-0.000 000 000 742 143 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 143 7(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
-0.000 000 000 742 143 7(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:

|-0.000 000 000 742 143 7| = 0.000 000 000 742 143 7


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: 0.000 000 000 742 143 7.

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) 0.000 000 000 742 143 7 × 2 = 0 + 0.000 000 001 484 287 4;
  • 2) 0.000 000 001 484 287 4 × 2 = 0 + 0.000 000 002 968 574 8;
  • 3) 0.000 000 002 968 574 8 × 2 = 0 + 0.000 000 005 937 149 6;
  • 4) 0.000 000 005 937 149 6 × 2 = 0 + 0.000 000 011 874 299 2;
  • 5) 0.000 000 011 874 299 2 × 2 = 0 + 0.000 000 023 748 598 4;
  • 6) 0.000 000 023 748 598 4 × 2 = 0 + 0.000 000 047 497 196 8;
  • 7) 0.000 000 047 497 196 8 × 2 = 0 + 0.000 000 094 994 393 6;
  • 8) 0.000 000 094 994 393 6 × 2 = 0 + 0.000 000 189 988 787 2;
  • 9) 0.000 000 189 988 787 2 × 2 = 0 + 0.000 000 379 977 574 4;
  • 10) 0.000 000 379 977 574 4 × 2 = 0 + 0.000 000 759 955 148 8;
  • 11) 0.000 000 759 955 148 8 × 2 = 0 + 0.000 001 519 910 297 6;
  • 12) 0.000 001 519 910 297 6 × 2 = 0 + 0.000 003 039 820 595 2;
  • 13) 0.000 003 039 820 595 2 × 2 = 0 + 0.000 006 079 641 190 4;
  • 14) 0.000 006 079 641 190 4 × 2 = 0 + 0.000 012 159 282 380 8;
  • 15) 0.000 012 159 282 380 8 × 2 = 0 + 0.000 024 318 564 761 6;
  • 16) 0.000 024 318 564 761 6 × 2 = 0 + 0.000 048 637 129 523 2;
  • 17) 0.000 048 637 129 523 2 × 2 = 0 + 0.000 097 274 259 046 4;
  • 18) 0.000 097 274 259 046 4 × 2 = 0 + 0.000 194 548 518 092 8;
  • 19) 0.000 194 548 518 092 8 × 2 = 0 + 0.000 389 097 036 185 6;
  • 20) 0.000 389 097 036 185 6 × 2 = 0 + 0.000 778 194 072 371 2;
  • 21) 0.000 778 194 072 371 2 × 2 = 0 + 0.001 556 388 144 742 4;
  • 22) 0.001 556 388 144 742 4 × 2 = 0 + 0.003 112 776 289 484 8;
  • 23) 0.003 112 776 289 484 8 × 2 = 0 + 0.006 225 552 578 969 6;
  • 24) 0.006 225 552 578 969 6 × 2 = 0 + 0.012 451 105 157 939 2;
  • 25) 0.012 451 105 157 939 2 × 2 = 0 + 0.024 902 210 315 878 4;
  • 26) 0.024 902 210 315 878 4 × 2 = 0 + 0.049 804 420 631 756 8;
  • 27) 0.049 804 420 631 756 8 × 2 = 0 + 0.099 608 841 263 513 6;
  • 28) 0.099 608 841 263 513 6 × 2 = 0 + 0.199 217 682 527 027 2;
  • 29) 0.199 217 682 527 027 2 × 2 = 0 + 0.398 435 365 054 054 4;
  • 30) 0.398 435 365 054 054 4 × 2 = 0 + 0.796 870 730 108 108 8;
  • 31) 0.796 870 730 108 108 8 × 2 = 1 + 0.593 741 460 216 217 6;
  • 32) 0.593 741 460 216 217 6 × 2 = 1 + 0.187 482 920 432 435 2;
  • 33) 0.187 482 920 432 435 2 × 2 = 0 + 0.374 965 840 864 870 4;
  • 34) 0.374 965 840 864 870 4 × 2 = 0 + 0.749 931 681 729 740 8;
  • 35) 0.749 931 681 729 740 8 × 2 = 1 + 0.499 863 363 459 481 6;
  • 36) 0.499 863 363 459 481 6 × 2 = 0 + 0.999 726 726 918 963 2;
  • 37) 0.999 726 726 918 963 2 × 2 = 1 + 0.999 453 453 837 926 4;
  • 38) 0.999 453 453 837 926 4 × 2 = 1 + 0.998 906 907 675 852 8;
  • 39) 0.998 906 907 675 852 8 × 2 = 1 + 0.997 813 815 351 705 6;
  • 40) 0.997 813 815 351 705 6 × 2 = 1 + 0.995 627 630 703 411 2;
  • 41) 0.995 627 630 703 411 2 × 2 = 1 + 0.991 255 261 406 822 4;
  • 42) 0.991 255 261 406 822 4 × 2 = 1 + 0.982 510 522 813 644 8;
  • 43) 0.982 510 522 813 644 8 × 2 = 1 + 0.965 021 045 627 289 6;
  • 44) 0.965 021 045 627 289 6 × 2 = 1 + 0.930 042 091 254 579 2;
  • 45) 0.930 042 091 254 579 2 × 2 = 1 + 0.860 084 182 509 158 4;
  • 46) 0.860 084 182 509 158 4 × 2 = 1 + 0.720 168 365 018 316 8;
  • 47) 0.720 168 365 018 316 8 × 2 = 1 + 0.440 336 730 036 633 6;
  • 48) 0.440 336 730 036 633 6 × 2 = 0 + 0.880 673 460 073 267 2;
  • 49) 0.880 673 460 073 267 2 × 2 = 1 + 0.761 346 920 146 534 4;
  • 50) 0.761 346 920 146 534 4 × 2 = 1 + 0.522 693 840 293 068 8;
  • 51) 0.522 693 840 293 068 8 × 2 = 1 + 0.045 387 680 586 137 6;
  • 52) 0.045 387 680 586 137 6 × 2 = 0 + 0.090 775 361 172 275 2;
  • 53) 0.090 775 361 172 275 2 × 2 = 0 + 0.181 550 722 344 550 4;
  • 54) 0.181 550 722 344 550 4 × 2 = 0 + 0.363 101 444 689 100 8;

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:


0.000 000 000 742 143 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1110 00(2)

6. Positive number before normalization:

0.000 000 000 742 143 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1110 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:


0.000 000 000 742 143 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1110 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1110 1110 00(2) × 20 =

1.1001 0111 1111 1111 0111 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 1111 0111 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 1111 1011 1000 =

100 1011 1111 1111 1011 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) =
100 1011 1111 1111 1011 1000


Decimal number -0.000 000 000 742 143 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1011 1000

Share

» Convert decimal -0.000 000 000 742 147 3 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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