = 0 + 0.128 254 410 227 712;
  • 42) 0.128 254 410 227 712 × 2 = 0 + 0.256 508 820 455 424;
  • 43) 0.256 508 820 455 424 × 2 = 0 + 0.513 017 640 910 848;
  • 44) 0.513 017 640 910 848 × 2 = 1 + 0.026 035 281 821 696;
  • 45) 0.026 035 281 821 696 × 2 = 0 + 0.052 070 563 643 392;
  • 46) 0.052 070 563 643 392 × 2 = 0 + 0.104 141 127 286 784;
  • 47) 0.104 141 127 286 784 × 2 = 0 + 0.208 282 254 573 568;
  • 48) 0.208 282 254 573 568 × 2 = 0 + 0.416 564 509 147 136;
  • 49) 0.416 564 509 147 136 × 2 = 0 + 0.833 129 018 294 272;
  • 50) 0.833 129 018 294 272 × 2 = 1 + 0.666 258 036 588 544;
  • 51) 0.666 258 036 588 544 × 2 = 1 + 0.332 516 073 177 088;
  • 52) 0.332 516 073 177 088 × 2 = 0 + 0.665 032 146 354 176;
  • 53) 0.665 032 146 354 176 × 2 = 1 + 0.330 064 292 708 352;
  • 54) 0.330 064 292 708 352 × 2 = 0 + 0.660 128 585 416 704;

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:


0.000 000 000 742 206(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0000 0110 10(2)

6. Positive number before normalization:

0.000 000 000 742 206(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0000 0110 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:


0.000 000 000 742 206(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0000 0110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0000 0110 10(2) × 20 =

1.1001 1000 0000 1000 0011 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 1000 0000 1000 0011 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 1100 0000 0100 0001 1010 =

100 1100 0000 0100 0001 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 1100 0000 0100 0001 1010


Decimal number -0.000 000 000 742 206 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0100 0001 1010

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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