-0.000 000 000 742 147 676 646 720 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 720 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
-0.000 000 000 742 147 676 646 720 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:

|-0.000 000 000 742 147 676 646 720 2| = 0.000 000 000 742 147 676 646 720 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: 0.000 000 000 742 147 676 646 720 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) 0.000 000 000 742 147 676 646 720 2 × 2 = 0 + 0.000 000 001 484 295 353 293 440 4;
  • 2) 0.000 000 001 484 295 353 293 440 4 × 2 = 0 + 0.000 000 002 968 590 706 586 880 8;
  • 3) 0.000 000 002 968 590 706 586 880 8 × 2 = 0 + 0.000 000 005 937 181 413 173 761 6;
  • 4) 0.000 000 005 937 181 413 173 761 6 × 2 = 0 + 0.000 000 011 874 362 826 347 523 2;
  • 5) 0.000 000 011 874 362 826 347 523 2 × 2 = 0 + 0.000 000 023 748 725 652 695 046 4;
  • 6) 0.000 000 023 748 725 652 695 046 4 × 2 = 0 + 0.000 000 047 497 451 305 390 092 8;
  • 7) 0.000 000 047 497 451 305 390 092 8 × 2 = 0 + 0.000 000 094 994 902 610 780 185 6;
  • 8) 0.000 000 094 994 902 610 780 185 6 × 2 = 0 + 0.000 000 189 989 805 221 560 371 2;
  • 9) 0.000 000 189 989 805 221 560 371 2 × 2 = 0 + 0.000 000 379 979 610 443 120 742 4;
  • 10) 0.000 000 379 979 610 443 120 742 4 × 2 = 0 + 0.000 000 759 959 220 886 241 484 8;
  • 11) 0.000 000 759 959 220 886 241 484 8 × 2 = 0 + 0.000 001 519 918 441 772 482 969 6;
  • 12) 0.000 001 519 918 441 772 482 969 6 × 2 = 0 + 0.000 003 039 836 883 544 965 939 2;
  • 13) 0.000 003 039 836 883 544 965 939 2 × 2 = 0 + 0.000 006 079 673 767 089 931 878 4;
  • 14) 0.000 006 079 673 767 089 931 878 4 × 2 = 0 + 0.000 012 159 347 534 179 863 756 8;
  • 15) 0.000 012 159 347 534 179 863 756 8 × 2 = 0 + 0.000 024 318 695 068 359 727 513 6;
  • 16) 0.000 024 318 695 068 359 727 513 6 × 2 = 0 + 0.000 048 637 390 136 719 455 027 2;
  • 17) 0.000 048 637 390 136 719 455 027 2 × 2 = 0 + 0.000 097 274 780 273 438 910 054 4;
  • 18) 0.000 097 274 780 273 438 910 054 4 × 2 = 0 + 0.000 194 549 560 546 877 820 108 8;
  • 19) 0.000 194 549 560 546 877 820 108 8 × 2 = 0 + 0.000 389 099 121 093 755 640 217 6;
  • 20) 0.000 389 099 121 093 755 640 217 6 × 2 = 0 + 0.000 778 198 242 187 511 280 435 2;
  • 21) 0.000 778 198 242 187 511 280 435 2 × 2 = 0 + 0.001 556 396 484 375 022 560 870 4;
  • 22) 0.001 556 396 484 375 022 560 870 4 × 2 = 0 + 0.003 112 792 968 750 045 121 740 8;
  • 23) 0.003 112 792 968 750 045 121 740 8 × 2 = 0 + 0.006 225 585 937 500 090 243 481 6;
  • 24) 0.006 225 585 937 500 090 243 481 6 × 2 = 0 + 0.012 451 171 875 000 180 486 963 2;
  • 25) 0.012 451 171 875 000 180 486 963 2 × 2 = 0 + 0.024 902 343 750 000 360 973 926 4;
  • 26) 0.024 902 343 750 000 360 973 926 4 × 2 = 0 + 0.049 804 687 500 000 721 947 852 8;
  • 27) 0.049 804 687 500 000 721 947 852 8 × 2 = 0 + 0.099 609 375 000 001 443 895 705 6;
  • 28) 0.099 609 375 000 001 443 895 705 6 × 2 = 0 + 0.199 218 750 000 002 887 791 411 2;
  • 29) 0.199 218 750 000 002 887 791 411 2 × 2 = 0 + 0.398 437 500 000 005 775 582 822 4;
  • 30) 0.398 437 500 000 005 775 582 822 4 × 2 = 0 + 0.796 875 000 000 011 551 165 644 8;
  • 31) 0.796 875 000 000 011 551 165 644 8 × 2 = 1 + 0.593 750 000 000 023 102 331 289 6;
  • 32) 0.593 750 000 000 023 102 331 289 6 × 2 = 1 + 0.187 500 000 000 046 204 662 579 2;
  • 33) 0.187 500 000 000 046 204 662 579 2 × 2 = 0 + 0.375 000 000 000 092 409 325 158 4;
  • 34) 0.375 000 000 000 092 409 325 158 4 × 2 = 0 + 0.750 000 000 000 184 818 650 316 8;
  • 35) 0.750 000 000 000 184 818 650 316 8 × 2 = 1 + 0.500 000 000 000 369 637 300 633 6;
  • 36) 0.500 000 000 000 369 637 300 633 6 × 2 = 1 + 0.000 000 000 000 739 274 601 267 2;
  • 37) 0.000 000 000 000 739 274 601 267 2 × 2 = 0 + 0.000 000 000 001 478 549 202 534 4;
  • 38) 0.000 000 000 001 478 549 202 534 4 × 2 = 0 + 0.000 000 000 002 957 098 405 068 8;
  • 39) 0.000 000 000 002 957 098 405 068 8 × 2 = 0 + 0.000 000 000 005 914 196 810 137 6;
  • 40) 0.000 000 000 005 914 196 810 137 6 × 2 = 0 + 0.000 000 000 011 828 393 620 275 2;
  • 41) 0.000 000 000 011 828 393 620 275 2 × 2 = 0 + 0.000 000 000 023 656 787 240 550 4;
  • 42) 0.000 000 000 023 656 787 240 550 4 × 2 = 0 + 0.000 000 000 047 313 574 481 100 8;
  • 43) 0.000 000 000 047 313 574 481 100 8 × 2 = 0 + 0.000 000 000 094 627 148 962 201 6;
  • 44) 0.000 000 000 094 627 148 962 201 6 × 2 = 0 + 0.000 000 000 189 254 297 924 403 2;
  • 45) 0.000 000 000 189 254 297 924 403 2 × 2 = 0 + 0.000 000 000 378 508 595 848 806 4;
  • 46) 0.000 000 000 378 508 595 848 806 4 × 2 = 0 + 0.000 000 000 757 017 191 697 612 8;
  • 47) 0.000 000 000 757 017 191 697 612 8 × 2 = 0 + 0.000 000 001 514 034 383 395 225 6;
  • 48) 0.000 000 001 514 034 383 395 225 6 × 2 = 0 + 0.000 000 003 028 068 766 790 451 2;
  • 49) 0.000 000 003 028 068 766 790 451 2 × 2 = 0 + 0.000 000 006 056 137 533 580 902 4;
  • 50) 0.000 000 006 056 137 533 580 902 4 × 2 = 0 + 0.000 000 012 112 275 067 161 804 8;
  • 51) 0.000 000 012 112 275 067 161 804 8 × 2 = 0 + 0.000 000 024 224 550 134 323 609 6;
  • 52) 0.000 000 024 224 550 134 323 609 6 × 2 = 0 + 0.000 000 048 449 100 268 647 219 2;
  • 53) 0.000 000 048 449 100 268 647 219 2 × 2 = 0 + 0.000 000 096 898 200 537 294 438 4;
  • 54) 0.000 000 096 898 200 537 294 438 4 × 2 = 0 + 0.000 000 193 796 401 074 588 876 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:


0.000 000 000 742 147 676 646 720 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 720 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 646 720 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 720 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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