-0.000 000 000 742 147 676 650 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 650 6(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
-0.000 000 000 742 147 676 650 6(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:

|-0.000 000 000 742 147 676 650 6| = 0.000 000 000 742 147 676 650 6


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: 0.000 000 000 742 147 676 650 6.

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) 0.000 000 000 742 147 676 650 6 × 2 = 0 + 0.000 000 001 484 295 353 301 2;
  • 2) 0.000 000 001 484 295 353 301 2 × 2 = 0 + 0.000 000 002 968 590 706 602 4;
  • 3) 0.000 000 002 968 590 706 602 4 × 2 = 0 + 0.000 000 005 937 181 413 204 8;
  • 4) 0.000 000 005 937 181 413 204 8 × 2 = 0 + 0.000 000 011 874 362 826 409 6;
  • 5) 0.000 000 011 874 362 826 409 6 × 2 = 0 + 0.000 000 023 748 725 652 819 2;
  • 6) 0.000 000 023 748 725 652 819 2 × 2 = 0 + 0.000 000 047 497 451 305 638 4;
  • 7) 0.000 000 047 497 451 305 638 4 × 2 = 0 + 0.000 000 094 994 902 611 276 8;
  • 8) 0.000 000 094 994 902 611 276 8 × 2 = 0 + 0.000 000 189 989 805 222 553 6;
  • 9) 0.000 000 189 989 805 222 553 6 × 2 = 0 + 0.000 000 379 979 610 445 107 2;
  • 10) 0.000 000 379 979 610 445 107 2 × 2 = 0 + 0.000 000 759 959 220 890 214 4;
  • 11) 0.000 000 759 959 220 890 214 4 × 2 = 0 + 0.000 001 519 918 441 780 428 8;
  • 12) 0.000 001 519 918 441 780 428 8 × 2 = 0 + 0.000 003 039 836 883 560 857 6;
  • 13) 0.000 003 039 836 883 560 857 6 × 2 = 0 + 0.000 006 079 673 767 121 715 2;
  • 14) 0.000 006 079 673 767 121 715 2 × 2 = 0 + 0.000 012 159 347 534 243 430 4;
  • 15) 0.000 012 159 347 534 243 430 4 × 2 = 0 + 0.000 024 318 695 068 486 860 8;
  • 16) 0.000 024 318 695 068 486 860 8 × 2 = 0 + 0.000 048 637 390 136 973 721 6;
  • 17) 0.000 048 637 390 136 973 721 6 × 2 = 0 + 0.000 097 274 780 273 947 443 2;
  • 18) 0.000 097 274 780 273 947 443 2 × 2 = 0 + 0.000 194 549 560 547 894 886 4;
  • 19) 0.000 194 549 560 547 894 886 4 × 2 = 0 + 0.000 389 099 121 095 789 772 8;
  • 20) 0.000 389 099 121 095 789 772 8 × 2 = 0 + 0.000 778 198 242 191 579 545 6;
  • 21) 0.000 778 198 242 191 579 545 6 × 2 = 0 + 0.001 556 396 484 383 159 091 2;
  • 22) 0.001 556 396 484 383 159 091 2 × 2 = 0 + 0.003 112 792 968 766 318 182 4;
  • 23) 0.003 112 792 968 766 318 182 4 × 2 = 0 + 0.006 225 585 937 532 636 364 8;
  • 24) 0.006 225 585 937 532 636 364 8 × 2 = 0 + 0.012 451 171 875 065 272 729 6;
  • 25) 0.012 451 171 875 065 272 729 6 × 2 = 0 + 0.024 902 343 750 130 545 459 2;
  • 26) 0.024 902 343 750 130 545 459 2 × 2 = 0 + 0.049 804 687 500 261 090 918 4;
  • 27) 0.049 804 687 500 261 090 918 4 × 2 = 0 + 0.099 609 375 000 522 181 836 8;
  • 28) 0.099 609 375 000 522 181 836 8 × 2 = 0 + 0.199 218 750 001 044 363 673 6;
  • 29) 0.199 218 750 001 044 363 673 6 × 2 = 0 + 0.398 437 500 002 088 727 347 2;
  • 30) 0.398 437 500 002 088 727 347 2 × 2 = 0 + 0.796 875 000 004 177 454 694 4;
  • 31) 0.796 875 000 004 177 454 694 4 × 2 = 1 + 0.593 750 000 008 354 909 388 8;
  • 32) 0.593 750 000 008 354 909 388 8 × 2 = 1 + 0.187 500 000 016 709 818 777 6;
  • 33) 0.187 500 000 016 709 818 777 6 × 2 = 0 + 0.375 000 000 033 419 637 555 2;
  • 34) 0.375 000 000 033 419 637 555 2 × 2 = 0 + 0.750 000 000 066 839 275 110 4;
  • 35) 0.750 000 000 066 839 275 110 4 × 2 = 1 + 0.500 000 000 133 678 550 220 8;
  • 36) 0.500 000 000 133 678 550 220 8 × 2 = 1 + 0.000 000 000 267 357 100 441 6;
  • 37) 0.000 000 000 267 357 100 441 6 × 2 = 0 + 0.000 000 000 534 714 200 883 2;
  • 38) 0.000 000 000 534 714 200 883 2 × 2 = 0 + 0.000 000 001 069 428 401 766 4;
  • 39) 0.000 000 001 069 428 401 766 4 × 2 = 0 + 0.000 000 002 138 856 803 532 8;
  • 40) 0.000 000 002 138 856 803 532 8 × 2 = 0 + 0.000 000 004 277 713 607 065 6;
  • 41) 0.000 000 004 277 713 607 065 6 × 2 = 0 + 0.000 000 008 555 427 214 131 2;
  • 42) 0.000 000 008 555 427 214 131 2 × 2 = 0 + 0.000 000 017 110 854 428 262 4;
  • 43) 0.000 000 017 110 854 428 262 4 × 2 = 0 + 0.000 000 034 221 708 856 524 8;
  • 44) 0.000 000 034 221 708 856 524 8 × 2 = 0 + 0.000 000 068 443 417 713 049 6;
  • 45) 0.000 000 068 443 417 713 049 6 × 2 = 0 + 0.000 000 136 886 835 426 099 2;
  • 46) 0.000 000 136 886 835 426 099 2 × 2 = 0 + 0.000 000 273 773 670 852 198 4;
  • 47) 0.000 000 273 773 670 852 198 4 × 2 = 0 + 0.000 000 547 547 341 704 396 8;
  • 48) 0.000 000 547 547 341 704 396 8 × 2 = 0 + 0.000 001 095 094 683 408 793 6;
  • 49) 0.000 001 095 094 683 408 793 6 × 2 = 0 + 0.000 002 190 189 366 817 587 2;
  • 50) 0.000 002 190 189 366 817 587 2 × 2 = 0 + 0.000 004 380 378 733 635 174 4;
  • 51) 0.000 004 380 378 733 635 174 4 × 2 = 0 + 0.000 008 760 757 467 270 348 8;
  • 52) 0.000 008 760 757 467 270 348 8 × 2 = 0 + 0.000 017 521 514 934 540 697 6;
  • 53) 0.000 017 521 514 934 540 697 6 × 2 = 0 + 0.000 035 043 029 869 081 395 2;
  • 54) 0.000 035 043 029 869 081 395 2 × 2 = 0 + 0.000 070 086 059 738 162 790 4;

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:


0.000 000 000 742 147 676 650 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 650 6(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:


0.000 000 000 742 147 676 650 6(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 -0.000 000 000 742 147 676 650 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 644 2 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: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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