-0.000 000 000 742 147 676 292 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 292(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
-0.000 000 000 742 147 676 292(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:

|-0.000 000 000 742 147 676 292| = 0.000 000 000 742 147 676 292


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: 0.000 000 000 742 147 676 292.

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) 0.000 000 000 742 147 676 292 × 2 = 0 + 0.000 000 001 484 295 352 584;
  • 2) 0.000 000 001 484 295 352 584 × 2 = 0 + 0.000 000 002 968 590 705 168;
  • 3) 0.000 000 002 968 590 705 168 × 2 = 0 + 0.000 000 005 937 181 410 336;
  • 4) 0.000 000 005 937 181 410 336 × 2 = 0 + 0.000 000 011 874 362 820 672;
  • 5) 0.000 000 011 874 362 820 672 × 2 = 0 + 0.000 000 023 748 725 641 344;
  • 6) 0.000 000 023 748 725 641 344 × 2 = 0 + 0.000 000 047 497 451 282 688;
  • 7) 0.000 000 047 497 451 282 688 × 2 = 0 + 0.000 000 094 994 902 565 376;
  • 8) 0.000 000 094 994 902 565 376 × 2 = 0 + 0.000 000 189 989 805 130 752;
  • 9) 0.000 000 189 989 805 130 752 × 2 = 0 + 0.000 000 379 979 610 261 504;
  • 10) 0.000 000 379 979 610 261 504 × 2 = 0 + 0.000 000 759 959 220 523 008;
  • 11) 0.000 000 759 959 220 523 008 × 2 = 0 + 0.000 001 519 918 441 046 016;
  • 12) 0.000 001 519 918 441 046 016 × 2 = 0 + 0.000 003 039 836 882 092 032;
  • 13) 0.000 003 039 836 882 092 032 × 2 = 0 + 0.000 006 079 673 764 184 064;
  • 14) 0.000 006 079 673 764 184 064 × 2 = 0 + 0.000 012 159 347 528 368 128;
  • 15) 0.000 012 159 347 528 368 128 × 2 = 0 + 0.000 024 318 695 056 736 256;
  • 16) 0.000 024 318 695 056 736 256 × 2 = 0 + 0.000 048 637 390 113 472 512;
  • 17) 0.000 048 637 390 113 472 512 × 2 = 0 + 0.000 097 274 780 226 945 024;
  • 18) 0.000 097 274 780 226 945 024 × 2 = 0 + 0.000 194 549 560 453 890 048;
  • 19) 0.000 194 549 560 453 890 048 × 2 = 0 + 0.000 389 099 120 907 780 096;
  • 20) 0.000 389 099 120 907 780 096 × 2 = 0 + 0.000 778 198 241 815 560 192;
  • 21) 0.000 778 198 241 815 560 192 × 2 = 0 + 0.001 556 396 483 631 120 384;
  • 22) 0.001 556 396 483 631 120 384 × 2 = 0 + 0.003 112 792 967 262 240 768;
  • 23) 0.003 112 792 967 262 240 768 × 2 = 0 + 0.006 225 585 934 524 481 536;
  • 24) 0.006 225 585 934 524 481 536 × 2 = 0 + 0.012 451 171 869 048 963 072;
  • 25) 0.012 451 171 869 048 963 072 × 2 = 0 + 0.024 902 343 738 097 926 144;
  • 26) 0.024 902 343 738 097 926 144 × 2 = 0 + 0.049 804 687 476 195 852 288;
  • 27) 0.049 804 687 476 195 852 288 × 2 = 0 + 0.099 609 374 952 391 704 576;
  • 28) 0.099 609 374 952 391 704 576 × 2 = 0 + 0.199 218 749 904 783 409 152;
  • 29) 0.199 218 749 904 783 409 152 × 2 = 0 + 0.398 437 499 809 566 818 304;
  • 30) 0.398 437 499 809 566 818 304 × 2 = 0 + 0.796 874 999 619 133 636 608;
  • 31) 0.796 874 999 619 133 636 608 × 2 = 1 + 0.593 749 999 238 267 273 216;
  • 32) 0.593 749 999 238 267 273 216 × 2 = 1 + 0.187 499 998 476 534 546 432;
  • 33) 0.187 499 998 476 534 546 432 × 2 = 0 + 0.374 999 996 953 069 092 864;
  • 34) 0.374 999 996 953 069 092 864 × 2 = 0 + 0.749 999 993 906 138 185 728;
  • 35) 0.749 999 993 906 138 185 728 × 2 = 1 + 0.499 999 987 812 276 371 456;
  • 36) 0.499 999 987 812 276 371 456 × 2 = 0 + 0.999 999 975 624 552 742 912;
  • 37) 0.999 999 975 624 552 742 912 × 2 = 1 + 0.999 999 951 249 105 485 824;
  • 38) 0.999 999 951 249 105 485 824 × 2 = 1 + 0.999 999 902 498 210 971 648;
  • 39) 0.999 999 902 498 210 971 648 × 2 = 1 + 0.999 999 804 996 421 943 296;
  • 40) 0.999 999 804 996 421 943 296 × 2 = 1 + 0.999 999 609 992 843 886 592;
  • 41) 0.999 999 609 992 843 886 592 × 2 = 1 + 0.999 999 219 985 687 773 184;
  • 42) 0.999 999 219 985 687 773 184 × 2 = 1 + 0.999 998 439 971 375 546 368;
  • 43) 0.999 998 439 971 375 546 368 × 2 = 1 + 0.999 996 879 942 751 092 736;
  • 44) 0.999 996 879 942 751 092 736 × 2 = 1 + 0.999 993 759 885 502 185 472;
  • 45) 0.999 993 759 885 502 185 472 × 2 = 1 + 0.999 987 519 771 004 370 944;
  • 46) 0.999 987 519 771 004 370 944 × 2 = 1 + 0.999 975 039 542 008 741 888;
  • 47) 0.999 975 039 542 008 741 888 × 2 = 1 + 0.999 950 079 084 017 483 776;
  • 48) 0.999 950 079 084 017 483 776 × 2 = 1 + 0.999 900 158 168 034 967 552;
  • 49) 0.999 900 158 168 034 967 552 × 2 = 1 + 0.999 800 316 336 069 935 104;
  • 50) 0.999 800 316 336 069 935 104 × 2 = 1 + 0.999 600 632 672 139 870 208;
  • 51) 0.999 600 632 672 139 870 208 × 2 = 1 + 0.999 201 265 344 279 740 416;
  • 52) 0.999 201 265 344 279 740 416 × 2 = 1 + 0.998 402 530 688 559 480 832;
  • 53) 0.998 402 530 688 559 480 832 × 2 = 1 + 0.996 805 061 377 118 961 664;
  • 54) 0.996 805 061 377 118 961 664 × 2 = 1 + 0.993 610 122 754 237 923 328;

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:


0.000 000 000 742 147 676 292(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 292(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:


0.000 000 000 742 147 676 292(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 -0.000 000 000 742 147 676 292 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