-0.000 000 000 742 150 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 150 3(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
-0.000 000 000 742 150 3(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:

|-0.000 000 000 742 150 3| = 0.000 000 000 742 150 3


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: 0.000 000 000 742 150 3.

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) 0.000 000 000 742 150 3 × 2 = 0 + 0.000 000 001 484 300 6;
  • 2) 0.000 000 001 484 300 6 × 2 = 0 + 0.000 000 002 968 601 2;
  • 3) 0.000 000 002 968 601 2 × 2 = 0 + 0.000 000 005 937 202 4;
  • 4) 0.000 000 005 937 202 4 × 2 = 0 + 0.000 000 011 874 404 8;
  • 5) 0.000 000 011 874 404 8 × 2 = 0 + 0.000 000 023 748 809 6;
  • 6) 0.000 000 023 748 809 6 × 2 = 0 + 0.000 000 047 497 619 2;
  • 7) 0.000 000 047 497 619 2 × 2 = 0 + 0.000 000 094 995 238 4;
  • 8) 0.000 000 094 995 238 4 × 2 = 0 + 0.000 000 189 990 476 8;
  • 9) 0.000 000 189 990 476 8 × 2 = 0 + 0.000 000 379 980 953 6;
  • 10) 0.000 000 379 980 953 6 × 2 = 0 + 0.000 000 759 961 907 2;
  • 11) 0.000 000 759 961 907 2 × 2 = 0 + 0.000 001 519 923 814 4;
  • 12) 0.000 001 519 923 814 4 × 2 = 0 + 0.000 003 039 847 628 8;
  • 13) 0.000 003 039 847 628 8 × 2 = 0 + 0.000 006 079 695 257 6;
  • 14) 0.000 006 079 695 257 6 × 2 = 0 + 0.000 012 159 390 515 2;
  • 15) 0.000 012 159 390 515 2 × 2 = 0 + 0.000 024 318 781 030 4;
  • 16) 0.000 024 318 781 030 4 × 2 = 0 + 0.000 048 637 562 060 8;
  • 17) 0.000 048 637 562 060 8 × 2 = 0 + 0.000 097 275 124 121 6;
  • 18) 0.000 097 275 124 121 6 × 2 = 0 + 0.000 194 550 248 243 2;
  • 19) 0.000 194 550 248 243 2 × 2 = 0 + 0.000 389 100 496 486 4;
  • 20) 0.000 389 100 496 486 4 × 2 = 0 + 0.000 778 200 992 972 8;
  • 21) 0.000 778 200 992 972 8 × 2 = 0 + 0.001 556 401 985 945 6;
  • 22) 0.001 556 401 985 945 6 × 2 = 0 + 0.003 112 803 971 891 2;
  • 23) 0.003 112 803 971 891 2 × 2 = 0 + 0.006 225 607 943 782 4;
  • 24) 0.006 225 607 943 782 4 × 2 = 0 + 0.012 451 215 887 564 8;
  • 25) 0.012 451 215 887 564 8 × 2 = 0 + 0.024 902 431 775 129 6;
  • 26) 0.024 902 431 775 129 6 × 2 = 0 + 0.049 804 863 550 259 2;
  • 27) 0.049 804 863 550 259 2 × 2 = 0 + 0.099 609 727 100 518 4;
  • 28) 0.099 609 727 100 518 4 × 2 = 0 + 0.199 219 454 201 036 8;
  • 29) 0.199 219 454 201 036 8 × 2 = 0 + 0.398 438 908 402 073 6;
  • 30) 0.398 438 908 402 073 6 × 2 = 0 + 0.796 877 816 804 147 2;
  • 31) 0.796 877 816 804 147 2 × 2 = 1 + 0.593 755 633 608 294 4;
  • 32) 0.593 755 633 608 294 4 × 2 = 1 + 0.187 511 267 216 588 8;
  • 33) 0.187 511 267 216 588 8 × 2 = 0 + 0.375 022 534 433 177 6;
  • 34) 0.375 022 534 433 177 6 × 2 = 0 + 0.750 045 068 866 355 2;
  • 35) 0.750 045 068 866 355 2 × 2 = 1 + 0.500 090 137 732 710 4;
  • 36) 0.500 090 137 732 710 4 × 2 = 1 + 0.000 180 275 465 420 8;
  • 37) 0.000 180 275 465 420 8 × 2 = 0 + 0.000 360 550 930 841 6;
  • 38) 0.000 360 550 930 841 6 × 2 = 0 + 0.000 721 101 861 683 2;
  • 39) 0.000 721 101 861 683 2 × 2 = 0 + 0.001 442 203 723 366 4;
  • 40) 0.001 442 203 723 366 4 × 2 = 0 + 0.002 884 407 446 732 8;
  • 41) 0.002 884 407 446 732 8 × 2 = 0 + 0.005 768 814 893 465 6;
  • 42) 0.005 768 814 893 465 6 × 2 = 0 + 0.011 537 629 786 931 2;
  • 43) 0.011 537 629 786 931 2 × 2 = 0 + 0.023 075 259 573 862 4;
  • 44) 0.023 075 259 573 862 4 × 2 = 0 + 0.046 150 519 147 724 8;
  • 45) 0.046 150 519 147 724 8 × 2 = 0 + 0.092 301 038 295 449 6;
  • 46) 0.092 301 038 295 449 6 × 2 = 0 + 0.184 602 076 590 899 2;
  • 47) 0.184 602 076 590 899 2 × 2 = 0 + 0.369 204 153 181 798 4;
  • 48) 0.369 204 153 181 798 4 × 2 = 0 + 0.738 408 306 363 596 8;
  • 49) 0.738 408 306 363 596 8 × 2 = 1 + 0.476 816 612 727 193 6;
  • 50) 0.476 816 612 727 193 6 × 2 = 0 + 0.953 633 225 454 387 2;
  • 51) 0.953 633 225 454 387 2 × 2 = 1 + 0.907 266 450 908 774 4;
  • 52) 0.907 266 450 908 774 4 × 2 = 1 + 0.814 532 901 817 548 8;
  • 53) 0.814 532 901 817 548 8 × 2 = 1 + 0.629 065 803 635 097 6;
  • 54) 0.629 065 803 635 097 6 × 2 = 1 + 0.258 131 607 270 195 2;

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:


0.000 000 000 742 150 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1011 11(2)

6. Positive number before normalization:

0.000 000 000 742 150 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1011 11(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:


0.000 000 000 742 150 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1011 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1011 11(2) × 20 =

1.1001 1000 0000 0000 0101 111(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 0000 0101 111


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 0000 0010 1111 =

100 1100 0000 0000 0010 1111


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 0000 0010 1111


Decimal number -0.000 000 000 742 150 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0010 1111

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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