-0.000 000 000 742 119 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 119 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
-0.000 000 000 742 119 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:

|-0.000 000 000 742 119 3| = 0.000 000 000 742 119 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: 0.000 000 000 742 119 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) 0.000 000 000 742 119 3 × 2 = 0 + 0.000 000 001 484 238 6;
  • 2) 0.000 000 001 484 238 6 × 2 = 0 + 0.000 000 002 968 477 2;
  • 3) 0.000 000 002 968 477 2 × 2 = 0 + 0.000 000 005 936 954 4;
  • 4) 0.000 000 005 936 954 4 × 2 = 0 + 0.000 000 011 873 908 8;
  • 5) 0.000 000 011 873 908 8 × 2 = 0 + 0.000 000 023 747 817 6;
  • 6) 0.000 000 023 747 817 6 × 2 = 0 + 0.000 000 047 495 635 2;
  • 7) 0.000 000 047 495 635 2 × 2 = 0 + 0.000 000 094 991 270 4;
  • 8) 0.000 000 094 991 270 4 × 2 = 0 + 0.000 000 189 982 540 8;
  • 9) 0.000 000 189 982 540 8 × 2 = 0 + 0.000 000 379 965 081 6;
  • 10) 0.000 000 379 965 081 6 × 2 = 0 + 0.000 000 759 930 163 2;
  • 11) 0.000 000 759 930 163 2 × 2 = 0 + 0.000 001 519 860 326 4;
  • 12) 0.000 001 519 860 326 4 × 2 = 0 + 0.000 003 039 720 652 8;
  • 13) 0.000 003 039 720 652 8 × 2 = 0 + 0.000 006 079 441 305 6;
  • 14) 0.000 006 079 441 305 6 × 2 = 0 + 0.000 012 158 882 611 2;
  • 15) 0.000 012 158 882 611 2 × 2 = 0 + 0.000 024 317 765 222 4;
  • 16) 0.000 024 317 765 222 4 × 2 = 0 + 0.000 048 635 530 444 8;
  • 17) 0.000 048 635 530 444 8 × 2 = 0 + 0.000 097 271 060 889 6;
  • 18) 0.000 097 271 060 889 6 × 2 = 0 + 0.000 194 542 121 779 2;
  • 19) 0.000 194 542 121 779 2 × 2 = 0 + 0.000 389 084 243 558 4;
  • 20) 0.000 389 084 243 558 4 × 2 = 0 + 0.000 778 168 487 116 8;
  • 21) 0.000 778 168 487 116 8 × 2 = 0 + 0.001 556 336 974 233 6;
  • 22) 0.001 556 336 974 233 6 × 2 = 0 + 0.003 112 673 948 467 2;
  • 23) 0.003 112 673 948 467 2 × 2 = 0 + 0.006 225 347 896 934 4;
  • 24) 0.006 225 347 896 934 4 × 2 = 0 + 0.012 450 695 793 868 8;
  • 25) 0.012 450 695 793 868 8 × 2 = 0 + 0.024 901 391 587 737 6;
  • 26) 0.024 901 391 587 737 6 × 2 = 0 + 0.049 802 783 175 475 2;
  • 27) 0.049 802 783 175 475 2 × 2 = 0 + 0.099 605 566 350 950 4;
  • 28) 0.099 605 566 350 950 4 × 2 = 0 + 0.199 211 132 701 900 8;
  • 29) 0.199 211 132 701 900 8 × 2 = 0 + 0.398 422 265 403 801 6;
  • 30) 0.398 422 265 403 801 6 × 2 = 0 + 0.796 844 530 807 603 2;
  • 31) 0.796 844 530 807 603 2 × 2 = 1 + 0.593 689 061 615 206 4;
  • 32) 0.593 689 061 615 206 4 × 2 = 1 + 0.187 378 123 230 412 8;
  • 33) 0.187 378 123 230 412 8 × 2 = 0 + 0.374 756 246 460 825 6;
  • 34) 0.374 756 246 460 825 6 × 2 = 0 + 0.749 512 492 921 651 2;
  • 35) 0.749 512 492 921 651 2 × 2 = 1 + 0.499 024 985 843 302 4;
  • 36) 0.499 024 985 843 302 4 × 2 = 0 + 0.998 049 971 686 604 8;
  • 37) 0.998 049 971 686 604 8 × 2 = 1 + 0.996 099 943 373 209 6;
  • 38) 0.996 099 943 373 209 6 × 2 = 1 + 0.992 199 886 746 419 2;
  • 39) 0.992 199 886 746 419 2 × 2 = 1 + 0.984 399 773 492 838 4;
  • 40) 0.984 399 773 492 838 4 × 2 = 1 + 0.968 799 546 985 676 8;
  • 41) 0.968 799 546 985 676 8 × 2 = 1 + 0.937 599 093 971 353 6;
  • 42) 0.937 599 093 971 353 6 × 2 = 1 + 0.875 198 187 942 707 2;
  • 43) 0.875 198 187 942 707 2 × 2 = 1 + 0.750 396 375 885 414 4;
  • 44) 0.750 396 375 885 414 4 × 2 = 1 + 0.500 792 751 770 828 8;
  • 45) 0.500 792 751 770 828 8 × 2 = 1 + 0.001 585 503 541 657 6;
  • 46) 0.001 585 503 541 657 6 × 2 = 0 + 0.003 171 007 083 315 2;
  • 47) 0.003 171 007 083 315 2 × 2 = 0 + 0.006 342 014 166 630 4;
  • 48) 0.006 342 014 166 630 4 × 2 = 0 + 0.012 684 028 333 260 8;
  • 49) 0.012 684 028 333 260 8 × 2 = 0 + 0.025 368 056 666 521 6;
  • 50) 0.025 368 056 666 521 6 × 2 = 0 + 0.050 736 113 333 043 2;
  • 51) 0.050 736 113 333 043 2 × 2 = 0 + 0.101 472 226 666 086 4;
  • 52) 0.101 472 226 666 086 4 × 2 = 0 + 0.202 944 453 332 172 8;
  • 53) 0.202 944 453 332 172 8 × 2 = 0 + 0.405 888 906 664 345 6;
  • 54) 0.405 888 906 664 345 6 × 2 = 0 + 0.811 777 813 328 691 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:


0.000 000 000 742 119 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 119 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0000 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:


0.000 000 000 742 119 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 0000 00(2) × 20 =

1.1001 0111 1111 1100 0000 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):
1.1001 0111 1111 1100 0000 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. 100 1011 1111 1110 0000 0000 =

100 1011 1111 1110 0000 0000


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 1111 1110 0000 0000


Decimal number -0.000 000 000 742 119 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 0000 0000

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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