-0.000 000 000 742 147 676 646 700 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 700 2(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
-0.000 000 000 742 147 676 646 700 2(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:

|-0.000 000 000 742 147 676 646 700 2| = 0.000 000 000 742 147 676 646 700 2


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: 0.000 000 000 742 147 676 646 700 2.

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) 0.000 000 000 742 147 676 646 700 2 × 2 = 0 + 0.000 000 001 484 295 353 293 400 4;
  • 2) 0.000 000 001 484 295 353 293 400 4 × 2 = 0 + 0.000 000 002 968 590 706 586 800 8;
  • 3) 0.000 000 002 968 590 706 586 800 8 × 2 = 0 + 0.000 000 005 937 181 413 173 601 6;
  • 4) 0.000 000 005 937 181 413 173 601 6 × 2 = 0 + 0.000 000 011 874 362 826 347 203 2;
  • 5) 0.000 000 011 874 362 826 347 203 2 × 2 = 0 + 0.000 000 023 748 725 652 694 406 4;
  • 6) 0.000 000 023 748 725 652 694 406 4 × 2 = 0 + 0.000 000 047 497 451 305 388 812 8;
  • 7) 0.000 000 047 497 451 305 388 812 8 × 2 = 0 + 0.000 000 094 994 902 610 777 625 6;
  • 8) 0.000 000 094 994 902 610 777 625 6 × 2 = 0 + 0.000 000 189 989 805 221 555 251 2;
  • 9) 0.000 000 189 989 805 221 555 251 2 × 2 = 0 + 0.000 000 379 979 610 443 110 502 4;
  • 10) 0.000 000 379 979 610 443 110 502 4 × 2 = 0 + 0.000 000 759 959 220 886 221 004 8;
  • 11) 0.000 000 759 959 220 886 221 004 8 × 2 = 0 + 0.000 001 519 918 441 772 442 009 6;
  • 12) 0.000 001 519 918 441 772 442 009 6 × 2 = 0 + 0.000 003 039 836 883 544 884 019 2;
  • 13) 0.000 003 039 836 883 544 884 019 2 × 2 = 0 + 0.000 006 079 673 767 089 768 038 4;
  • 14) 0.000 006 079 673 767 089 768 038 4 × 2 = 0 + 0.000 012 159 347 534 179 536 076 8;
  • 15) 0.000 012 159 347 534 179 536 076 8 × 2 = 0 + 0.000 024 318 695 068 359 072 153 6;
  • 16) 0.000 024 318 695 068 359 072 153 6 × 2 = 0 + 0.000 048 637 390 136 718 144 307 2;
  • 17) 0.000 048 637 390 136 718 144 307 2 × 2 = 0 + 0.000 097 274 780 273 436 288 614 4;
  • 18) 0.000 097 274 780 273 436 288 614 4 × 2 = 0 + 0.000 194 549 560 546 872 577 228 8;
  • 19) 0.000 194 549 560 546 872 577 228 8 × 2 = 0 + 0.000 389 099 121 093 745 154 457 6;
  • 20) 0.000 389 099 121 093 745 154 457 6 × 2 = 0 + 0.000 778 198 242 187 490 308 915 2;
  • 21) 0.000 778 198 242 187 490 308 915 2 × 2 = 0 + 0.001 556 396 484 374 980 617 830 4;
  • 22) 0.001 556 396 484 374 980 617 830 4 × 2 = 0 + 0.003 112 792 968 749 961 235 660 8;
  • 23) 0.003 112 792 968 749 961 235 660 8 × 2 = 0 + 0.006 225 585 937 499 922 471 321 6;
  • 24) 0.006 225 585 937 499 922 471 321 6 × 2 = 0 + 0.012 451 171 874 999 844 942 643 2;
  • 25) 0.012 451 171 874 999 844 942 643 2 × 2 = 0 + 0.024 902 343 749 999 689 885 286 4;
  • 26) 0.024 902 343 749 999 689 885 286 4 × 2 = 0 + 0.049 804 687 499 999 379 770 572 8;
  • 27) 0.049 804 687 499 999 379 770 572 8 × 2 = 0 + 0.099 609 374 999 998 759 541 145 6;
  • 28) 0.099 609 374 999 998 759 541 145 6 × 2 = 0 + 0.199 218 749 999 997 519 082 291 2;
  • 29) 0.199 218 749 999 997 519 082 291 2 × 2 = 0 + 0.398 437 499 999 995 038 164 582 4;
  • 30) 0.398 437 499 999 995 038 164 582 4 × 2 = 0 + 0.796 874 999 999 990 076 329 164 8;
  • 31) 0.796 874 999 999 990 076 329 164 8 × 2 = 1 + 0.593 749 999 999 980 152 658 329 6;
  • 32) 0.593 749 999 999 980 152 658 329 6 × 2 = 1 + 0.187 499 999 999 960 305 316 659 2;
  • 33) 0.187 499 999 999 960 305 316 659 2 × 2 = 0 + 0.374 999 999 999 920 610 633 318 4;
  • 34) 0.374 999 999 999 920 610 633 318 4 × 2 = 0 + 0.749 999 999 999 841 221 266 636 8;
  • 35) 0.749 999 999 999 841 221 266 636 8 × 2 = 1 + 0.499 999 999 999 682 442 533 273 6;
  • 36) 0.499 999 999 999 682 442 533 273 6 × 2 = 0 + 0.999 999 999 999 364 885 066 547 2;
  • 37) 0.999 999 999 999 364 885 066 547 2 × 2 = 1 + 0.999 999 999 998 729 770 133 094 4;
  • 38) 0.999 999 999 998 729 770 133 094 4 × 2 = 1 + 0.999 999 999 997 459 540 266 188 8;
  • 39) 0.999 999 999 997 459 540 266 188 8 × 2 = 1 + 0.999 999 999 994 919 080 532 377 6;
  • 40) 0.999 999 999 994 919 080 532 377 6 × 2 = 1 + 0.999 999 999 989 838 161 064 755 2;
  • 41) 0.999 999 999 989 838 161 064 755 2 × 2 = 1 + 0.999 999 999 979 676 322 129 510 4;
  • 42) 0.999 999 999 979 676 322 129 510 4 × 2 = 1 + 0.999 999 999 959 352 644 259 020 8;
  • 43) 0.999 999 999 959 352 644 259 020 8 × 2 = 1 + 0.999 999 999 918 705 288 518 041 6;
  • 44) 0.999 999 999 918 705 288 518 041 6 × 2 = 1 + 0.999 999 999 837 410 577 036 083 2;
  • 45) 0.999 999 999 837 410 577 036 083 2 × 2 = 1 + 0.999 999 999 674 821 154 072 166 4;
  • 46) 0.999 999 999 674 821 154 072 166 4 × 2 = 1 + 0.999 999 999 349 642 308 144 332 8;
  • 47) 0.999 999 999 349 642 308 144 332 8 × 2 = 1 + 0.999 999 998 699 284 616 288 665 6;
  • 48) 0.999 999 998 699 284 616 288 665 6 × 2 = 1 + 0.999 999 997 398 569 232 577 331 2;
  • 49) 0.999 999 997 398 569 232 577 331 2 × 2 = 1 + 0.999 999 994 797 138 465 154 662 4;
  • 50) 0.999 999 994 797 138 465 154 662 4 × 2 = 1 + 0.999 999 989 594 276 930 309 324 8;
  • 51) 0.999 999 989 594 276 930 309 324 8 × 2 = 1 + 0.999 999 979 188 553 860 618 649 6;
  • 52) 0.999 999 979 188 553 860 618 649 6 × 2 = 1 + 0.999 999 958 377 107 721 237 299 2;
  • 53) 0.999 999 958 377 107 721 237 299 2 × 2 = 1 + 0.999 999 916 754 215 442 474 598 4;
  • 54) 0.999 999 916 754 215 442 474 598 4 × 2 = 1 + 0.999 999 833 508 430 884 949 196 8;

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:


0.000 000 000 742 147 676 646 700 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 700 2(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:


0.000 000 000 742 147 676 646 700 2(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 -0.000 000 000 742 147 676 646 700 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 691 8 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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