-0.000 000 000 742 147 688 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 688 1(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
-0.000 000 000 742 147 688 1(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:

|-0.000 000 000 742 147 688 1| = 0.000 000 000 742 147 688 1


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: 0.000 000 000 742 147 688 1.

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) 0.000 000 000 742 147 688 1 × 2 = 0 + 0.000 000 001 484 295 376 2;
  • 2) 0.000 000 001 484 295 376 2 × 2 = 0 + 0.000 000 002 968 590 752 4;
  • 3) 0.000 000 002 968 590 752 4 × 2 = 0 + 0.000 000 005 937 181 504 8;
  • 4) 0.000 000 005 937 181 504 8 × 2 = 0 + 0.000 000 011 874 363 009 6;
  • 5) 0.000 000 011 874 363 009 6 × 2 = 0 + 0.000 000 023 748 726 019 2;
  • 6) 0.000 000 023 748 726 019 2 × 2 = 0 + 0.000 000 047 497 452 038 4;
  • 7) 0.000 000 047 497 452 038 4 × 2 = 0 + 0.000 000 094 994 904 076 8;
  • 8) 0.000 000 094 994 904 076 8 × 2 = 0 + 0.000 000 189 989 808 153 6;
  • 9) 0.000 000 189 989 808 153 6 × 2 = 0 + 0.000 000 379 979 616 307 2;
  • 10) 0.000 000 379 979 616 307 2 × 2 = 0 + 0.000 000 759 959 232 614 4;
  • 11) 0.000 000 759 959 232 614 4 × 2 = 0 + 0.000 001 519 918 465 228 8;
  • 12) 0.000 001 519 918 465 228 8 × 2 = 0 + 0.000 003 039 836 930 457 6;
  • 13) 0.000 003 039 836 930 457 6 × 2 = 0 + 0.000 006 079 673 860 915 2;
  • 14) 0.000 006 079 673 860 915 2 × 2 = 0 + 0.000 012 159 347 721 830 4;
  • 15) 0.000 012 159 347 721 830 4 × 2 = 0 + 0.000 024 318 695 443 660 8;
  • 16) 0.000 024 318 695 443 660 8 × 2 = 0 + 0.000 048 637 390 887 321 6;
  • 17) 0.000 048 637 390 887 321 6 × 2 = 0 + 0.000 097 274 781 774 643 2;
  • 18) 0.000 097 274 781 774 643 2 × 2 = 0 + 0.000 194 549 563 549 286 4;
  • 19) 0.000 194 549 563 549 286 4 × 2 = 0 + 0.000 389 099 127 098 572 8;
  • 20) 0.000 389 099 127 098 572 8 × 2 = 0 + 0.000 778 198 254 197 145 6;
  • 21) 0.000 778 198 254 197 145 6 × 2 = 0 + 0.001 556 396 508 394 291 2;
  • 22) 0.001 556 396 508 394 291 2 × 2 = 0 + 0.003 112 793 016 788 582 4;
  • 23) 0.003 112 793 016 788 582 4 × 2 = 0 + 0.006 225 586 033 577 164 8;
  • 24) 0.006 225 586 033 577 164 8 × 2 = 0 + 0.012 451 172 067 154 329 6;
  • 25) 0.012 451 172 067 154 329 6 × 2 = 0 + 0.024 902 344 134 308 659 2;
  • 26) 0.024 902 344 134 308 659 2 × 2 = 0 + 0.049 804 688 268 617 318 4;
  • 27) 0.049 804 688 268 617 318 4 × 2 = 0 + 0.099 609 376 537 234 636 8;
  • 28) 0.099 609 376 537 234 636 8 × 2 = 0 + 0.199 218 753 074 469 273 6;
  • 29) 0.199 218 753 074 469 273 6 × 2 = 0 + 0.398 437 506 148 938 547 2;
  • 30) 0.398 437 506 148 938 547 2 × 2 = 0 + 0.796 875 012 297 877 094 4;
  • 31) 0.796 875 012 297 877 094 4 × 2 = 1 + 0.593 750 024 595 754 188 8;
  • 32) 0.593 750 024 595 754 188 8 × 2 = 1 + 0.187 500 049 191 508 377 6;
  • 33) 0.187 500 049 191 508 377 6 × 2 = 0 + 0.375 000 098 383 016 755 2;
  • 34) 0.375 000 098 383 016 755 2 × 2 = 0 + 0.750 000 196 766 033 510 4;
  • 35) 0.750 000 196 766 033 510 4 × 2 = 1 + 0.500 000 393 532 067 020 8;
  • 36) 0.500 000 393 532 067 020 8 × 2 = 1 + 0.000 000 787 064 134 041 6;
  • 37) 0.000 000 787 064 134 041 6 × 2 = 0 + 0.000 001 574 128 268 083 2;
  • 38) 0.000 001 574 128 268 083 2 × 2 = 0 + 0.000 003 148 256 536 166 4;
  • 39) 0.000 003 148 256 536 166 4 × 2 = 0 + 0.000 006 296 513 072 332 8;
  • 40) 0.000 006 296 513 072 332 8 × 2 = 0 + 0.000 012 593 026 144 665 6;
  • 41) 0.000 012 593 026 144 665 6 × 2 = 0 + 0.000 025 186 052 289 331 2;
  • 42) 0.000 025 186 052 289 331 2 × 2 = 0 + 0.000 050 372 104 578 662 4;
  • 43) 0.000 050 372 104 578 662 4 × 2 = 0 + 0.000 100 744 209 157 324 8;
  • 44) 0.000 100 744 209 157 324 8 × 2 = 0 + 0.000 201 488 418 314 649 6;
  • 45) 0.000 201 488 418 314 649 6 × 2 = 0 + 0.000 402 976 836 629 299 2;
  • 46) 0.000 402 976 836 629 299 2 × 2 = 0 + 0.000 805 953 673 258 598 4;
  • 47) 0.000 805 953 673 258 598 4 × 2 = 0 + 0.001 611 907 346 517 196 8;
  • 48) 0.001 611 907 346 517 196 8 × 2 = 0 + 0.003 223 814 693 034 393 6;
  • 49) 0.003 223 814 693 034 393 6 × 2 = 0 + 0.006 447 629 386 068 787 2;
  • 50) 0.006 447 629 386 068 787 2 × 2 = 0 + 0.012 895 258 772 137 574 4;
  • 51) 0.012 895 258 772 137 574 4 × 2 = 0 + 0.025 790 517 544 275 148 8;
  • 52) 0.025 790 517 544 275 148 8 × 2 = 0 + 0.051 581 035 088 550 297 6;
  • 53) 0.051 581 035 088 550 297 6 × 2 = 0 + 0.103 162 070 177 100 595 2;
  • 54) 0.103 162 070 177 100 595 2 × 2 = 0 + 0.206 324 140 354 201 190 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:


0.000 000 000 742 147 688 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 688 1(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:


0.000 000 000 742 147 688 1(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 -0.000 000 000 742 147 688 1 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