-0.000 000 000 742 147 678 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 678 4(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
-0.000 000 000 742 147 678 4(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:

|-0.000 000 000 742 147 678 4| = 0.000 000 000 742 147 678 4


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: 0.000 000 000 742 147 678 4.

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) 0.000 000 000 742 147 678 4 × 2 = 0 + 0.000 000 001 484 295 356 8;
  • 2) 0.000 000 001 484 295 356 8 × 2 = 0 + 0.000 000 002 968 590 713 6;
  • 3) 0.000 000 002 968 590 713 6 × 2 = 0 + 0.000 000 005 937 181 427 2;
  • 4) 0.000 000 005 937 181 427 2 × 2 = 0 + 0.000 000 011 874 362 854 4;
  • 5) 0.000 000 011 874 362 854 4 × 2 = 0 + 0.000 000 023 748 725 708 8;
  • 6) 0.000 000 023 748 725 708 8 × 2 = 0 + 0.000 000 047 497 451 417 6;
  • 7) 0.000 000 047 497 451 417 6 × 2 = 0 + 0.000 000 094 994 902 835 2;
  • 8) 0.000 000 094 994 902 835 2 × 2 = 0 + 0.000 000 189 989 805 670 4;
  • 9) 0.000 000 189 989 805 670 4 × 2 = 0 + 0.000 000 379 979 611 340 8;
  • 10) 0.000 000 379 979 611 340 8 × 2 = 0 + 0.000 000 759 959 222 681 6;
  • 11) 0.000 000 759 959 222 681 6 × 2 = 0 + 0.000 001 519 918 445 363 2;
  • 12) 0.000 001 519 918 445 363 2 × 2 = 0 + 0.000 003 039 836 890 726 4;
  • 13) 0.000 003 039 836 890 726 4 × 2 = 0 + 0.000 006 079 673 781 452 8;
  • 14) 0.000 006 079 673 781 452 8 × 2 = 0 + 0.000 012 159 347 562 905 6;
  • 15) 0.000 012 159 347 562 905 6 × 2 = 0 + 0.000 024 318 695 125 811 2;
  • 16) 0.000 024 318 695 125 811 2 × 2 = 0 + 0.000 048 637 390 251 622 4;
  • 17) 0.000 048 637 390 251 622 4 × 2 = 0 + 0.000 097 274 780 503 244 8;
  • 18) 0.000 097 274 780 503 244 8 × 2 = 0 + 0.000 194 549 561 006 489 6;
  • 19) 0.000 194 549 561 006 489 6 × 2 = 0 + 0.000 389 099 122 012 979 2;
  • 20) 0.000 389 099 122 012 979 2 × 2 = 0 + 0.000 778 198 244 025 958 4;
  • 21) 0.000 778 198 244 025 958 4 × 2 = 0 + 0.001 556 396 488 051 916 8;
  • 22) 0.001 556 396 488 051 916 8 × 2 = 0 + 0.003 112 792 976 103 833 6;
  • 23) 0.003 112 792 976 103 833 6 × 2 = 0 + 0.006 225 585 952 207 667 2;
  • 24) 0.006 225 585 952 207 667 2 × 2 = 0 + 0.012 451 171 904 415 334 4;
  • 25) 0.012 451 171 904 415 334 4 × 2 = 0 + 0.024 902 343 808 830 668 8;
  • 26) 0.024 902 343 808 830 668 8 × 2 = 0 + 0.049 804 687 617 661 337 6;
  • 27) 0.049 804 687 617 661 337 6 × 2 = 0 + 0.099 609 375 235 322 675 2;
  • 28) 0.099 609 375 235 322 675 2 × 2 = 0 + 0.199 218 750 470 645 350 4;
  • 29) 0.199 218 750 470 645 350 4 × 2 = 0 + 0.398 437 500 941 290 700 8;
  • 30) 0.398 437 500 941 290 700 8 × 2 = 0 + 0.796 875 001 882 581 401 6;
  • 31) 0.796 875 001 882 581 401 6 × 2 = 1 + 0.593 750 003 765 162 803 2;
  • 32) 0.593 750 003 765 162 803 2 × 2 = 1 + 0.187 500 007 530 325 606 4;
  • 33) 0.187 500 007 530 325 606 4 × 2 = 0 + 0.375 000 015 060 651 212 8;
  • 34) 0.375 000 015 060 651 212 8 × 2 = 0 + 0.750 000 030 121 302 425 6;
  • 35) 0.750 000 030 121 302 425 6 × 2 = 1 + 0.500 000 060 242 604 851 2;
  • 36) 0.500 000 060 242 604 851 2 × 2 = 1 + 0.000 000 120 485 209 702 4;
  • 37) 0.000 000 120 485 209 702 4 × 2 = 0 + 0.000 000 240 970 419 404 8;
  • 38) 0.000 000 240 970 419 404 8 × 2 = 0 + 0.000 000 481 940 838 809 6;
  • 39) 0.000 000 481 940 838 809 6 × 2 = 0 + 0.000 000 963 881 677 619 2;
  • 40) 0.000 000 963 881 677 619 2 × 2 = 0 + 0.000 001 927 763 355 238 4;
  • 41) 0.000 001 927 763 355 238 4 × 2 = 0 + 0.000 003 855 526 710 476 8;
  • 42) 0.000 003 855 526 710 476 8 × 2 = 0 + 0.000 007 711 053 420 953 6;
  • 43) 0.000 007 711 053 420 953 6 × 2 = 0 + 0.000 015 422 106 841 907 2;
  • 44) 0.000 015 422 106 841 907 2 × 2 = 0 + 0.000 030 844 213 683 814 4;
  • 45) 0.000 030 844 213 683 814 4 × 2 = 0 + 0.000 061 688 427 367 628 8;
  • 46) 0.000 061 688 427 367 628 8 × 2 = 0 + 0.000 123 376 854 735 257 6;
  • 47) 0.000 123 376 854 735 257 6 × 2 = 0 + 0.000 246 753 709 470 515 2;
  • 48) 0.000 246 753 709 470 515 2 × 2 = 0 + 0.000 493 507 418 941 030 4;
  • 49) 0.000 493 507 418 941 030 4 × 2 = 0 + 0.000 987 014 837 882 060 8;
  • 50) 0.000 987 014 837 882 060 8 × 2 = 0 + 0.001 974 029 675 764 121 6;
  • 51) 0.001 974 029 675 764 121 6 × 2 = 0 + 0.003 948 059 351 528 243 2;
  • 52) 0.003 948 059 351 528 243 2 × 2 = 0 + 0.007 896 118 703 056 486 4;
  • 53) 0.007 896 118 703 056 486 4 × 2 = 0 + 0.015 792 237 406 112 972 8;
  • 54) 0.015 792 237 406 112 972 8 × 2 = 0 + 0.031 584 474 812 225 945 6;

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:


0.000 000 000 742 147 678 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 678 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 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:


0.000 000 000 742 147 678 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 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 1000 0000 0000 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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 678 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 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