-0.000 000 000 742 147 676 542 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 542(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
-0.000 000 000 742 147 676 542(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:

|-0.000 000 000 742 147 676 542| = 0.000 000 000 742 147 676 542


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: 0.000 000 000 742 147 676 542.

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) 0.000 000 000 742 147 676 542 × 2 = 0 + 0.000 000 001 484 295 353 084;
  • 2) 0.000 000 001 484 295 353 084 × 2 = 0 + 0.000 000 002 968 590 706 168;
  • 3) 0.000 000 002 968 590 706 168 × 2 = 0 + 0.000 000 005 937 181 412 336;
  • 4) 0.000 000 005 937 181 412 336 × 2 = 0 + 0.000 000 011 874 362 824 672;
  • 5) 0.000 000 011 874 362 824 672 × 2 = 0 + 0.000 000 023 748 725 649 344;
  • 6) 0.000 000 023 748 725 649 344 × 2 = 0 + 0.000 000 047 497 451 298 688;
  • 7) 0.000 000 047 497 451 298 688 × 2 = 0 + 0.000 000 094 994 902 597 376;
  • 8) 0.000 000 094 994 902 597 376 × 2 = 0 + 0.000 000 189 989 805 194 752;
  • 9) 0.000 000 189 989 805 194 752 × 2 = 0 + 0.000 000 379 979 610 389 504;
  • 10) 0.000 000 379 979 610 389 504 × 2 = 0 + 0.000 000 759 959 220 779 008;
  • 11) 0.000 000 759 959 220 779 008 × 2 = 0 + 0.000 001 519 918 441 558 016;
  • 12) 0.000 001 519 918 441 558 016 × 2 = 0 + 0.000 003 039 836 883 116 032;
  • 13) 0.000 003 039 836 883 116 032 × 2 = 0 + 0.000 006 079 673 766 232 064;
  • 14) 0.000 006 079 673 766 232 064 × 2 = 0 + 0.000 012 159 347 532 464 128;
  • 15) 0.000 012 159 347 532 464 128 × 2 = 0 + 0.000 024 318 695 064 928 256;
  • 16) 0.000 024 318 695 064 928 256 × 2 = 0 + 0.000 048 637 390 129 856 512;
  • 17) 0.000 048 637 390 129 856 512 × 2 = 0 + 0.000 097 274 780 259 713 024;
  • 18) 0.000 097 274 780 259 713 024 × 2 = 0 + 0.000 194 549 560 519 426 048;
  • 19) 0.000 194 549 560 519 426 048 × 2 = 0 + 0.000 389 099 121 038 852 096;
  • 20) 0.000 389 099 121 038 852 096 × 2 = 0 + 0.000 778 198 242 077 704 192;
  • 21) 0.000 778 198 242 077 704 192 × 2 = 0 + 0.001 556 396 484 155 408 384;
  • 22) 0.001 556 396 484 155 408 384 × 2 = 0 + 0.003 112 792 968 310 816 768;
  • 23) 0.003 112 792 968 310 816 768 × 2 = 0 + 0.006 225 585 936 621 633 536;
  • 24) 0.006 225 585 936 621 633 536 × 2 = 0 + 0.012 451 171 873 243 267 072;
  • 25) 0.012 451 171 873 243 267 072 × 2 = 0 + 0.024 902 343 746 486 534 144;
  • 26) 0.024 902 343 746 486 534 144 × 2 = 0 + 0.049 804 687 492 973 068 288;
  • 27) 0.049 804 687 492 973 068 288 × 2 = 0 + 0.099 609 374 985 946 136 576;
  • 28) 0.099 609 374 985 946 136 576 × 2 = 0 + 0.199 218 749 971 892 273 152;
  • 29) 0.199 218 749 971 892 273 152 × 2 = 0 + 0.398 437 499 943 784 546 304;
  • 30) 0.398 437 499 943 784 546 304 × 2 = 0 + 0.796 874 999 887 569 092 608;
  • 31) 0.796 874 999 887 569 092 608 × 2 = 1 + 0.593 749 999 775 138 185 216;
  • 32) 0.593 749 999 775 138 185 216 × 2 = 1 + 0.187 499 999 550 276 370 432;
  • 33) 0.187 499 999 550 276 370 432 × 2 = 0 + 0.374 999 999 100 552 740 864;
  • 34) 0.374 999 999 100 552 740 864 × 2 = 0 + 0.749 999 998 201 105 481 728;
  • 35) 0.749 999 998 201 105 481 728 × 2 = 1 + 0.499 999 996 402 210 963 456;
  • 36) 0.499 999 996 402 210 963 456 × 2 = 0 + 0.999 999 992 804 421 926 912;
  • 37) 0.999 999 992 804 421 926 912 × 2 = 1 + 0.999 999 985 608 843 853 824;
  • 38) 0.999 999 985 608 843 853 824 × 2 = 1 + 0.999 999 971 217 687 707 648;
  • 39) 0.999 999 971 217 687 707 648 × 2 = 1 + 0.999 999 942 435 375 415 296;
  • 40) 0.999 999 942 435 375 415 296 × 2 = 1 + 0.999 999 884 870 750 830 592;
  • 41) 0.999 999 884 870 750 830 592 × 2 = 1 + 0.999 999 769 741 501 661 184;
  • 42) 0.999 999 769 741 501 661 184 × 2 = 1 + 0.999 999 539 483 003 322 368;
  • 43) 0.999 999 539 483 003 322 368 × 2 = 1 + 0.999 999 078 966 006 644 736;
  • 44) 0.999 999 078 966 006 644 736 × 2 = 1 + 0.999 998 157 932 013 289 472;
  • 45) 0.999 998 157 932 013 289 472 × 2 = 1 + 0.999 996 315 864 026 578 944;
  • 46) 0.999 996 315 864 026 578 944 × 2 = 1 + 0.999 992 631 728 053 157 888;
  • 47) 0.999 992 631 728 053 157 888 × 2 = 1 + 0.999 985 263 456 106 315 776;
  • 48) 0.999 985 263 456 106 315 776 × 2 = 1 + 0.999 970 526 912 212 631 552;
  • 49) 0.999 970 526 912 212 631 552 × 2 = 1 + 0.999 941 053 824 425 263 104;
  • 50) 0.999 941 053 824 425 263 104 × 2 = 1 + 0.999 882 107 648 850 526 208;
  • 51) 0.999 882 107 648 850 526 208 × 2 = 1 + 0.999 764 215 297 701 052 416;
  • 52) 0.999 764 215 297 701 052 416 × 2 = 1 + 0.999 528 430 595 402 104 832;
  • 53) 0.999 528 430 595 402 104 832 × 2 = 1 + 0.999 056 861 190 804 209 664;
  • 54) 0.999 056 861 190 804 209 664 × 2 = 1 + 0.998 113 722 381 608 419 328;

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:


0.000 000 000 742 147 676 542(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 542(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:


0.000 000 000 742 147 676 542(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 -0.000 000 000 742 147 676 542 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