-0.000 000 000 742 147 676 226 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 226(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
-0.000 000 000 742 147 676 226(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:

|-0.000 000 000 742 147 676 226| = 0.000 000 000 742 147 676 226


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: 0.000 000 000 742 147 676 226.

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) 0.000 000 000 742 147 676 226 × 2 = 0 + 0.000 000 001 484 295 352 452;
  • 2) 0.000 000 001 484 295 352 452 × 2 = 0 + 0.000 000 002 968 590 704 904;
  • 3) 0.000 000 002 968 590 704 904 × 2 = 0 + 0.000 000 005 937 181 409 808;
  • 4) 0.000 000 005 937 181 409 808 × 2 = 0 + 0.000 000 011 874 362 819 616;
  • 5) 0.000 000 011 874 362 819 616 × 2 = 0 + 0.000 000 023 748 725 639 232;
  • 6) 0.000 000 023 748 725 639 232 × 2 = 0 + 0.000 000 047 497 451 278 464;
  • 7) 0.000 000 047 497 451 278 464 × 2 = 0 + 0.000 000 094 994 902 556 928;
  • 8) 0.000 000 094 994 902 556 928 × 2 = 0 + 0.000 000 189 989 805 113 856;
  • 9) 0.000 000 189 989 805 113 856 × 2 = 0 + 0.000 000 379 979 610 227 712;
  • 10) 0.000 000 379 979 610 227 712 × 2 = 0 + 0.000 000 759 959 220 455 424;
  • 11) 0.000 000 759 959 220 455 424 × 2 = 0 + 0.000 001 519 918 440 910 848;
  • 12) 0.000 001 519 918 440 910 848 × 2 = 0 + 0.000 003 039 836 881 821 696;
  • 13) 0.000 003 039 836 881 821 696 × 2 = 0 + 0.000 006 079 673 763 643 392;
  • 14) 0.000 006 079 673 763 643 392 × 2 = 0 + 0.000 012 159 347 527 286 784;
  • 15) 0.000 012 159 347 527 286 784 × 2 = 0 + 0.000 024 318 695 054 573 568;
  • 16) 0.000 024 318 695 054 573 568 × 2 = 0 + 0.000 048 637 390 109 147 136;
  • 17) 0.000 048 637 390 109 147 136 × 2 = 0 + 0.000 097 274 780 218 294 272;
  • 18) 0.000 097 274 780 218 294 272 × 2 = 0 + 0.000 194 549 560 436 588 544;
  • 19) 0.000 194 549 560 436 588 544 × 2 = 0 + 0.000 389 099 120 873 177 088;
  • 20) 0.000 389 099 120 873 177 088 × 2 = 0 + 0.000 778 198 241 746 354 176;
  • 21) 0.000 778 198 241 746 354 176 × 2 = 0 + 0.001 556 396 483 492 708 352;
  • 22) 0.001 556 396 483 492 708 352 × 2 = 0 + 0.003 112 792 966 985 416 704;
  • 23) 0.003 112 792 966 985 416 704 × 2 = 0 + 0.006 225 585 933 970 833 408;
  • 24) 0.006 225 585 933 970 833 408 × 2 = 0 + 0.012 451 171 867 941 666 816;
  • 25) 0.012 451 171 867 941 666 816 × 2 = 0 + 0.024 902 343 735 883 333 632;
  • 26) 0.024 902 343 735 883 333 632 × 2 = 0 + 0.049 804 687 471 766 667 264;
  • 27) 0.049 804 687 471 766 667 264 × 2 = 0 + 0.099 609 374 943 533 334 528;
  • 28) 0.099 609 374 943 533 334 528 × 2 = 0 + 0.199 218 749 887 066 669 056;
  • 29) 0.199 218 749 887 066 669 056 × 2 = 0 + 0.398 437 499 774 133 338 112;
  • 30) 0.398 437 499 774 133 338 112 × 2 = 0 + 0.796 874 999 548 266 676 224;
  • 31) 0.796 874 999 548 266 676 224 × 2 = 1 + 0.593 749 999 096 533 352 448;
  • 32) 0.593 749 999 096 533 352 448 × 2 = 1 + 0.187 499 998 193 066 704 896;
  • 33) 0.187 499 998 193 066 704 896 × 2 = 0 + 0.374 999 996 386 133 409 792;
  • 34) 0.374 999 996 386 133 409 792 × 2 = 0 + 0.749 999 992 772 266 819 584;
  • 35) 0.749 999 992 772 266 819 584 × 2 = 1 + 0.499 999 985 544 533 639 168;
  • 36) 0.499 999 985 544 533 639 168 × 2 = 0 + 0.999 999 971 089 067 278 336;
  • 37) 0.999 999 971 089 067 278 336 × 2 = 1 + 0.999 999 942 178 134 556 672;
  • 38) 0.999 999 942 178 134 556 672 × 2 = 1 + 0.999 999 884 356 269 113 344;
  • 39) 0.999 999 884 356 269 113 344 × 2 = 1 + 0.999 999 768 712 538 226 688;
  • 40) 0.999 999 768 712 538 226 688 × 2 = 1 + 0.999 999 537 425 076 453 376;
  • 41) 0.999 999 537 425 076 453 376 × 2 = 1 + 0.999 999 074 850 152 906 752;
  • 42) 0.999 999 074 850 152 906 752 × 2 = 1 + 0.999 998 149 700 305 813 504;
  • 43) 0.999 998 149 700 305 813 504 × 2 = 1 + 0.999 996 299 400 611 627 008;
  • 44) 0.999 996 299 400 611 627 008 × 2 = 1 + 0.999 992 598 801 223 254 016;
  • 45) 0.999 992 598 801 223 254 016 × 2 = 1 + 0.999 985 197 602 446 508 032;
  • 46) 0.999 985 197 602 446 508 032 × 2 = 1 + 0.999 970 395 204 893 016 064;
  • 47) 0.999 970 395 204 893 016 064 × 2 = 1 + 0.999 940 790 409 786 032 128;
  • 48) 0.999 940 790 409 786 032 128 × 2 = 1 + 0.999 881 580 819 572 064 256;
  • 49) 0.999 881 580 819 572 064 256 × 2 = 1 + 0.999 763 161 639 144 128 512;
  • 50) 0.999 763 161 639 144 128 512 × 2 = 1 + 0.999 526 323 278 288 257 024;
  • 51) 0.999 526 323 278 288 257 024 × 2 = 1 + 0.999 052 646 556 576 514 048;
  • 52) 0.999 052 646 556 576 514 048 × 2 = 1 + 0.998 105 293 113 153 028 096;
  • 53) 0.998 105 293 113 153 028 096 × 2 = 1 + 0.996 210 586 226 306 056 192;
  • 54) 0.996 210 586 226 306 056 192 × 2 = 1 + 0.992 421 172 452 612 112 384;

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:


0.000 000 000 742 147 676 226(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 226(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:


0.000 000 000 742 147 676 226(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 -0.000 000 000 742 147 676 226 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