-0.000 000 000 742 147 676 775 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 775(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
-0.000 000 000 742 147 676 775(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:

|-0.000 000 000 742 147 676 775| = 0.000 000 000 742 147 676 775


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: 0.000 000 000 742 147 676 775.

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) 0.000 000 000 742 147 676 775 × 2 = 0 + 0.000 000 001 484 295 353 55;
  • 2) 0.000 000 001 484 295 353 55 × 2 = 0 + 0.000 000 002 968 590 707 1;
  • 3) 0.000 000 002 968 590 707 1 × 2 = 0 + 0.000 000 005 937 181 414 2;
  • 4) 0.000 000 005 937 181 414 2 × 2 = 0 + 0.000 000 011 874 362 828 4;
  • 5) 0.000 000 011 874 362 828 4 × 2 = 0 + 0.000 000 023 748 725 656 8;
  • 6) 0.000 000 023 748 725 656 8 × 2 = 0 + 0.000 000 047 497 451 313 6;
  • 7) 0.000 000 047 497 451 313 6 × 2 = 0 + 0.000 000 094 994 902 627 2;
  • 8) 0.000 000 094 994 902 627 2 × 2 = 0 + 0.000 000 189 989 805 254 4;
  • 9) 0.000 000 189 989 805 254 4 × 2 = 0 + 0.000 000 379 979 610 508 8;
  • 10) 0.000 000 379 979 610 508 8 × 2 = 0 + 0.000 000 759 959 221 017 6;
  • 11) 0.000 000 759 959 221 017 6 × 2 = 0 + 0.000 001 519 918 442 035 2;
  • 12) 0.000 001 519 918 442 035 2 × 2 = 0 + 0.000 003 039 836 884 070 4;
  • 13) 0.000 003 039 836 884 070 4 × 2 = 0 + 0.000 006 079 673 768 140 8;
  • 14) 0.000 006 079 673 768 140 8 × 2 = 0 + 0.000 012 159 347 536 281 6;
  • 15) 0.000 012 159 347 536 281 6 × 2 = 0 + 0.000 024 318 695 072 563 2;
  • 16) 0.000 024 318 695 072 563 2 × 2 = 0 + 0.000 048 637 390 145 126 4;
  • 17) 0.000 048 637 390 145 126 4 × 2 = 0 + 0.000 097 274 780 290 252 8;
  • 18) 0.000 097 274 780 290 252 8 × 2 = 0 + 0.000 194 549 560 580 505 6;
  • 19) 0.000 194 549 560 580 505 6 × 2 = 0 + 0.000 389 099 121 161 011 2;
  • 20) 0.000 389 099 121 161 011 2 × 2 = 0 + 0.000 778 198 242 322 022 4;
  • 21) 0.000 778 198 242 322 022 4 × 2 = 0 + 0.001 556 396 484 644 044 8;
  • 22) 0.001 556 396 484 644 044 8 × 2 = 0 + 0.003 112 792 969 288 089 6;
  • 23) 0.003 112 792 969 288 089 6 × 2 = 0 + 0.006 225 585 938 576 179 2;
  • 24) 0.006 225 585 938 576 179 2 × 2 = 0 + 0.012 451 171 877 152 358 4;
  • 25) 0.012 451 171 877 152 358 4 × 2 = 0 + 0.024 902 343 754 304 716 8;
  • 26) 0.024 902 343 754 304 716 8 × 2 = 0 + 0.049 804 687 508 609 433 6;
  • 27) 0.049 804 687 508 609 433 6 × 2 = 0 + 0.099 609 375 017 218 867 2;
  • 28) 0.099 609 375 017 218 867 2 × 2 = 0 + 0.199 218 750 034 437 734 4;
  • 29) 0.199 218 750 034 437 734 4 × 2 = 0 + 0.398 437 500 068 875 468 8;
  • 30) 0.398 437 500 068 875 468 8 × 2 = 0 + 0.796 875 000 137 750 937 6;
  • 31) 0.796 875 000 137 750 937 6 × 2 = 1 + 0.593 750 000 275 501 875 2;
  • 32) 0.593 750 000 275 501 875 2 × 2 = 1 + 0.187 500 000 551 003 750 4;
  • 33) 0.187 500 000 551 003 750 4 × 2 = 0 + 0.375 000 001 102 007 500 8;
  • 34) 0.375 000 001 102 007 500 8 × 2 = 0 + 0.750 000 002 204 015 001 6;
  • 35) 0.750 000 002 204 015 001 6 × 2 = 1 + 0.500 000 004 408 030 003 2;
  • 36) 0.500 000 004 408 030 003 2 × 2 = 1 + 0.000 000 008 816 060 006 4;
  • 37) 0.000 000 008 816 060 006 4 × 2 = 0 + 0.000 000 017 632 120 012 8;
  • 38) 0.000 000 017 632 120 012 8 × 2 = 0 + 0.000 000 035 264 240 025 6;
  • 39) 0.000 000 035 264 240 025 6 × 2 = 0 + 0.000 000 070 528 480 051 2;
  • 40) 0.000 000 070 528 480 051 2 × 2 = 0 + 0.000 000 141 056 960 102 4;
  • 41) 0.000 000 141 056 960 102 4 × 2 = 0 + 0.000 000 282 113 920 204 8;
  • 42) 0.000 000 282 113 920 204 8 × 2 = 0 + 0.000 000 564 227 840 409 6;
  • 43) 0.000 000 564 227 840 409 6 × 2 = 0 + 0.000 001 128 455 680 819 2;
  • 44) 0.000 001 128 455 680 819 2 × 2 = 0 + 0.000 002 256 911 361 638 4;
  • 45) 0.000 002 256 911 361 638 4 × 2 = 0 + 0.000 004 513 822 723 276 8;
  • 46) 0.000 004 513 822 723 276 8 × 2 = 0 + 0.000 009 027 645 446 553 6;
  • 47) 0.000 009 027 645 446 553 6 × 2 = 0 + 0.000 018 055 290 893 107 2;
  • 48) 0.000 018 055 290 893 107 2 × 2 = 0 + 0.000 036 110 581 786 214 4;
  • 49) 0.000 036 110 581 786 214 4 × 2 = 0 + 0.000 072 221 163 572 428 8;
  • 50) 0.000 072 221 163 572 428 8 × 2 = 0 + 0.000 144 442 327 144 857 6;
  • 51) 0.000 144 442 327 144 857 6 × 2 = 0 + 0.000 288 884 654 289 715 2;
  • 52) 0.000 288 884 654 289 715 2 × 2 = 0 + 0.000 577 769 308 579 430 4;
  • 53) 0.000 577 769 308 579 430 4 × 2 = 0 + 0.001 155 538 617 158 860 8;
  • 54) 0.001 155 538 617 158 860 8 × 2 = 0 + 0.002 311 077 234 317 721 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:


0.000 000 000 742 147 676 775(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 775(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:


0.000 000 000 742 147 676 775(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 -0.000 000 000 742 147 676 775 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 844 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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