-0.000 000 000 742 147 676 640 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 640 5(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
-0.000 000 000 742 147 676 640 5(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:

|-0.000 000 000 742 147 676 640 5| = 0.000 000 000 742 147 676 640 5


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: 0.000 000 000 742 147 676 640 5.

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) 0.000 000 000 742 147 676 640 5 × 2 = 0 + 0.000 000 001 484 295 353 281;
  • 2) 0.000 000 001 484 295 353 281 × 2 = 0 + 0.000 000 002 968 590 706 562;
  • 3) 0.000 000 002 968 590 706 562 × 2 = 0 + 0.000 000 005 937 181 413 124;
  • 4) 0.000 000 005 937 181 413 124 × 2 = 0 + 0.000 000 011 874 362 826 248;
  • 5) 0.000 000 011 874 362 826 248 × 2 = 0 + 0.000 000 023 748 725 652 496;
  • 6) 0.000 000 023 748 725 652 496 × 2 = 0 + 0.000 000 047 497 451 304 992;
  • 7) 0.000 000 047 497 451 304 992 × 2 = 0 + 0.000 000 094 994 902 609 984;
  • 8) 0.000 000 094 994 902 609 984 × 2 = 0 + 0.000 000 189 989 805 219 968;
  • 9) 0.000 000 189 989 805 219 968 × 2 = 0 + 0.000 000 379 979 610 439 936;
  • 10) 0.000 000 379 979 610 439 936 × 2 = 0 + 0.000 000 759 959 220 879 872;
  • 11) 0.000 000 759 959 220 879 872 × 2 = 0 + 0.000 001 519 918 441 759 744;
  • 12) 0.000 001 519 918 441 759 744 × 2 = 0 + 0.000 003 039 836 883 519 488;
  • 13) 0.000 003 039 836 883 519 488 × 2 = 0 + 0.000 006 079 673 767 038 976;
  • 14) 0.000 006 079 673 767 038 976 × 2 = 0 + 0.000 012 159 347 534 077 952;
  • 15) 0.000 012 159 347 534 077 952 × 2 = 0 + 0.000 024 318 695 068 155 904;
  • 16) 0.000 024 318 695 068 155 904 × 2 = 0 + 0.000 048 637 390 136 311 808;
  • 17) 0.000 048 637 390 136 311 808 × 2 = 0 + 0.000 097 274 780 272 623 616;
  • 18) 0.000 097 274 780 272 623 616 × 2 = 0 + 0.000 194 549 560 545 247 232;
  • 19) 0.000 194 549 560 545 247 232 × 2 = 0 + 0.000 389 099 121 090 494 464;
  • 20) 0.000 389 099 121 090 494 464 × 2 = 0 + 0.000 778 198 242 180 988 928;
  • 21) 0.000 778 198 242 180 988 928 × 2 = 0 + 0.001 556 396 484 361 977 856;
  • 22) 0.001 556 396 484 361 977 856 × 2 = 0 + 0.003 112 792 968 723 955 712;
  • 23) 0.003 112 792 968 723 955 712 × 2 = 0 + 0.006 225 585 937 447 911 424;
  • 24) 0.006 225 585 937 447 911 424 × 2 = 0 + 0.012 451 171 874 895 822 848;
  • 25) 0.012 451 171 874 895 822 848 × 2 = 0 + 0.024 902 343 749 791 645 696;
  • 26) 0.024 902 343 749 791 645 696 × 2 = 0 + 0.049 804 687 499 583 291 392;
  • 27) 0.049 804 687 499 583 291 392 × 2 = 0 + 0.099 609 374 999 166 582 784;
  • 28) 0.099 609 374 999 166 582 784 × 2 = 0 + 0.199 218 749 998 333 165 568;
  • 29) 0.199 218 749 998 333 165 568 × 2 = 0 + 0.398 437 499 996 666 331 136;
  • 30) 0.398 437 499 996 666 331 136 × 2 = 0 + 0.796 874 999 993 332 662 272;
  • 31) 0.796 874 999 993 332 662 272 × 2 = 1 + 0.593 749 999 986 665 324 544;
  • 32) 0.593 749 999 986 665 324 544 × 2 = 1 + 0.187 499 999 973 330 649 088;
  • 33) 0.187 499 999 973 330 649 088 × 2 = 0 + 0.374 999 999 946 661 298 176;
  • 34) 0.374 999 999 946 661 298 176 × 2 = 0 + 0.749 999 999 893 322 596 352;
  • 35) 0.749 999 999 893 322 596 352 × 2 = 1 + 0.499 999 999 786 645 192 704;
  • 36) 0.499 999 999 786 645 192 704 × 2 = 0 + 0.999 999 999 573 290 385 408;
  • 37) 0.999 999 999 573 290 385 408 × 2 = 1 + 0.999 999 999 146 580 770 816;
  • 38) 0.999 999 999 146 580 770 816 × 2 = 1 + 0.999 999 998 293 161 541 632;
  • 39) 0.999 999 998 293 161 541 632 × 2 = 1 + 0.999 999 996 586 323 083 264;
  • 40) 0.999 999 996 586 323 083 264 × 2 = 1 + 0.999 999 993 172 646 166 528;
  • 41) 0.999 999 993 172 646 166 528 × 2 = 1 + 0.999 999 986 345 292 333 056;
  • 42) 0.999 999 986 345 292 333 056 × 2 = 1 + 0.999 999 972 690 584 666 112;
  • 43) 0.999 999 972 690 584 666 112 × 2 = 1 + 0.999 999 945 381 169 332 224;
  • 44) 0.999 999 945 381 169 332 224 × 2 = 1 + 0.999 999 890 762 338 664 448;
  • 45) 0.999 999 890 762 338 664 448 × 2 = 1 + 0.999 999 781 524 677 328 896;
  • 46) 0.999 999 781 524 677 328 896 × 2 = 1 + 0.999 999 563 049 354 657 792;
  • 47) 0.999 999 563 049 354 657 792 × 2 = 1 + 0.999 999 126 098 709 315 584;
  • 48) 0.999 999 126 098 709 315 584 × 2 = 1 + 0.999 998 252 197 418 631 168;
  • 49) 0.999 998 252 197 418 631 168 × 2 = 1 + 0.999 996 504 394 837 262 336;
  • 50) 0.999 996 504 394 837 262 336 × 2 = 1 + 0.999 993 008 789 674 524 672;
  • 51) 0.999 993 008 789 674 524 672 × 2 = 1 + 0.999 986 017 579 349 049 344;
  • 52) 0.999 986 017 579 349 049 344 × 2 = 1 + 0.999 972 035 158 698 098 688;
  • 53) 0.999 972 035 158 698 098 688 × 2 = 1 + 0.999 944 070 317 396 197 376;
  • 54) 0.999 944 070 317 396 197 376 × 2 = 1 + 0.999 888 140 634 792 394 752;

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:


0.000 000 000 742 147 676 640 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 640 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 640 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 1111 111


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 1111 1111 =

100 1011 1111 1111 1111 1111


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 1111 1111


Decimal number -0.000 000 000 742 147 676 640 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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